MODULE AND MORE Matematik Tambahan Tingkatan 5

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 98 BAB 5 (b) Tinggi seorang pelajar dipilih secara rawak dari kelas tingkatan lima The height of a student is chosen at random from a form five class. Pemboleh ubah ialah tinggi pelajar itu. The variable is the height of a student. (c) Jumlah nilai apabila dua biji dadu dilambungkan serentak. The sum of the numbers on two dice when they are tossed simultaneously. Pemboleh ubah ialah {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. The variable is 2. Tulis semua kesudahan yang mungkin dalam tata-tanda set bagi yang berikut. Nyatakan sama ada pemboleh ubah ialah pemboleh ubah rawak diskret atau pemboleh ubah rawak selanjar. Beri alasan anda. Write all the possible outcomes in set notations for the following. State whether the variable is a random discrete variable or a random continuous variable. Give your reasons. TP 1 (a) X ialah pemboleh ubah yang mewakili bilangan pelajar yang memakai cermin mata di dalam satu kelas yang terdiri daripada 20 orang murid. X is a variable that represents the number of students wearing spectacles in a class of 20 students. X = {0, 1, 2, 3, 4,… 20} adalah diskret kerana dapat dikira. X = {0, 1, 2, 3, 4,… 20} is discrete because they can be counted. (b) Y ialah pemboleh ubah yang mewakili masa yang diambil untuk membuat sebiji kek. Masa diambil adalah di antara 1 1 2 jam hingga 3 jam. Y is a variable that represents the time taken to make a piece of cake. The time taken is between 1 1 2 hours to 3 hours. Y = {y: 1 1 2 ≤ y ≤ 3} adalah selanjar kerana sudah menentukan dengan tepat masa yang sewajar. Y = {y: 1 1 2 ≤ y ≤ 3} is continuous because it has determined by the time accordingly. (c) Z ialah pemboleh ubah yang mewakili masa seorang pesakit menjadi sembuh selepas dijangkiti kuman flu. Biasanya 3 sehingga 8 hari. Z is the variable that represents the time taken by a patient to get well after catching a flu. Normally it is between 3 to 8 days. Z = {z: 3 ≤ z ≤ 8} adalah selanjar kerana sudah menentukan dengan tepat masa yang sewajarnya. Z = {z: 3 ≤ z ≤ 8} is continuous because it has determind by the time accordingly. CONTOH Sekeping syiling dilambungkan empat kali. X ialah pemboleh ubah rawak yang mewakili bilangan kali mendapat bunga. A piece of coin is tossed four times. X is a random variable that represents the number of times of getting the flower. Penyelesaian: X = {0, 1, 2, 3, 4} adalah diskret kerana dapat dikira. X = {0, 1, 2, 3, 4} is discrete because they can be counted.

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 99 5 (d) X ialah pemboleh ubah yang mewakili bilangan buah durian yang dihasilkan dari setiap pokok di kebun Rahman. X is the variable that represents the number of durians produced by a tree in Rahman’s orchard. X = {0, 1, 2, 3, ... , n} adalah diskret kerana dapat dikira. X = {0, 1, 2, 3, ... , n} is discrete because they can be counted. 3. Lukis gambar rajah pokok untuk mewakili semua kesudahan, X yang mungkin bagi setiap yang berikut. Jadualkan taburan kebarangkalian bagi X. Draw tree diagrams to represent all the possible outcomes, X for each of the following. Tabulate the probability distribution of X in a table. TP 3 CONTOH Sebiji dadu dilambungkan tiga kali dan X ialah pemboleh ubah diskret yang mewakili bilangan kali mendapat nombor 3. A dice is tossed three times and X is a discrete variable that represents the number of times of getting the number 3. Penyelesaian: X = {0, 1, 2, 3} Tip Kebarangkalian peristiwa A berlaku, P(A) = n(A) n(S) , dengan keadaan n(A) ialah bilangan kesudahan bagi peristiwa A dan n(S) ialah bilangan kesudahan dalam ruang sampel. The probability of an event A happening, P(A) = n(A) n(S) , where n(A) is the number of outcomes of event A and n(S) is the number of outcomes of the sample space. 1 6 3 3 3 3 3 3 3 3’ 3’ 3’ 3’ 3’ 3’ 3’ 1 6 1 6 1 6 1 6 1 6 1 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 L1 L2 L3 P[X = 2] = P[3, 3, 3’] + P[3, 3’, 3] + P[3’, 3, 3] = 3 5 6 × 1 6 × 1 6  = 5 72 P[X = 3] = P[3, 3, 3] = 1 6 × 1 6 × 1 6 = 1 216 X = x 0 1 2 3 P[X = x] 125 216 25 72 5 72 1 216 P[X = 0] = P[3’, 3’, 3’] = 5 6 × 5 6 × 5 6 = 125 216 P[X = 1] = P[3, 3’, 3’] + P[3’, 3, 3’] + P[3’, 3’, 3] = 3 5 6 × 5 6 × 1 6  = 75 216 = 25 72

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 100 BAB 5 (a) Sebuah beg mengandungi 2 biji guli merah dan 4 biji guli hijau. Sebiji guli dicabut dan selepas mencatatkan warnanya, guli itu dikembalikan ke dalam beg sebelum sebiji guli dikeluarkan lagi. X mewakili bilangan kali mendapat guli hijau dalam dua cabutan berturut-turut. A bag contains 2 red marbles and 4 green marbles. A marble is drawn at random from the bag and after the colour is noted, the marble is returned into the bag before another one marble is drawn. X is the number of times of getting a green marble in the two consecutive draws. (b) 30% pelajar menggunakan bas sekolah ke sekolah dari suatu sekolah tertentu. X mewakili bilangan pelajar menggunakan bas sekolah jika 3 pelajar dipilih secara rawak. 30% of the students take bus to the school from a certain school. X represents the number of students who take school bus if 3 students are chosen at random. X = {0, 1, 2} P[X = 0] = P[M, M] = 1 3 × 1 3 = 1 9 P[X = 1] = P[H, M] + P[M, H] = 2 3 × 1 3 + 1 3 × 2 3 = 4 9 P[X = 2] = P[H, H] = 2 3 × 2 3 = 4 9 X = x 0 1 2 P[X = x] 1 9 4 9 4 9 X = {0, 1, 2, 3} P[X = 0] = P[B’, B’, B’] =  7 10 3 = 343 1000 P[X = 1] = 3 C1 3 10 7 10 2 = 441 1000 P[X = 2] = 3 C21 3 10 2  7 10 = 189 1000 P[X = 3] = 1 3 10 3 = 27 1000 X = x 0 1 2 3 P[X = x] 343 1000 441 1000 189 1000 27 1000 H H H M M M 2 3 2 3 2 1 3 3 1 3 1 3 B B B B B B B B’ B’ B’ B’ B’ B’ B’ 7 10 7 10 7 10 7 10 7 10 7 10 7 10 3 10 3 10 3 10 3 10 3 10 3 10 3 10

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 101 5 CONTOH 4. Pemboleh ubah X mempunyai taburan kebarangkalian yang berikut. The random variable X has the following probability distributions. (i) Bina jadual taburan kebarangkalian bagi X, dan tunjukkan X ialah satu pemboleh ubah rawak diskret. Construct a probability distribution table for X, and show that X is a discrete random variable. (ii) Lukis satu graf untuk taburan kebarangkalian X. Draw a graph for the probability distribution of X. TP 4 (a) P(X = x) = x 6 untuk/for x = {0, 1, 2, 3} (i) P(X = 0) = 0 6 = 0 P(X = 1) = 1 6 P(X = 2) = 2 6 = 1 3 P(X = 3) = 3 6 = 1 2 X = x 0 1 2 3 P(X = x) 0 1 6 1 3 1 2 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 1 6 + 1 3 + 1 2 = 1 (Terbukti)/(Proved) (ii) x 12 36 10 36 8 36 6 36 0 0 2 4 6 P(X = x) x 1 3 1 2 1 6 0 0 1 2 3 P(X = x) P(X = x) = 12 – x 36 untuk/for x = {0, 2, 4, 6}. Penyelesaian: (i) P(X = 0) = 12 – 0 36 = 1 3 P(X = 2) = 12 – 2 36 = 5 18 P(X = 4) = 12 – 4 36 = 2 9 P(X = 6) = 12 – 6 36 = 1 6 Jadual taburan kebarangkalian bagi X. Probability distribution table for X. X = x 0 2 4 6 P(X = x) 1 3 5 18 2 9 1 6 Untuk menunjukkan X ialah satu pemboleh ubah rawak diskret, kita mesti tunjukkan ∑n 0 P[X = x] = 1. To show that X is a random discrete variable, we must show ∑n 0 P[X = x] = 1. P(X = 0) + P(X = 2) + P(X = 4) + P(X = 6) = 1 3 + 5 18 + 2 9 + 1 6 = 1 (Terbukti)/(Proved) (ii)

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 102 BAB 5 (b) P(X = x) = (x + 1)2 54 untuk/for x = {1, 2, 3, 4} 5. Selesaikan setiap soalan yang berikut. Solve each of the following questions. TP 4 (a) X ialah suatu pemboleh ubah rawak diskret dengan taburan kebarangkalian yang berikut X is a random discrete variable with the following probability distribution. X 2 4 6 8 P(X = x) 0.14 0.21 0.38 0.27 Cari/Find (i) P(X ≤ 6) ii) P(2 ≤ X ≤ 6) (i) P(X ≤ 6) = 1 − P(X = 8) = 1 − 0.27 = 0.73 (ii) P(2 ≤ X ≤ 6) = 1 − P(X = 8) = 0.73 x 9 54 16 54 25 54 4 54 0 1 2 3 4 P(X = x) (i) P(X = 1) = (1 + 1)2 54 = 2 27 P(X = 2) = (2 + 1)2 54 = 1 6 P(X = 3) = (3 + 1)2 54 = 8 27 P(X = 4) = (4 + 1)2 54 = 25 54 X = x 1 2 3 4 P(X = x) 2 27 1 6 8 27 25 54 P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 2 27 + 1 6 + 8 27 + 25 54 = 1 (Terbukti)/(Proved) (ii) CONTOH X ialah suatu pemboleh ubah rawak diskret dengan taburan kebarangkalian yang berikut. X is a random discrete variable with the following probability distribution. X 0 1 2 3 P(X = x) 0.24 0.31 0.28 0.17 Cari/Find (i) P(X > 1) (ii) P(0 < X < 3) Penyelesaian: (i) P(X > 1) = P(X = 2) + P(X = 3) = 0.28 + 0.17 = 0.45 (ii) P(0 < X < 3) = P(X = 1) + P(X = 2) = 0.31 + 0.28 = 0.59 Kaedah Alternatif P(X > 1) = 1 – P(X ≤ 1) = 1 – [P(X = 0) + P(X = 1)] = 1 – 0.24 – 0.31 = 0.45

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 103 5 (b) Y ialah suatu pemboleh ubah rawak diskret dengan taburan kebarangkalian yang berikut. Y is a random discrete variable with the following probability distribution. {0.1 untuk y = 1,3,5,7 P(Y = y) =      0.2 untuk y = 2,4,6     0 untuk y yang lain  Tunjukkan bahawa Y ialah suatu pemboleh ubah rawak diskret. Seterusnya, cari Show that Y is a random discrete variable. Then, find (i) P(Y < 4) (ii) P(3 < Y < 5) Y 1 2 3 4 5 6 7 P(Y = y) 0.1 0.2 0.1 0.2 0.1 0.2 0.1 ∑7 0 P(Y = y) = 0.1 × 4 + 0.2 × 3 = 0.4 + 0.6 = 1 Maka, Y ialah suatu pemboleh ubah rawak diskret. Hence, y is a random discrete variable. (i) P(Y < 4) = P(Y = 1) + P(Y = 2) + P(Y = 3) = 0.1 + 0.2 + 01 = 0.4 (ii) P(3 < Y < 5) = P(Y = 4) = 0.2 (c) Z ialah suatu pemboleh ubah rawak diskret dengan taburan kebarangkalian yang berikut. Z is a random discrete variable with the following probability distribution. p untuk z = 0,1 P(Z = z) =      p + q untuk z = 2     q untuk z = 3,4}  (i) Jika P(Z > 2) = 4 9 , cari nilai p dan q. If P(Z > 2) = 4 9 , find the values of p and q. (ii) Bina satu jadual taburan kebarangkalian Z. Construct a table of the probability distribution of Z. (iii) Lukis satu graf taburan kebarangkalian Z. Draw a graph of the probability distribution of Z. (i) P(Z > 2) = P(Z = 3) + P(Z = 4) = 4 9 q + q = 4 9 q = 2 9 P(Z = 0) + P(Z = 1) + P(Z = 2) = 5 9 p + p + p + q = 5 9 3p + 2 9 = 5 9 3p = 3 9 = 1 9 (ii) z 0 1 2 3 4 P(Z = z) 1 9 1 9 1 3 2 9 2 9 (iii) 0 0 1 2 3 4 z P(Z = z) 1 9 2 9 3 9

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 104 BAB 5 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 5.2 Taburan Binomial Binomial Distribution 1. Suatu eksperimen dengan 2 kesudahan sahaja iaitu kejayaan dan kegagalan dikenali sebagai eksperimen Bernoulli. An experiment with only two outcomes that is success and failure is a Bernoulli experiment. 2. Jika eksperimen Bernoulli diulangi n kali secara tak bersandar, maka eksperimen tersebut dikenali sebagai eksperimen Binomial. When the Bernoulli experiment is repeated n times independently, then the experiment is called a Binomial experiment 3. Katakan X adalah pemboleh ubah rawak diskret yang menunjukkan bilangan kejayaan dalam suatu eksperimen Binomial, maka X dikatakan bertaburan Binomial dengan simbol B(n, p), dengan n ialah bilangan percubaan yang tak bersandar dan p ialah kebarangkalian mendapat kejayaan. Suppose X is the discrete random variable that shows the number of success in the Binomial experiment, then X is said to have a Binomial distribution with the notation B(n, p), where n is the number of trials and p is the probability of each success. 4. Oleh itu, kebarangkalian mendapat x kejayaan ialah Hence, the probability of getting x successes is given by P(X = x) = n Cr pr (1 − p) n − r , x = 0, 1, 2, 3 …. n. (Kadang-kala, kita menggunakan q sebagai tanda kegagalan, iaitu q = 1 – p.) (Sometimes, we use q for the probability of failure, which is q = 1 – p.) 5. Katakan X adalah pemboleh ubah rawak diskret binomial dengan parameter n dan p, maka pengiraan min, varians dan sisihan piawai ialah When X is the binomial discrete random variable with parameters n and p, then to find the mean, the variance and the standard deviation, we have Min / mean, E(X) = np Varians / variance, Var(X) = np(1 − p) Sisihan piawai = np(1− p), dengan n ialah bilangan percubaan dan p ialah kebarangkalian bagi kejayaan. Standard deviation = np(1− p) , where n is the number of trials and p is the probability of success. 6. Nyatakan sama ada percubaan yang berikut ialah percubaan Bernoulli. Jelaskan jawapan anda. State whether the following are Bernoulli trials or not. Explain your answers. TP 2 (a) Sebiji bola dikeluarkan daripada satu kotak yang mengandungi 5 bola merah dan 3 bola biru. Selepas mencatatkan warna, bola itu dikembali ke dalam kotak sebelum sebiji bola dikeluarkan lagi. Proses ini berulang 4 kali dan X ialah bilangan kali mendapat bola merah. A ball is drawn from a box which contains 5 red and 3 blue balls. After noting the colour, the ball is returned into the box before another ball is drawn. This process is repeated 4 times and X is the number of times of getting a red ball. Ini ialah percubaan Bernoulli, kerana kebarangkalian mendapat bola merah setiap kali ialah tak bersandar, iaitu tetap = 5 8 dan percubaan ini berlaku 4 kali, n = 4. This is Bernoulli’s trial, because the probability of getting a red ball every time is not dependent, that is fixed = 5 8 and the trial occurs 4 times, n = 4. CONTOH X ialah keputusan perlawanan suatu pertandingan bola sepak. Keputusan terdiri daripada menang, tewas atau seri. X is the result of a football game. the results can be win, loss or draw. Penyelesaian: Bukan, kerana terdapat tiga kesudahan, iaitu menang, tewas dan seri. Binomial hanya mempunyai dua jenis kesudahan. This is not because there are three outcomes, that are win, lose and draw. Binomial has only two types of outcomes.

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 105 5 (b) 85% orang pelajar dari kelas A mempunyai telefon pintar. X ialah bilangan kali mendapat seorang pelajar mempunyai telefon pintar jika proses diulangi 10 kali. 85% of the students from class A has smartphones. X is the number of times of choosing a student with a smartphone if this process is repeated 10 times Ini ialah percubaan Bernoulli, kerana terdapat dua kesudahan sahaja, dengan p = 0.85 dan percubaan ini berlaku 10 kali, n = 10. This is a Bernoulli trial because there are two outcomes, where p = 0.85 and this trial occurs 10 times, n = 10. (c) Satu huruf dipilih secara rawak daripada perkataan BACAAN. X ialah bilangan kali memilih huruf A jika semua huruf tidak boleh diulangi dan proses ini dibuat 3 kali. A letter is chosen at random from the word BACAAN. X is the number of times of choosing A if all the letters cannot be repeated and this process is done three times. Ini bukan percubaan Bernoulli, kerana kebarangkalian mendapat A pertama kali, p = 3 6 , kedua kali p = 2 5 dan ketiga kali p = 1 4 dengan setiap kali p berubah, iaitu bersandar. This is not Bernoulli trial, because the probability to get A the first time, p = 3 6 , the second time p = 2 5 and the third time p = 1 4 , where each time p changes, that is dependent. 7. Tentukan kebarangkalian suatu peristiwa bagi taburan binomial yang berikut. Determine the distribution of an event with the following binomial distribution. TP 4 CONTOH Satu syiling dilambungkan 5 kali berturut-turut. Apakah kebarangakalian mendapat A coin is tossed 5 times continuously. What is the probability of getting (i) 2 gambar dalam 5 lambungan ini? 2 heads in the 5 tosses? (ii) kurang daripada 2 gambar? less than 2 heads? Penyelesaian: (i) Kebarangkalian mendapat gambar dalam satu lambungan, p = 1 2 . Katakan X ialah bilangan kali mendapat gambar, maka X = {0, 1, 2, 3, 4, 5}. The probability of getting a head in one toss, p = 1 2 . Let X be the number of times of getting heads, then X = {0, 1, 2, 3, 4, 5}. Kebarangakalian mendapat 2 gambar dalam 5 lambungan ini The probability of getting 2 heads in 5 tosses = P(X = 2) = 5 C2 1 2 2 2 1 – 1 2 2 3 = 5 16 (ii) P(X < 2) = P(X = 0) + P(X = 1) = 5 C0 1 2 2 0 1 − 1 2 2 5 + 5 C1 1 2 2 1 1 − 1 2 2 4 = 3 16 Sudut Kalkulator Tekan Press 5 ( 2 1 ( 1 1 2 2 5 16 ) ) x2 x2 = × – SHIFT SHIFT a b/c a b/c n Cr

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 106 BAB 5 (a) Di sepanjang suatu jalan raya, kebarangkalian kemalangan berlaku pada suatu hari ialah 5%. Cari kebarangkalian dalam dua minggu berlakunya kemalangan jalan raya Along a certain highway, the probability of road accident in a certain day is 5%. Find the probability in two weeks of a road accident (i) tepat 4 kali./exactly 4 times. (ii) kurang daripada 12 kali. less than 12 times. (iii) 5 hingga 7 kali./5 to 7 times. n = 14, p = 0.05, q = 0.95 (i) P(X = 4) = 14C4 (0.05)4 (0.95)10 = 0.0037 (ii) P(X < 12) = 1 − P[X = 12] − P(X = 13) + P(X = 14) = 1 − (14C12(0.05)12(0.95)2 + 14C13(0.05)13(0.95)1 + 14C14(0.05)14) = 1 (iii)P(5 ≤ X ≤ 7) = P(X = 5) + P[X = 6) + P(X = 7) = [14C5 (0.05)5 (0.95)9 + 14C6 (0.05)6 (0.95)8 + 14C7 ( 0.57)(0.95)7 ] = 0.00043 (b) Dalam suatu kaji selidik, 90 daripada 100 orang yang didedahkan kepada orang yang mempunyai penyakit Covid 19 akan diuji positif. Jika 8 orang daripada kumpulan tersebut dipilih secara rawak, cari peratus bahawa In a research, 90 out of 100 people exposed to the Covid-19 sickness are tested positive. If 8 people are chosen at random from this group of people, find the percentage that (i) semua dijangkiti Covid-19. all contracted Covid-19. (ii) lebih daripada 6 orang dijangkiti covid-19. more than 6 of them contracted covid-19. n = 8, p = 0.9, q = 0.1 (i) P(X = 8) = 8 C8 (0.9)8 = 0.4305 Maka, 43.05% dijangkiti Covid-19. Hence, 43.05% contracted Covid-19. (ii) P(X > 6) = P(X = 7) + P(X = 8) = 8 C7 (0.9)7 (0.1)1 + 8 C8 (0.9)8 = 0.8131 Maka, 81.31% dijangkiti Covid-19. Hence, 81.31% contracted Covid-19. (c) 95% daripada murid di sebuah sekolah mematuhi undang-undang sekolah. Jika 10 orang murid dipilih secara rawak dari sekolah itu, cari kebarangkalian bahawa 95% of the students in a school comply with school rules. If 10 students are chosen at random from the school, find the probability that (i) lebih daripada 8 orang murid mematuhi undang-undang sekolah. more than 8 students comply with school rules. (ii) 2 hingga 8 orang murid mematuhi undangundang sekolah. 2 to 8 of them comply with school rules. n = 10, p = 0.95, q = 0.05 (i) P(X > 8) = P(X = 9] + P(X = 10) = 10C9 (0.95)9 (0.05)1 + 10C10(0.95)10 = 0.9139 (ii) P(2 ≤ X ≤ 8) = 1 − P(X = 0) − P(X = 1) − P(X = 9) − P(X = 10) = 0.0861 (d) Kebarangkalian Karol akan menang dalam pertandingan tenis ialah 2 3 . Jika dia mengambil bahagian dalam n pertandingan, kebarangkalian dia tidak menang semua pertandingan ialah 0.0014. Cari The probability that Karol will win in the tennis competition is 2 3 . If he takes part in n competitions, the probability that he does not win all the competitions is 0.0014. Find (i) nilai n./the value of n. (ii) kebarangkalian dia akan menang lebih daripada empat kali dalam n pertandingan ini. the probability he will win more than four times in n competitions. p = 2 3 , q = 1 3 (i) P(X = 0) = 0.0014 = n C0 2 3 2 0  1 3 2 n = 0.0014 n log  1 3 2 = log 0.0014 n = log0.0014 log 1 3 = 6 (ii) P(X > 4) = P(X = 5) + P(X = 6) = 6 C5 2 3 2 5  1 3 2 + 6 C6 2 3 2 6 = 0.3512

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 107 5 8. Selesaikan masalah yang berikut. Solve the following problems. TP 4 CONTOH Diketahui bahawa 20% daripada murid di sebauh sekolah mempunyai iPad. Jika 4 orang murid dipilih secara rawak dari sekolah itu, tentukan taburan binomial dengan menggunakan graf. It is known that 20% of the students in a school have iPad. If 4 students are chosen at random from the school, determine the binomial distribution using graph. Penyelesaian: Katakan X mewakili bilangan murid mempunyai iPad. Maka, X ialah suatu pemboleh ubah rawak binomial dengan n = 4, p = 0.2, q = 1 – 0.2 = 0.8 Let X represents the number of students have iPad Hence, X is binomial random variable with n = 4, p = 0.2. P(X = 0) = 4 C0 (0.2)0 (0.8)4 = 0.4096 P(X = 1) = 4 C1 (0.2)1 (0.8)3 = 0.4096 P(X = 2) = 4 C2 (0.2)2 (0.8)2 = 0.1536 P(X = 3) = 4 C3 (0.2)3 (0.8)1 = 0.0256 P(X = 4) = 4 C4 (0.2)4 (0.8)0 = 0.0016 (a) Graf di sebelah menunjukkan taburan binomial bagi suatu pemboleh ubah rawak X dengan n = 5. The graph shows a binomial distribution for a random variable X with n = 5. Tentukan nilai p. Determine the value of p. 0.20 + 0.25 + p + p + 0.15 + 0.05 = 1 0.65 + 2p = 1 2p = 0.35 p = 0.175 0 0 1 2 3 4 0.1 0.2 0.3 0.4 0.5 P (X = x) x P (X = x) x 0 0 1 2 3 4 5 0.05 0.10 0.15 0.20 0.25 0.30 p Tip Hasil tambah semua kesudahan yang mungkin ialah 1. The sum of all possible outcomes is 1. P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 108 BAB 5 (b) Sekeping duit syiling dilambangkan sebanyak 4 kali. Jika X mewakili bilangan kali mendapat kepala, cari taburan binomial bagi X dan plot grafnya. A coin is tossed 4 times. If X represents the number of times of getting a head, find the binomial distribution for X and plot the graph. n = 4, p = 1 2 , q = 1 2 P(X = 0) = 4 C0 1 2 2 0  1 2 2 4 = 0.0625 P(X = 1) = 4 C1 1 2 2 1  1 2 2 3 = 0.2500 P(X = 2) = 4 C2 1 2 2 2  1 2 2 2 = 0.3750 P(X = 3) = 4 C3 1 2 2 3  1 2 2 1 = 0.2500 P(X = 4) = 4 C4 1 2 2 4  1 2 2 0 = 0.0625 9. Selesaikan masalah berikut yang melibatkan nilai min, varians dan sisihan piawai bagi suatu taburan binomial. Solve the following problems involving mean, variance and standard deviation of the binomial distributions. TP 3 (a) 3 biji dadu dilambungkan serentak. 3 dices are thrown simultaneously. (i) Hitung kebarangkalian untuk mendapat dua ‘5’. Calculate the probability of getting two ‘5’. (ii) Jika 3 biji dadu itu dilambungkan 216 kali, berapa kalikah dijangka untuk mendapat dua ‘5’ ? If the 3 dices are thrown 216 times, estimate the number of times of getting two ‘5’? (i) n = 3, p = 1 6 P(mendapat dua ‘5’) = 3 C2 1 6 2 2  5 6 2 = 5 72 (ii) E(X) = np = 216 × 5 72 = 15 kali P (X = x) 0 x 0 1 2 3 4 0.1 0.2 0.3 0.4 CONTOH 1 20 daripada ayam yang dijual dalam pasar mempunyai berat yang kurang memuaskan untuk dijual. Cari min dan sisihan piawai bagi bilangan ayam yang kurang memuaskan untuk dijual daripada 160 ekor ayam pada suatu hari tertentu. 1 20 of the chickens sold in the market have the masses which are not suitable to be sold. Find the mean and standard deviation of the number of chickens which are not suitable to be sold from 160 chickens on a particular day. Penyelesaian: Katakan X mewakili bilangan ayam yang kurang memuaskan untuk dijual. Let X be the number of chickens which are not suitable to be sold. n = 160, p = 1 20, q = 19 20 Min/Mean = E(X) = np = 160 × 1 20 = 8 Sisihan piawai = 160 1 202 19 202 = 2.76 Standard deviation

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 109 5 (b) Satu pemboleh ubah rawak diskret X bertaburan binomial Bn, 3 142. Jika min ialah 3, cari A random discrete variable X has a binomial distribution Bn, 3 14 2. If the mean is 3, find (i) nilai n. the value of n. (ii) varians dan sisihan piawai. the variance and the standard deviation. (i) E(X) = np = 3 n 3 14 2 = 3 n = 14 (ii) Varians = npq = 14 × 3 14 × 11 14 = 2 5 14 Sisihan piawai = 1.54 Standard deviation (c) X bertaburan binomial dengan B(n, p). Jika q ialah 0.35 dan sisihan piawai ialah 1.5, cari X is a binomial distribution with B(n, p). If q is 0.35 and the standard deviation is 1.5, find (i) nilai n. the value of n. (ii) min / mean (i) q = 0.35, p = 0.65 Sisihan piawai = 1.5 = n(0.65)(0.35) Standard deviation n = 1.52 (0.65)(0.35) = 10 (ii) Min/Mean = np = 10(0.65) = 6.5 10. Selesaikan setiap yang berikut. Solve each of the following. TP 5 (a) 2% daripada lampu yang dihasilkan oleh sebuah kilang adalah rosak. Cari bilangan lampu yang perlu disemak supaya kebarangkalian bahawa sekurang-kurangnya satu lampu yang rosak adalah lebih daripada 0.95. 2% of the bulbs produced by a factory is faulty. Find the number of bulbs which must be inspected so that the probability that at least one bulb is faulty is 0.95. p = 0.02, q = 0.98 P(X ≥ 1) > 0.95 1 – P(X = 0) > 0.95 1 − n C0 (0.02)0 (0.98)n > 0.95 0.98n < 0.05 n log 0.98 < log 0.05 n > log0.05 log0.98 n > 148.28 n = 149 CONTOH Dalam suatu temasya sukan sekolah, kebarangkalian seorang murid mendapat markah untuk rumah sukannya ialah 3 8 . Cari bilangan minimum murid yang mendapat sekurangkurangnya satu markah jika kebarangkalian adalah lebih daripada 0.65. In a school sports meet, the probability of a student getting a mark for the sports house is 3 8 . Find the minimum number of students getting at least one mark if the probability is more than 0.65. Katakan X mewakili bilangan murid yang mendapat markah. Let X represents the number of students that get marks P(X ≥ 1) > 0.65 1 – P(X = 0) > 0.65 P(X = 0) < 0.35 n C0 3 8 2 0  5 8 2 n < 0.35 n log 5 8 < log 0.35 n > log 0.35 log 5 8 n > 2.23 ∴n = 3 Tip n < log0.35 log 5 8 2 n < –0.4559 –0.2041 n > 2.23 n = 3

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 110 BAB 5 (b) Lima dadu dilambungkan serentak. Five dice are thrown simultaneously. (i) Apakah kebarangkalian untuk mendapat tiga ‘2’? What is the probability of getting three ‘2’s? (ii) Jika lima dadu itu dilambungkan 550 kali, berapa kali mendapat tiga ‘2’? If the five dice are thrown 550 times, how many times can get three ‘2’s? (i) P(X = 3) = 5 C3 1 6 2 3  5 6 2 2 = 0.0322 (ii) E(X)= np = 550 × 0.0322 = 17.68 (c) Kebarangkalian pelanggan yang membuat tempahan meja di sebuah restoran tetapi tidak hadir ialah 0.2. Restoran itu cuma boleh muat 50 meja. Jika terdapat tempahan sebanyak 5 meja pada suatu hari, cari kebarangkalian bahawa The probability of a customer who booked the table reservation at a restaurant but not present is 0.2. The restaurant can only accommodate 50 tables. If there are 5 tables booked on a day, find the probability that (i) tepat 3 pelanggan hadir. exactly 3 customers present. (ii) kurang daripada 2 tempahan meja hadir. less than 2 table reservations are present. (iii) Cari min hadir jika semua meja ditempahkan. Find the mean if all the tables are booked. n = 5, p = 0.8, q = 0.2 (i) P(X = 3) = 5 C3 (0.8)3 (0.2)2 = 0.2048 (ii) P(X < 2) = P(X = 0) + P(X = 1) = 0.00672 (iii)E(X)= np = 50 × 0.8 = 40 NOTA IMBASAN NOTA IMBASAN 5.3 Taburan Normal Normal Distribution 1. Suatu pemboleh ubah rawak selanjar, X adalah bertaburan normal jika graf fungsi kebarangkaliannya mempunyai ciri-ciri berikut: A continuous random variable, X is normally distributed if the probability of function graph has the following characteristics: (i) Berbentuk loceng dan bersimetri pada X = µ, dengan µ ialah min, iaitu X ~ N(µ, σ2 ). It is bell-shaped and symmetrical at X = μ, where µ is a mean, which is X ~ N(μ, σ2 ). (ii) Jumlah luas di antara lengkung dan paksi-x ialah 1. The total area between the curve and the x-axis is 1. (iii) Kebarangkalian bagi X yang mempunyai nilai di antara a dan b, ditulis sebagai P(a < X < b) = luas rantau berlorek dalam rajah. The probability of X that have values between a and b, is written as P(a < X < b) = area of shaded region in the diagram x = fi Lengkung ini simetri pada x = fi. The curve is symmetry at x = fi. P(a < X < b) x Kekerapan Frequency 0 a b 2. Jika suatu pemboleh ubah rawak normal, X mempunyai min, µ = 0 dan sisihan piawai, σ = 1, X dikatakan bertaburan normal piawai, iaitu X ~ N(0, 1). If the normal random variable, X has a mean, µ = 0 and standard deviation, σ = 1, X is said to be the standard normal distribution, that is X ~ N(0, 1). 3. Taburan normal boleh ditukar kepada taburan normal piawai melalui rumus Z = x – µ σ , dengan Z dinamakan skor piawai atau skor-z.

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 111 5 The normal distribution can be converted into the standard normal distribution by the formula x – µ σ , where Z is called the standard score or z-score. 4. Kebarangkalian untuk suatu pemboleh ubah rawak normal X boleh ditentukan daripada sifir taburan normal melalui langkah-langkah berikut: The probability of the random normal variable X can be determined from the normal distribution table by using the steps below: (i) (ii) Dapatkan nilai P(Z > z) daripada jadual kebarangkalian hujung atas bagi taburan normal dalam buku sifir. Find the value of P(Z > z) from the probability table of the upper end of the normal distribution table. f(z) 0 z Z fi x 0 f(x) Proses pemiawaian Standardising process Z = x – fi ff 11. Selesaikan yang berikut. Solve the following. TP 2 (a) Rajah menunjukkan graf taburan normal bagi suatu pemboleh ubah rawak selanjar X. The diagram shows the normal distribution of a continuous random variable X. (i) Nyatakan nilai µ. State the value of µ. (ii) Nyatakan nilai P(X > µ). State the value of P(X > µ). (iii) Cari P(23 < X < 32). Find P(23 < X < 32). (i) µ = 32 (ii) P(X > 32) = 0.5 (iii)P(23 < X < 32) = 0.5 – 0.16 = 0.34 CONTOH Rajah menunjukkan graf taburan normal bagi suatu pemboleh ubah rawak selanjar X. The diagram shows the normal distribution of a continuous random variable X. (i) Nyatakan nilai µ. State the value of µ. (ii) Lorek rantau P(3 < x < µ). Shade the region of P(3 < x < µ). (iii) Ungkapkan rantau berlorek dalam tatatanda kebarangkalian. Express the shaded region in probability notation. Penyelesaian: (i) µ = 5 (iii) P(5 < X < 8) f(x) 0 3 (ii) 5 8 x x f(x) 0 23 32 38 Tip Graf bersimetri pada min µ The graph is symmetrical at the mean µ. Luas di bawah graf lebih daripada µ = 0.5 kerana jumlah luas di bawah graf dengan paksi-x = 1 (kebarangkalian keseluruhan = 1) The area under the graph is larger than µ = 0.5 because the total area under the graph with the x-axis = 1. (The total probability is 1)

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 112 BAB 5 CONTOH (b) Rajah menunjukkan graf taburan normal bagi suatu pemboleh ubah rawak selanjar X. The diagram shows the normal distribution of a continuous random variable X. Jika P(X < 76) = 0.26, cari If P(X < 76) = 0.26, find (i) P(X > 54). (ii) P(54 ≤ X ≤ 76). (i) P(X > 54) = 0.26 (ii) P(54 ≤ X ≤ 76) = 1 − 2(0.26) = 0.48 (c) Suatu pemboleh ubah rawak selanjar X bertaburan normal. Lakarkan satu graf taburan normal untuk menunjukkan yang berikut. A continuous random variable X has a normal distribution. Sketch the normal graph to show the following. (i) min, µ = 2.5. mean, µ = 2.5. (ii) Lorekkan P(2.8 < X < 3.0). Shade P(2.8 < X < 3.0). (iii) Lorekan secara anggaran P(X < x) = 0.6. Shade the estimate region of P(X < x) = 0.6. 12. Suatu pemboleh ubah rawak selanjar, X bertaburan normal dengan min, µ dan sisihan piawai, σ, boleh ditukar kepada taburan normal piawai. Lengkapkan yang berikut. A continuous random variable X has a normal distribution with mean, µ and standard deviation, σ, can be converted into standard normal distribution. Complete the following. TP 4 Pemboleh ubah X Variable X Min, µ Mean, µ Sisihan piawai, σ Standard deviation, σ Skor-z z-score 35 40 10 Z = 35− 40 10 = −0.5 (a) 0.8 = X − 67 15 X = 15(0.8) + 67 = 79 67 15 0.8 (b) 4.6 −2.1 = 4.6 – µ 0.9 µ = 4.6 + 2.1(0.9) = 6.49 0.9 −2.1 (c) 130 143 1.85 = 130 − 143 σ σ = –13 1.85 = 7.03 1.85 x f(x) 0 54 65 76 x f(x) 0 2.5 2.8 3.0

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 113 5 CONTOH 13. Dengan menggunakan jadual taburan normal piawai, cari nilai bagi setiap yang berikut diberi bahawa Z ialah pemboleh ubah taburan normal piawai. Using the standard normal distribution table, find the following values for each of the following given that Z is the standardised variable of the normal distribution. TP 2 (a) P(Z < –0.4) P(Z < –0.4) = P(Z > 0.4) = 0.3446 Graf ialah simetri pada f(z) = 0. Graph is symmetrical about f(z) = 0. (b) P(Z > –1.2) P(Z > –1.2) = 1 – P(Z ≥ 1.2) = 1 – 0.1151 = 0.8849 Jumlah luas di bawah graf ialah 1. Total area under the graph is 1. (c) P(–1.5 < Z < 2.1) 1 – P(Z > 1.5) – P(Z > 2.1) = 1 – 0.0668 – 0.0179 = 0.9153 14. Cari skor-z bagi setiap yang berikut. Find the z-score for each of the following. TP 3 (a) P(Z < z) = 0.0375 skor-z = –1.78 CONTOH P(Z > 0.3) Penyelesaian: P(Z > 0.3) = 0.3821 z 0 0.3 f(z) P (Z > 0.3) z –0.4 0 f(z) z –1.2 0 f(z) z –1.5 0 2.1 f(z) z 0 z f(z) Luas Area = 0.2451 z 0 z f(z) P(Z > z) = 0.2451 Penyelesaian: skor-z = 0.69 Sudut Kalkulator P(Z > 0.3) Tekan: MODE 1 3 3 0 . 3 = 0.38209 MODE SHIFT

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 114 BAB 5 (b) P(Z > z) = 0.9126 1 – P(Z ≤ z) = 0.9126 P(Z ≤ z) = 0.0874 skor-z = –1.357 Luas adalah lebih daripada 0.5, maka skor-z ialah negatif. Area is more than 0.5, so z-score is negative. (c) P(Z ≤ z) = 0.8251 1 – P(Z > z) = 0.8251 P(Z > z) = 0.1749 skor-z = 0.935 Luas lebih daripada 0.5, maka skor-z ialah positif. Area is more than 0.5, so z-score is positive. 15. Cari nilai k bagi setiap yang berikut. Find the value of k in each of the following. TP 3 (a) µ = 15, σ = 22, P(X > k) = 0.4152 P(Z > k − 15 22 ) = 0.4152 k − 15 22 = 0.214 k = 19.71 (b) µ = 250, σ = 16, P(X < k) = 0.32 PZ < k − 250 16 2 = 0.32 k − 250 16 = –0.468 k = 242.51 z 0 z f(z) z 0 z f(z) z 0 0.214 f(z) 0.4152 z –0.468 0 f(z) 0.32 CONTOH z 0 f(z) 0.35 0.65 k–350 38 µ = 350, σ = 38, P(X > k) = 0.65 Penyelesaian: P(X > k) = 0.65 P X − µ σ > k – 350 38 2 = 0.65 PZ > k – 350 38 2 = 0.65 1 – PZ > k – 350 38 2 = 0.65 P(Z < k – 350 38 2 = 0.35 skor-z = –0.385 k – 350 38 = –0.385 ∴k = 335.37 Tip Luas lebih daripada 0.5, maka skor-z ialah negatif. Area is more than 0.5, then z-score is negative

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 115 5 (c) µ = 40.2, σ = 12.5, P(35 < X < k) = 0.458 (d) µ = 35, σ = k, P(30 < X < 40) = 0.578 16. Selesaikan yang berikut. Solve the following. TP 5 P 35 – 40.2 12.5 < Z < k – 40.2 12.5  = 0.458 P–0.416 < Z < k – 40.2 12.5 2 = 0.458 1 – P(Z < –0.416) – P1Z > k – 40.2 12.5 2 = 0.458 P1Z < k – 40.2 12.5 2 = 0.2033 ∴ k – 40.2 12.5 = 0.83 k = 50.58 P 30 – 35 k < Z < 40 – 35 k  = 0.578 P– 5 k < Z < 5 k 2 = 0.578 1 – 2P(Z ≥ 5 k ) = 0.578 P1Z ≥ 5 k 2 = 0.211 5 k = 0.803 k = 6.23 z –0.416 0 0.83 f(z) 0.2033 0.3387 z – 0 f(z) 0.211 5 k 5 k 0.578 CONTOH Ketinggian pelajar di sebuah kelas tertentu bertaburan normal dengan min 135.6 cm dan sisihan piawai 5.8 cm. The height of students in a class is normally distributed with a mean of 135.6 cm and a standard deviation of 5.8 cm. (i) Hitung kebarangkalian seorang pelajar dipilih secara rawak akan mempunyai ketinggian antara 130 cm dengan 140 cm. Calculate the probability of a student chosen at random will have a height between 130 cm and 140 cm. (ii) Apakah ketinggian minimum bagi 80% pelajar? What is the minimum height of 80% of the students? (iii) Cari julat antara kuartil bagi taburan ketinggian ini. Find the range between the quartiles for the distribution of the heights. Penyelesaian: (i) Diberi/Given X ~ N(135.6, 5.82 ) P(130 < X < 140) = P 130 − 135.6 5.8 < X – µ σ < 140 − 135.6 5.8 2 = P(−0.9655 < Z < 0.7586) = 1 − P(Z > 0.7586) – P(Z > 0.9655) = 1 – [P(Z > 0.7586) + P(Z > 0.9655)] = 1 – 0.3912 = 0.6088 (ii) P(X ≥ x) = 0.8 P X – µ σ ≥ x − 135.6 5.8 2 = 0.8 PZ ≥ x − 135.6 5.8 2 = 0.8 1 − PZ ≤ x − 135.6 5.8 2 = 0.8 PZ ≤ x − 135.6 5.8 2 = 0.2 x − 135.6 5.8 = − 0.842 z 0 –0.9655 0.7586 f(z) Tip Lakar graf untuk menentukan rantau yang betul dahulu. Sketch the graph to determine the correct region first. z x–135.6 0 5.8 Sudut Kalkulator Guna kalkulator saintifik untuk mencari P(Z > 0.7586) + P(Z > 0.9655). Using scientific calculator to find P(Z > 0.7586) + P(Z > 0.9655). Tekan Press 1 3 ) ) + 3 = MODE MODE SHIFT DISTR 0.7586 SHIFT DISTR 0.9655 0.3912

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 116 BAB 5 x = 135.6 – 5.8(0.842) = 130.72 (iii) Kuartil bawah ialah 25% dan kuartil atas ialah 75%, perlu cari x1 dan x2 . The lower quartile is 25% and the upper quartile is 75%, need to find x1and x2. x1 x2 0 (a) Skor suatu ujian bulanan bagi sebuah kelas yang mengandungi 155 orang bertaburan normal dengan min 53 dan sisihan piawai 9. The scores of the monthly test of a class with 155 students are normally distributed with mean 53 and standard deviation 9. (i) Cari kebarangkalian skor calon yang mendapat antara 42 dengan 58. Find the probability of the score of a candidate to get between 42 dengan 58. (ii) Sekiranya calon gagal jika mendapat kurang daripada 40, berapa peratus pelajar gagal? A candidate fails if he gets less than 40, what is the percentage of the students fail? (iii) Guru yang menyemak ingin meluluskan 74% pelajarnya, cari skor lulus paling rendah yang dapat mencapai matlamat guru. The reviewing teacher wants to approve 74% of his students, find the lowest passing score that can achieve the teacher’s goal. (b) Suatu hutan simpan yang terdiri daripada 10 000 batang pokok perlu ditebang. Bilangan pokok yang perlu ditebang adalah bertaburan normal dengan min μ dan sisihan piawai σ. Kebarangkalian kurang daripada 60 batang pokok ditebang adalah 0.8 dan kebarangkalian kurang daripada 70 batang pokok ditebang adalah 0.9. Hitung nilai min dan varians bagi taburan ini. A reserve forest has 10 000 trees need to be cut down. The number of trees that need to be cut down are normally distributed with mean μ and standard deviation σ. The probability of less than 60 trees to be cut down is 0.8 and the probability of less than 70 trees to be cut down is 0.9. Calculate the value of mean and variance. P(X < 60) = 0.8 P1Z < 60 – µ σ 2 = 0.8 60 – µ σ = 0.842 60 – µ = 0.842σ ...➀ P(X < 70) = 0.9 P1Z < 70 – µ σ 2 = 0.9 70 – µ σ = 1.282 70 – µ = 1.282σ ...➁ ➁ – ➀ 10 = 0.44 σ σ = 22.73 varians/variance = 516.53 µ = 60 – 0.842(22.73) = 40.86 (i) µ = 53, σ = 9 P[42 < X < 58] = P 42 – 53 9 < Z < 58 – 53 9 4 = P– 11 9 < Z < 5 9 4 = 0.5999 (ii) P(X < 40] = PZ < 40 – 53 9 4 = 0.0743 0.0743 × 100% = 7.43% (iii) P[X > m] = 0.74 P3Z > m – 53 9 4 = 0.74 m – 53 9 = –0.643 m = 53 – 9(0.643) = 47.21 P(X > x2 ) = P(X < x1 ) = 0.25 P3 X – µ σ > x2 − 135.6 5.8 4 = 0.25 P1Z > x2 − 135.6 5.8 2 = 0.25 x2 − 135.6 5.8 = 0.674 x2 = 139.51 Oleh sebab graf bersimetri pada min, maka Since the graph is symmetry about the mean, hence x1 = 135.6 – (139.51 − 135.6) = 131.69

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 117 5 PRAKTIS PRAKTIS SPM SPM 15 Kertas 1 1. Rajah menunjukkan graf taburan normal piawai dengan min μ dan sisihan piawai σ. The diagram shows a standard normal distribution graph with mean µ and the standard deviation σ. (a) Nyatakan nilai min dan sisihan piawai State the value of the mean and the standard deviation. (b) Cari luas rantau berlorek. Find the area of the shaded region. (a) min = 0 dan sisihan piawai = 1 Mean = 0 and standard deviation = 1 (b) P(0 < Z < 1) = 0.3413 2. Pemboleh ubah X bertaburan normal dengan min 32 dan sisihan piawai σ. Diberi skor-z ialah 1.8 apabila X = 38. Cari The variable X is normally distributed with mean 32 and the standard deviation σ. Given the z-score is 1.8 when X = 38. Find (a) nilai σ. the value of σ. (b) nilai k jika P(X < k) = 0.4213. the value of k if P(X < k) = 0.4213. (a) 1.8 = 38 −32 σ , σ = 3.33 (b) P(X < k) = P(Z < k −32 3.33 ) = 0.4213 k −32 3.33 = −0.199 k = 31.34 3. Rajah menunjukkan graf taburan binomial X ~ B(4, p). the diagram shows the binomial distribution graph X ~ B(4, p). (a) Ungkapkan P(X = 0) + P(X > 2) dalam sebutan h dan k. Express P(X = 0) + P(X > 2) in terms of h and k. (b) Cari nilai p. Find the value of p. (a) P(X = 0) + P(X > 2) = 1 – h – k (b) P(X = 0)= 4 C0 p0 (1 − p)4 = 4 27 1 − p = 0.037 p = 0.963 Kertas 2 1. (a) Didapati bahawa 22% daripada murid membawa makanan tengah hari ke sekolah. Jika 7 orang murid itu dipilih secara rawak, cari kebarangkalian tepat 4 orang murid membawa bekal makanan tengah hari ke sekolah. It is found that 22% of the students bring packed lunch to school. If 7 students are chosen at random, find the probability that exactly 4 students bring packed lunch to school. (b) Jisim satu bungkus mi goreng dari sebuah kedai makanan bertaburan normal dengan min 0.46 kg dan sisihan piawai m kg. Diberi bahawa 18% daripada bungkusan mi goreng mempunyai jisim lebih daripada 0.52 kg. 2014 2015 2017 2016 f(z) 0 fi z 0 0 1 2 3 4 1 9 4 27 5 27 P[X = x] x k h

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 118 BAB 5 The mass of a packed fried mee from a restaurant is normally distributed with the mean of 0.46 kg and the standard deviation of m kg. Given that 18% of the packed fried mee have a mass of more than 0.52 kg. (i) Hitung nilai m. Calculate the value of m. (ii) Diberi bahawa 375 bungkus mi goreng dijual pada suatu hari, cari bilangan bungkus mi goreng yang mempunyai jisim antara 0.4 kg dengan 0.5 kg. Given that 375 packed fried mee are sold on a certain day, find the number of the packed fried mee that have the masses between 0.4 kg and 0.5 kg. (a) n = 7, p = 0.22 P(X = 4) = 7 C4 (0.22)4 (0.78)3 = 0.0389 (b) (i) Min 0.46 kg dan sisihan piawai m kg. Mean 0.46 kg and the standard deviation of m kg. P(X > 0.52) = 0.18 PZ > 0.52 – 0.46 m 2 = 0.18 0.52 – 0.46 m = 0.916 m = 0.52 – 0.46 0.916 = 0.066 kg (ii) P(0.4 < X < 0.5) = P1 0.4 – 0.46 0.066 < Z < 0.5 – 0.46 0.066 2 = 0.5461 0.5461 × 375 = 205 bungkus 2. (a) Jisim bagi nanas yang dihasilkan di sebuah ladang bertaburan normal dengan min 0.75 kg dan sisihan piawai 0.3 kg. Nanas itu dikelaskan kepada tiga gred, A, B dan C mengikut jisimnya, gred A > gred B > gred C. The masses of pineapples produced in a farm is normally distributed with the mean of 0.75 kg and the standard deviation of 0.3 kg. The pineapples are classified into 3 grades, A, B and C according to the weights, grade A > grade B > grade C. (i) Jisim minimum gred A ialah 1.2 kg. Jika sebuah nanas dipilih secara rawak, cari kebarangkalian bahawa nanas itu adalah gred A. The minimum mass of grade A is 1.2 kg. If a pineapple is chosen at random, find the probability that it is of grade A. (ii) Cari jisim minimum gred B jika 15% daripada nanas adalah gred C. Find the minimum mass of grade B if 15% of the pineapples are grade C. (b) Dalam perlawanan bola keranjang antara sekolah, kebarangkalian sekolah W menang ialah 40%. Pasukan sekolah W menyertai dalam n perlawanan. Kebarangkalian untuk menang sekali adalah 8 kali kebarangkalian kalah dalam semua perlawanan. In an inter-school basketball tournament, the probability that school W wins is 40%. The team of school W participated in n tournaments. The probability of winning once is 8 times the probability of losing all the games. (i) Cari nilai n. Find the value of n. (ii) Hitung sisihan piawai bagi bilangan kemenangan. Calculate the standard deviation for the number of wins. (a) (i) P(X > 1.2) = PZ > 1.2 – 0.75 0.3 2 = 0.0668 (ii) P(X < m) = PZ < m – 0.75 0.3 2 = 0.15 m – 0.75 0.3 = −1.036 m = 0.4392 kg (b) (i) p = 0.4 P(X = 1) = 8 × P(X = 0) n C1 (0.4)(0.6)n − 1 = 8 × n C0 (0.6)n n × 0.4 0.6 = 8 n = 8 × 0.6 0.4 = 12 (ii) Var(X) = npq = 12(0.4)(0.6) Sisihan piawai = 1.7 Standard deviation 2017

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 119 5 3. (a) Satu kajian menunjukkan bahawa perbelanjaan pelanggan di sebuah pasar raya bertaburan normal seperti ditunjukkan dalam rajah A survey shows that the expenditure of customers at a supermarket is normally distributed as shown in the diagram. 90 140 (RM) f(x) 50% 12% (i) Cari sisihan piawai Find the standard deviation. (ii) Jika 45 orang pelanggan dipilih secara rawak, cari bilangan pelanggan yang membelanja antara RM50 dengan RM100. If 45 customers are chosen at random, find the number of customers who spend between RM50 and RM100. (b) Didapati bahawa 23% pelanggan membelanja kurang daripada RMy. Cari nilai y. Given that 23% of the customers spend less than RMy. Find the value of y. (a) (i) Min/Mean = 90 P(X > 140) = 0.12 P(X > 140) = PZ > 140 – 90 σ 2 = 0.12 140 – 90 σ = 1.175 σ = 42.55 (ii) P(50 < X < 100) = P 50 – 90 42.55 < Z < 100 – 90 42.55 2 = 0.4193 0.4193 × 45 = 18.87 = 19 (b) P(X < y) = PZ < y – 90 42.55 2 = 0.23 y – 90 42.55 = −0.739 y = RM58.56 4. Rajah menunjukkan susunan tin bagi suatu permainan di pesta ria sekolah. The diagram shows the arrangement of the tins for a game in a school bazaar. Gerai itu menawarkan 3 lontaran bagi setiap permainan. Pelanggan perlu membayar RM4 bagi setiap permainan. Patung beruang akan diberi jika pelanggan dapat menjatuhkan semua tin bagi tiga lontaran itu dalam satu permainan. Awang ingin bermain, kebarangkalian dia dapat menjatuhkan semua tin bagi setiap lontaran ialah 0.7. The stall offers 3 throws for each game. A customer needs to pay RM4 for each game. A teddy bear will be given if the player can topple all the tins in all the three throws in a game. Awang wants to play the game, the probability that he can topple all the tins for each throw is 0.7. (a) Awang akan bermain jika dia mempunyai sekurang-kurangnya 91% peluang memenangi sekurang-kurangnya satu patung beruang dengan perbelanjaan RM24. Daripada pengiraan anda, cadangkan kepada Awang sama ada dia patut bermain atau tidak. Awang will play if he has at least 91% chance of winning at least a teddy bear with the spending of RM24. From your calculation, suggest to Awang whether he should play or not. (b) Apakah bilangan minimum permainan yang perlu jika dia ingin memenangi tiga patung beruang? What is the minimum number of games needed if he wants to win three teddy bears? (a) RM24 boleh bermain 6 set permainan RM24 can play 6 sets of game. n = 6 P(menjatuhkan semua tin dalam tiga lontaran berturut-turut) P(topple all the tins in three consecutive throws) = (0.7)3 = 0.343 P(X ≥ 1) = 1 − P(X = 0) = 1 − 6 C0 (0.343)0 (0.657)6 = 0.9196 Peratus menang ialah 91.96%. Awang patut bermain. Percentage to win is 91.96%. Awang should play (b) E(x) = np 3 = n(0.3430) n = 8.75 Dia perlu bermain 9 set. He needs to play 9 sets. 2019 2018 Praktis SPM Ekstra

Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 120 BAB 5 Sudut Sudut KBAT KBAT Tinggi 1 000 pokok di suatu hutan simpan diukur dan dicatat. Apabila graf dilukis, suatu lengkung seperti berikut didapati. Semua ketinggian adalah antara 3 sisihan piawai daripada min. The height of 1 000 trees in the reserved forest is measured and recorded. When the graph is drawn, a curve as follows is shown. All the heights are between 3 standard deviations from the mean. x m f(x) 14.0 15.2 15.8 17.0 17.6 18.2 16.4 34%34% 13.5% 13.5% 2.5% 2.5% (a) Apakah min tinggi dan sisihan piawai bagi taburan ini? What is the mean and standard deviation of the distribution? (b) Berapakah peratus pokok yang tingginya antara 15.2 m dengan 17.0 m? What is the percentage of the trees that are between 15.2 m and 17.0 m? (c) Berapa banyak pokok yang kurang daripada 15.2 m? How many trees are less than 15.2 m? KBAT Ekstra Quiz 5 (a) min/mean = 16.4 P[X > 17] = 0.16 P3Z > 17 – 16.4 σ 4 = 0.16 17 – 16.4 σ = 0.994 σ = 0.6036 m (b) P[15.2 < X < 17] = 13.5 + 34 + 34 = 81.5% (c) P[X < 15.2] = 0.025 P3Z < 15.2 – 16.4 0.6036 4 = 0.01563 0.01563 × 1000 = 15.63 pokok

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 121 6 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN BAB 6 Fungsi Trigonometri Trigonometric Functions 6.1 Sudut Positif dan Sudut Negatif Positive Angles and Negative Angles 1. Rajah di bawah menunjukkan garis lurus OP berputar pada pusat O di atas satah Cartes dan membentuk sudut θ1 dan θ2 bermula dengan paksi-x yang positif. The diagram shows the line OP is rotated about the centre O on a cartesian plane and forms the angles of θ1 and θ2 starting with the positive x-axis. 2. Sudut θ1 diukur daripada paksi-x positif dalam arah lawan jam dikenali sebagai sudut positif manakala θ2 bermula dengan paksi-x yang positif pada arah ikut jam dikenali sebagai sudut negatif. The angle θ1 is measured from the positive x-axis in the anticlockwise direction is defined as positive angle while the angle θ2 is measured from the positive x-axis in the clockwise direction is defined as a negative angle. 3. Satah Cartes ini dibahagikan kepada empat sukuan. The Cartesian plane is divided into four quadrants. y x P 0 fi1 fi2 Sudut positif Positive angle Sudut negatif Negative angle I fi ff 0 P x y II fi ff x y ff = fi III fi ff x y ff = 180ffi – fi ff = fi – 180ffi ff = 360ffi – fi IV fi ff x y 0 0 0 90° 0° 360° III IV I 180° Sukuan Pertama First Quadrant 0° < θ < 90° atau/or 0° < θ < π 2 Sukuan Keempat Fourth Quadrant 270°< θ < 360° 3π 2 < θ < 2π Sukuan Kedua Second Quadrant 90° < θ < 180° atau/or π 2 < θ < π Sukuan Ketiga Third Quadrant 180° < θ < 270° atau/or π < θ < 3 2 π II 4. Jika garis OP berputar lebih daripada 360° atau 2π, sudut yang dibentuk akan melebihi 360° atau 2π. If the line OP rotates more than 360° or 2π, the angle formed is more than 360° or 2π. 5. Sudut rujukan ditandakan sebagai α ialah sudut tirus positif yang dibentuk oleh jejari pemutaran OP terhadap paksi-x sahaja. The reference angle marked as α is the positive acute angle formed by the rotating radius of the OP against the x-axis only. (a) (b) (c) (d) 6. Sudut tirus adalah digunakan untuk mencari nisbah trigonometri asas. The acute angle is used to find the basic trigonometric ratios.

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 122 BAB 6 (a) 121° (b) –316° (c) 520° 1. Tempatkan setiap sudut yang berikut dalam sukuan yang betul dan seterusnya nyatakan sudut rujukan masing-masing. Place each of the following angles in the correct quadrant and then state its reference angle respectively. TP 1 CONTOH (i) 295° (ii) –139° (iii) 3π 5 Penyelesaian: (i) (ii) (iii) Tip –139° ialah sudut negatif. − 139° is a negative angle fi = 65ffl 295ffl y x 0 IV fi –139ff y x 0 III fi 108ff y x 0 II Sukuan keempat 4th quadrant Sudut rujukan/Reference angle α = 360° − 295° = 65° (dengan paksi-x) (with x-axis) Perhatian: α = 295° − 270° = 25° SALAH! WRONG! Sudut rujukan Reference angle = 180° − 139° = 41° Sudut rujukan Reference angle = 180° − 108° =72° fi = 25ffl 295ffl y x α = 180° − 121° = 59° α = 360° − 316° = 44° α = 180° − 160° = 20° fi 121ff y x II 0 fi 316ff y x 0 I fi 520ff y x II 0

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 123 6 (d) –412° (e) 2 – 3 π (f) 9 8 π 2. Nyatakan sudut sebenar, θ daripada rajah yang diberi dengan sudut rujukannya. State the correct angle, θ from the diagram with the given reference angle. TP 2 (a) (b) α = 412° − 360° = 52° θ = 360° + (180° – 42°) = 498° θ = –3π – π 6  = – 17 6 π kerana arah ikut jam. because follows clockwise direction. fi –412ff y x 0 IV –120fi 60fi y x 0 III fi 9ff 8 y x 0 9 8 p = 9 × 180° 8 = 202.5° α = 9 8 p – p = π 8 2 – 3 π = –2 × 180° 3 = –120° α = 60° CONTOH Penyelesaian: θ = 2 – 3 π kerana mengikut arah jam. θ = 2 – 3 π because follows clockwise direction. y x 0 P fi 3 ff 42fi y x P 0 fi 6 y x 0

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 124 BAB 6 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 6.2 Nisbah Trigonometri bagi Sebarang Sudut Trigonometric Ratios of any Angle 1. Apabila sudut θ diletak dalam sukuan pertama seperti ditunjukkan dalam rajah, OQ = x, OY = y dan OP = r = x2 + y2 , maka rajah When the angle θ is at the first quadrant as shown in the diagram, OQ = x, OY = y and OP = r = x2 + y2 , then the diagram sin θ = y x kos/ cos θ = x r tan θ = y x = sin θ kos/cos θ 2. Rajah yang berikut menunjukkan tanda-tanda nisbah trigonometri mengikut sukuan. The following diagram shows the signs for the trigonometric ratios according to the quadrants. 3. Salingan bagi nisbah trigonometri asas adalah seperti berikut. The inverse of the basic trigonometric ratios are as follows. kosekan/ cosecan θ = 1 sin θ = r y , sekan/ secan θ = 1 kos / cos θ = r x kotangen/ cotangen θ = 1 tan θ = x y = kos θ sin θ 4. Pertimbangkan sudut θ dengan pelengkapnya (90° − θ) dalam rajah di bawah, didapati bahawa Consider the angle θ with its complement (90° − θ) in the diagram below, it is found that (a) sin θ = kos (90°− θ) = y r (b) kos/ cos θ = sin (90° − θ) = x r (c) tan θ = kot/ cot (90° − θ) = y x (d) kosek/ cosec θ = sek/ sec (90°− θ) = r y (e) sek/ sec θ = kosek/ cosec (90°− θ) = r x (f ) kot/ cot θ = tan (90° − θ) = x y 5. Sudut istimewa 30°, 45°, 60°, 90°, 180°, 270° dan 360°. Special angles 30°, 45°, 60°, 90°, 180°, 270° and 360°. Merujuk kepada rajah, jadual di bawah didapati. Referring to the diagrams, the table below is found. θ 30° 45° 60° 90° 180° 270° 360° sin θ 1 2 1 2 3 2 1 0 −1 0 kos/ cos θ 3 2 1 2 1 2 0 −1 0 1 tan θ 1 3 1 3 ∞ 0 ∞ 0 0 P(x , y) y x y x fi r y x Sukuan II Quadrant II sinus (+) sin θ = sin (180° – θ) kos θ = –kos (180° – θ) tan θ = –tan (180° – θ) Sukuan III Quadrant III sin θ = –sin(θ – 180°) kos θ = –kos(θ – 180°) tan θ = tan(θ – 180°) tangen (+) Sukuan IV Quadrant IV sin θ = –sin(360° – θ) kos θ = kos(360° – θ) tan θ = –tan(360° – θ) kosinus (+) Sukuan I Quadrant I All (+) Semua (+) 0 P y x y x fi r (90ff–fi) 2 1 30fi 60fi √3 1 1 √2 45fi 45fi

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 125 6 3. Ungkapkan setiap fungsi trigonometri berikut dalam sudut rujukan yang sepadan. Express each of the following trigonometric function in the corresponding reference angle. TP 2 (a) sin 249° (b) sek (−200°) sec (−200°) (c) kot 219° cot 219° (d) kos (−152°) cos (−152°) (e) −kosek (−32°) −cosec (−32°) (f) –tan 3 4 π2 sin 249° = –sin 69° kot/ cot 219° = 1 tan 219° = kot/ cot 39° –kosek(–32°) = – 1 sin(–32°) = – 1 –sin 32° = kosek/ cosec 32° –tan 3 4 π2 = – tan  3 × 180° 4 2 = –tan 135° = −(−tan 45°) = tan 45° kos/ cos (−152°) = kos/ cos 28° sek (–200°)= 1 kos(–200°) 1 –kos 20° = –sek 20° CONTOH (i) tan 315° (ii) kosek (–15°) cosec (–15°) Penyelesaian: (i) tan 315° = –tan 45° kosek (−15°) cosec (−15°) = 1 sin(−15°) = 1 –sin15° = −kosek 15° −cosec 15° x y 315fi 45fi = ffl 0 y x 0 15fi Tip – Tentukan sudut dalam sukuan yang betul dahulu. Determine the angle in the right quadrant first. – Cari sudut rujukan. Find the reference angle. – Hanya kosinus positif dalam sukuan IV Only positive cosine in quadrant IV. Tip Tukar kosek θ = 1 sin θ Change cosec θ = 1 sin θ y x 69fi 0 y x 20fi 0 y x 39fi 0 y x 28fi 152fi 0 y x 0 –32fi y x 45fi 135fi 0

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 126 BAB 6 (g) –sin 8 – 3 π2 (h) –kot 7 4 π2 –cot  7 4 π2 4. Tentukan nilai nisbah trigonometri bagi sebarang sudut yang diberi. Determine the value of the trigonometric ratio for any given angles. TP 3 (a) Diberi/Given tan θ = 5 12 , 180° ≤ θ ≤ 270°. Cari / Find (i) sin θ (ii) sek/ sec θ (iii) kot/ cot θ –sin 8 – 3 π2 = –sin(–480°) = –(–sin 60°) = sin 60° –kot( 7 4 π) = – 1 tan 315° = – 1 –tan 45° = kot/ cot 45° y x 60fi 0 y x 0 45fi CONTOH Tip – Tempatkan sudut dalam sukuan betul. Place the angle in the correct quadrant. – Lengkapkan sisi dengan teoram Pythagoras. Complete the sides with Pythagoras theorem. y x –4 3 5 fi 0 y x –5 –12 13 θ 0 Diberi/Given sin θ = 3 5 , 90° ≤ θ ≤ 180°. Cari / Find (i) tan θ (ii) kosek/ cosec θ (iii)sek/ sec θ Penyelesaian: (i) tan θ = 3 –4 (ii) kosek/ cosec θ = 1 sin θ = 1 3 5 = 5 3 (iii) sek/ sec θ = 1 kos θ = 1 4 – 5 = 5 – 4 (i) sin θ = –5 13 (ii) sek/ sec θ = 1 kos/ cos θ = 1 –12 13 = 13 – 12 (iii) kot/ cot θ = 1 tan θ = 12 5

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 127 6 (b) Diberi sek θ = − 2 , 90° ≤ θ ≤ 180°. Given sec θ = − 2 , 90° ≤ θ ≤ 180°. Cari / Find (i) kot/ cot θ (ii) kosek/ cosec θ (iii)sin θ (c) Diberi kot θ = − 3 , 270° ≤ θ ≤ 360°. Given cot θ = − 3 , 270° ≤ θ ≤ 360°. Cari / Find (i) kos/ cos θ (ii) kosek/ cosec θ (iii)sek/ sec θ 5. Ungkapkan yang berikut dalam sebutan q. Express the following in terms of q. TP 3 (a) Diberi/Given tan θ = −p, 90° ≤ θ ≤ 180°. Cari/Find (i) kos/ cos θ (ii) kosek/ cosec θ (iii) sek/ sec θ sek/sec θ = – 2 = 1 cos θ kos/cos θ = –1 2 (i) kot/cot θ = –1 (ii) kosek/cosec θ = 1 sin θ = 2 (iii)sin θ = 1 2 kot/cot θ = – 3 tan θ = –1 3 (i) kos/cos θ = 3 2 (ii) kosek/cosec θ = 1 sin θ = 1 1 – 2 = –2 (iii)sek/sec θ = 1 kos/cos θ = 1 3 2 = 2 3 (i) kos/cos θ = –1 1 + p2 (ii) kosek/cosec θ = 1 sin θ = 1 + p2 p (iii)sek/sec θ = 1 kos/cos θ = – 1 + p2 y x √2 –1 1 fi 0 CONTOH Diberi kos θ = p, 0° ≤ θ ≤ 90°. Given cos θ = p, 0° ≤ θ ≤ 90°. Cari/Find (i) tan θ (ii) sek/ sec θ (iii)sin θ Penyelesaian: (i) tan θ = 1 – p2 p (ii) sek/ sec θ = 1 kos/cos θ = 1 p (iii)sin θ = 1 – p2 y x √1–p2 p 1 fi 0 y x –1 √1+p2 p fi 0 y x √3 –1 2 0 fi

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 128 BAB 6 (b) Diberi/Given sin θ = 2 – p , 180° ≤ θ ≤ 270°. Cari/Find (i) kot/ cot θ (ii) sek/ sec θ (iii) kos/ cos θ (c) Diberi kot θ = 1 + p, 0° ≤ θ ≤ 90°. Given cot θ = 1 + p, 0° ≤ θ ≤ 90°. Cari/Find (i) sin θ (ii) kosek/ cosec θ (iii) kos/ cos θ 6. Cari nilai θ untuk 0° ≤ θ ≤ 90° tanpa menggunakan kalkulator. Find the value of θ for 0° ≤ θ ≤ 90° without using calculator. TP 3 (a) kos θ = sin 13° 42’ cos θ = sin 13° 42' (b) kot θ = tan 82° 15’ cot θ = tan 82° 15' CONTOH (i) sin θ = kos 63° (ii) sek θ = kosek 54° sin θ = cos 63° sec θ = cosec 54° Penyelesaian: (i) sin θ = kos 63° = sin(90° – 63°) = sin 27° ∴θ = 27° (i) kot/cot θ = p2 – 4 2 (ii) sek/sec θ = 1 kos θ = –p p2 – 4 (iii) kos/cos θ = – p2 – 4 2 (i) sin θ = 1 2 + 2p + p2 (ii) kosek/cosec θ = 1 sin θ = 2 + 2p + p2 (iii) kos/cos θ = 1 + p 2 + 2p + p2 x y –2 p fi –√p2 –4 0 x y 1 (1+p) fi √2+2p+p2 0 y y r r x x 63fi 27fi ff 90fi–ff 0 0 (ii) sek θ = kosek 54° sec θ = cosec 54° 1 kos θ = 1 sin 54° kos θ = sin 54° = y r = kos 36° ∴θ = 36° y x r 54fi 36fi 0 Tip Tempatkan sudut dalam rajah. Gunakan hubungan sudut pelengkap. sin θ = kos (90° – θ) Place the angle in the right quadrant. Use the relation of complementary angle. sin θ = cos (90° – θ) y x 13fi42ff 76fi18ff 0 kos/cos θ = sin 13° 42’ = kos 76° 18’ θ = 76° 18’ kot/cot θ = tan 82° 15’ = kot 7° 45’ ∴θ = 7° 45’ y x 82fi15ff 7fi45ff 0

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 129 6 (c) kosek θ = sek p° cosec θ = sec p° (d) sin θ = kos 34° 54’ sin θ = cos 34° 54' (e) tan (90° − θ) = kot 68° 34’ tan (90° − θ) = cot 68° 34' 7. Diberi sin 67° = 0.9205, kos 67° = 0.3907 dan tan 67° = 2.3559, dengan 0° ≤ θ ≤ 90°, cari nilai yang berikut tanpa menggunakan kalkulator. Given sin 67° = 0.9205, cos 67° = 0.3907 and tan 67° = 2.3559, where 0° ≤ θ ≤ 90°, find the following values without using a calculator. TP 3 (a) kos/ cos 23° (b) tan 23° (c) sek 23° = kosek (90° – 23°) sec 23° = cosec (90° – 23°) y x pfi 90fi–pfi 0 y x 55fi6ff 34fi54ff 0 y x 21fi26ff 68fi34ff 0 kosek/ cosec θ = sek/ sec p° 1 sin θ = 1 kos/ cos p° sin θ = kos/ cos p° = sin 90° − p° θ = 90° − p tan 23°= kot/ cot = (90 – 23°) = kot/ cot 67° = 1 tan 67° = 1 2.3559 = 0.4245 sek/ sec 23° = kosek/ kosec 67° = 1 sin 67° = 1 0.9205 = 1.0864 kos/ cos 23° = sin(90° − 23°) = sin 67° = 0.9205 sin θ = kos/ cos 34° 54’ = sin 55° 6’ θ = 55° 6’ tan (90° − θ) = kot/ cot 68° 34’ = tan 21° 26’ 90° − θ = 21° 26’ 90° − 21° 26’ = θ θ = 68° 34’ CONTOH sin 23° Penyelesaian: sin 23° = kos/ cos (90° − 23°) = kos/ cos 67° = 0.3907 Tip Gunakan hubungan sudut pelengkapnya Use the complementary angle relationship.

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 130 BAB 6 CONTOH 8. Gunakan nilai nisbah bagi sudut khas 30°, 45° dan 60° untuk mencari nilai bagi setiap yang berikut. Use the value of the ratios of the special angles 30°, 45° and 60° to find the value of each of the following. TP 3 (a) tan 225° (b) sek/ sec (−240°) (c) – kosek/ cosec (330°) (d) – sin (−300°) (i) kos/ cos 210° (ii) sek/ sec (−135°) Penyelesaian: (i) kos/ cos 210° = −kos/ cos 30° = 3 – 2 (ii) sek/ sec (−135°) = 1 kos/ cos (–135°) = 1 –kos/ cos 45° = 1 – 2 2 –1 –√3 210fi 30fi y x 0 √2 –1 –1 45fi y x 0 Tip Lukis dan lengkapkan sisi segi tiga yang mengandungi sudut khas yang tertentu dalam sukuan yang dikehendaki. Draw and complete the sides of the triangle which contains certain special angle in the correct quadrant. tan 225° = tan 45° = 1 – kosek/ cosec (330°) = – 1 sin 330° = – 1 –sin 30° = 1 1 2 = 2 sek/ sec (−240°)= 1 kos/ cos(–240°) = 1 –kos/ cos 60° = 1 1 – 2 = –2 – sin (−300°) = – sin 60° = 3 – 2 y x –1 –1 √2 45fi 225fi 0 y x –1 60fi –240fi √3 2 0 y x –1 30fi √3 2 0 y x 1 60fi √3 2 0

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 131 6 NOTA IMBASAN NOTA IMBASAN (e) sin 315° + kos/ cos 135° (f) sek 420° + kosek (−120°) sec 420° + cosec (−120°) y x 1 1 –1 –1 45fi 45fi √2 √2 y x 2 60fi 1 60fi √3 sin 315° + kos/ cos 135° = 1 – 2 +  1 – 2  = 2 – 2 sek/sec 420° + kosek/cosec (−120°) = 1 kos/ cos 420° + 1 sin (–120°) = 1 kos/ cos 60° + 1 –sin 60° = 1 1 2 – 2 3 = 2 – 2 3 = 2 3 – 1 3 2 NOTA IMBASAN NOTA IMBASAN 6.3 Graf Fungsi Sinus, Kosinus dan Tangen Graphs of Sine, Cosine and Tangent Functions 1. Graf fungsi sinus y = a sin bx + c, dengan keadaan a, b dan c ialah pemalar dan b > 0. Graph of sine function y = a sin bx + c, where a, b and c are constants and b > 0. a = amplitud/ amplitude b = bilangan kala dalam 360° atau 2π/ number of complete cycles in 360° or 2π c = anjakan mencancang ke atas atau ke bawah bagi graf vertical displacement up or down of the graph 2. Graf/Graph y = a sin bx + c c berubah c changes y = sin x + 1 –1 1 2 –2 –1 1 –1 y x b berubah b changes Amplitud/Amplitude y x y = sin x 0 0 fi 0 2fi fi 2fi fi 2fi fi 2fi a berubah a changes 1 kala/cycle y x y = – sin x 3 2 y = 2 sin x y = sin x y = sin x y = sin 2x y x 0 1 2 1 1 1 2 1 2 1 2 –1 –1 1 2 – 1 2 y = sin x – 0 1 –1 y x y = │sin x │ fi 2fi 0 1 –1 y x y = –│sin x │

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 132 BAB 6 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 3. Graf y = a kos bx + c Graph y = a cos bx + c 4. Graf/ Graph y = a tan bx + c –1 1 0 –1 1 fi 2fi 0 fi 2fi y y x –1 1 0 fi 2fi y x –1 1 0 fi 2fi y x x 1 kala 1 cycle y = kos x – 1 y = cos x – 1 y = –kos x y = –cos x y = kos x y = cos x y = kos x y = cos x b berubah b changes y = kos 2x y = cos 2x y = │kos x│ y = │cos x│ 0 y fi 2fi x 0 y 2fi fi x 0 y fi 2fi x 0 y fi 2fi x 1 y = tan x y = tan 2x y = tan x +1 y = │tan x│ y = –│tan x│

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 133 6 9. Lengkapkan jadual dan seterusnya lukis graf y berdasarkan kepada nilai-nilai dalam jadual. Complete the table and then draw the graph y based on the values in the table. TP 3 (a) y = kos/ cos x x 0 π 4 π 2 3 4 π π 5 4 π 3 2 π 7 4 π 2π 0° 45° 90° 135° 180° 225° 270° 315° 360° kos/cos x 1 0.71 0 –0.71 –1 –0.71 0 0.71 1 CONTOH y = sin x x 0 π 4 π 2 3 4 π π 5 4 π 3 2 π 7 4 π 2π 0° 45° 90° 135° 180° 225° 270° 315° 360° sin x 0 0.71 1 0.71 0 –0.71 –1 –0.71 0 x sin x 1 –1 0 90fi 180fi 270fi 360fi x kos/ cos x 1 –1 0 90fi 180fi 270fi 360fi

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 134 BAB 6 (b) y = tan x x 0 π 4 π 2 3 4 π π 5 4 π 3 2 π 7 4 π 2π 0° 45° 90° 135° 180° 225° 270° 315° 360° tan x 0 1 ∞ –1 0 1 ∞ –1 0 10. Lakar graf fungsi trigonometri yang berikut. Sketch the following trigonometric function graph. TP 3 CONTOH (i) y = 1 – 2 sin 2x untuk/for 0 ≤ x ≤ 2π. Penyelesaian: Tip – Tentukan nilai a = amplitud (di sini ialah a = 1 – 2 ) Determine the value of a = amplitude (a = 1 – 2 ) – Tentukan berapa kala dalam 2π. (b = 2 kala) Determine the number of cycles in 2π. (b = 2 cycles) – Ada c untuk anjakan atas atau bawah (c = 0) Is there any c of vertical up or down movement (c = 0) x y – 0 1 2 1 2 fi 2fi 1 2 y = – sin 2x x tan x 4 2 –4 –2 0 90fi 180fi 270fi 360fi

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 135 6 (a) y = 3 4 kos 2x untuk 0 ≤ x ≤ 2π. y = 3 4 cos 2x for 0 ≤ x ≤ 2π. (b) y = 1 + kos 2x untuk 0 ≤ x ≤ π. y = 1 + cos 2x for 0 ≤ x ≤ π. (c) y = tan 2x − 1 untuk/for 0 ≤ x ≤ π. (d) y = 3  kos 1 2 x untuk 0 ≤ x ≤ 2π. y = 3  cos 1 2 x for 0 ≤ x ≤ 2π. (e) y = − sin 3 4 x − 1 untuk/for 0 ≤ x ≤ 2π. (f) y = 2 − |tan x| untuk/for 0 ≤ x ≤ 2π. x y – 0 3 4 3 4 fi 2fi 3 4 y = kos 2x 3 4 y = cos 2x x y –2 –1 0 2 1 fi y = 1+ kos 2x y = 1+ cos 2x x y –1 fi y = tan 2x–1 0 x y –3 –2 –1 0 1 2 3 fi 2fi 1 2 y = 3│kos x│ 1 2 y = 3│cos x│ x y –2 –1 0 1 fi 2fi y = –│sin x│–1 3 4 x y 0 2 fi 2fi y = 2 – │tan x│

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 136 BAB 6 11. Cari nilai-nilai a, b dan c bagi setiap lakaran graf yang berikut. Find the values of a, b and c for each of the following graphs. TP 3 (a) y = a kos bx + c y = a cos bx + c (b) y = a | kos bx | + c y = a | cos bx | + c (c) y = |a sin bx + c| a = −2 ; b = 12 kala/cycle ; c = 0 CONTOH y = a sin bx + c Penyelesaian: Amplitud/ Amplitude, a = 1 2 b = 1 2 kala dalam 2π / 1 2 cycle in 2π b = 1 2 Anjakan ke atas dari 0 Upward shift from 0 c = 1 2 y = a kos bx + c a = −2 1 1 2 kala dalam 2π 1 1 2 cycle in 2π b = 3 2 ; c = 0 a = −2 ; b = a = 2 ; b = 1 kala/cycle ; c = 1 1 2 kala/cycle ; c = 0 y x 0 1 1 2 2fi y x 0 2 –2 2fi y x 0 –2 fi 2fi y x 0 3 2 1 fi 2fi

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 137 6 12. Selesaikan persamaan trigonometri dengan menggunakan graf. Solve the trigonometric equations by using graphs. TP 4 π(1 – tan x) = 2x 1 – tan x = 2x π 1 – 2x π = tan x Maka/Hence y = 1 – 2x π Apabila/When x = 0 y = 1 x = π y = –1 Terdapat 3 penyelesaian There are 3 solution CONTOH Dengan menggunakan skala dan paksi yang sama, lakarkan graf lengkung y = kos 2x dan garis lurus y = 3 4 1 − 2x π  untuk 0 ≤ x ≤ 3 2 π. Nyatakan bilangan penyelesaian yang memuaskan 6x π + 4 kos 2x = 3. With the same scale and on the same axes, sketch the curve y = cos 2x and the straight line y = 3 4 1 − 2x π 2 for 0 ≤ x ≤ 3 2 π. State the number of solutions which satisfies 6x π + 4 cos 2x = 3. Penyelesaian Apabila/When x = 0, y = 3 4 x = π, y = 3 – 4 6x π + 4 kos 2x = 3 / 6x π + 4 cos 2x = 3 kos 2x = 3 4 − 6x 4π y = 3 4 1 − 2x π 2 Maka, ada dua penyelesaian. Hence, there is two solutions. Tip Titik persilangan ialah penyelesaian. The intersection point is a solution. (a) (i) Lakar graf lengkung y = tan x untuk 0 ≤ x ≤ 2π. Sketch the curve of y = tan x for 0 ≤ x ≤ 2π. (ii) Pada paksi yang sama, lakarkan satu graf yang sesuai untuk menyelesaikan persamaan π(1 – tan x) = 2x. Nyatakan bilangan penyelesaian yang memuaskan π(1 – tan x) = 2x. On the same axes, sketch a suitable graph to solve the equation π(1 – tan x) = 2x. State the number of solutions which satisfies π(1 – tan x) = 2x. y x 0 1 –1 fi 2fi y = kos 2x y = cos 2x 3fi 2 y = (1– ) 2x fi 3 4 y x 0 1 fi 2fi y = tan x

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 138 BAB 6 (b) (i) Lakar graf y = |2 sin 2x| untuk 0 ≤ x ≤ 2π. Sketch the graph of y = |2 sin 2x| for 0 ≤ x ≤ 2π. (ii) Pada paksi yang sama, lakar satu graf yang sesuai untuk menyelesaikan persamaan 2|2 sin 2x| − 3 = − 2x π . Nyatakan bilangan penyelesaian yang memuaskan 2|2 sin 2x| − 3 = − 2x π . On the same axes, sketch a suitable graph to solve the equation 2|2 sin 2x| − 3 = − 2x π . State the number of solutions which satisfies 2|2 sin 2x| − 3 = − 2x π . (c) (i) Lakar graf y = −2 sin 3 2 x untuk 0 ≤ x ≤ 2π. Sketch the curve y = −2 sin 3 2 x for 0 ≤ x ≤ 2π. (ii) Pada paksi yang sama, lakar satu graf yang sesuai, y = k supaya terdapat satu penyelesaian sahaja. Cari nilai yang mungkin bagi k. On the same axes, sketch a suitable graph, y = k so that there is only one solution. Find the possible values of k (i) y = –2 sin 3 2 x 0 ≤ x ≤ 2π (ii) y = k = −2 atau/or 0 < k ≤ 2 (i) (ii) π 2x = –2 kos 3 4 x ∴ y = π 2x = π 2  1 x  2 penyelesaian/solutions (i) (ii) 2|2 sin 2x| – 3 = –2x π |2 sin 2x| = 3 2 – x π ∴ y = 3 2 – x π Apabila/When x = 0, y = 3 2 x = 2π y = 3 2 – 2 = – 1 2 5 penyelesaian/solutions (d) (i) Lakar graf y = −2 kos 3 4 x untuk 0 ≤ x ≤ 2π. Sketch the curve y = −2 cos 3 4 x for 0 ≤ x ≤ 2π. (ii) Pada paksi yang sama, lakar satu graf yang sesuai untuk menyelesaikan persamaan π 2x + 2 kos 3 4 x = 0 untuk 0 ≤ x ≤ 2π. Nyatakan bilangan penyelesaian itu. On the same axes, sketch a suitable graph to solve the equation π 2x + 2 cos 3 4 x = 0 for 0 ≤ x ≤ 2π. State the number of solutions. y x 0 2 –2 fi 2fi y = │2 sin 2x│ y x 0 2 –2 2fi y = k y = k 3 2 y = –2 sin x y x 0 2 –2 2fi fi 2x y = y = –2 kos x 3 4 y = –2 cos x 3 4

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 139 6 (e) (i) Lakar graf y = 2 sin 2x + 1 untuk 0 ≤ x ≤ 3 2 π. Sketch the curve y = 2 sin 2x + 1 for 0 ≤ x ≤ 3 2 π. (ii) Seterusnya, pada paksi yang sama, lakar satu graf yang sesuai untuk menyelesaikan persamaan 2 sin 2x + 2 kos x + 1 = 0. Nyatakan bilangan penyelesaian. On the same axes, sketch a suitable graph to solve the equation 2 sin 2x + 2 cos x + 1 = 0. State the number of solutions. (i) (ii) 2 sin 2x + 2 kos x + 1 = 0 2 sin 2x + 1 = –2 kos x ∴ y = –2 kos x 2 penyelesaian/solutions NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 6.4 Identiti Asas Basic Identities 1. Identiti asas / Basic identities: (a) sin2 A + kos2 A = 1 sin2 A + cos2 A = 1 (b) 1 + tan2 A = sek2 A 1 + tan2 A = sec2 A (c) 1 + kot2 A = kosek2 A 1 + cot2 A = cosec2 A CONTOH Diberi satu titik P(x, y) dan OP membuat sudut A dengan paksi-x, buktikan sin2 A + kos2 A = 1. Given a point P(x, y) and OP makes an angle A with the x-axis, prove that sin2 A + cos2 A = 1. Penyelesaian: sin A = y r ... ➀ kos/ cos A = x r ... ➁ Menurut teorem Pythagoras, According to Pythagoras’ theorem, x2 + y2 = r2 x2 r2 + y2 r2 = 1 Dari ➀ dan ➁, sin2 A = y2 r2 , kos2 A = x2 r2 . From ➀ and ➁, sin2A = y2 r2 , cos2A = x2 r2 . Maka, sin2 A + kos2 A = 1 (Terbukti) Hence, sin2A + cos2A = 1 (Proved) 13. Dengan rajah yang diberi, buktikan rumus yang berikut. With the diagram given, prove the following formula. TP 3 y x 0 2 1 3 –2 –1 fi 2 3 2 fi fi y = 2 sin 2x+1 x y A 0 x P(x, y) r y

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 140 BAB 6 (a) Dengan kaedah yang serupa, buktikan 1 + tan2 A = sek2 A. With the similar method, prove that 1 + tan2 A = sec2 A. (b) Dengan kaedah yang serupa, buktikan 1 + kot2 A = kosek2 A With the similar method, prove that 1 + cot2 A = cosec2 A. 14. Buktikan identiti trigonometri yang berikut dengan menggunakan identiti asas. Prove the following trigonometric identities with the basic identities. TP 4 CONTOH Tip Sentiasa bermula dengan sebelah yang lebih kompleks. Always starts with a more complex side. Sebelah kanan: sek2 A = 1 kos2 A = r2 x2 Right: sec2A = 1 cos2A = r2 x2 Sebelah kiri: 1 + tan2 A = 1 + y2 x2 = (x2 + y2 ) x2 Left: 1 + tan2 A = 1 + y2 x2 = (x2 + y2) x2 Tetapi/But x2 + y2 = r2 Maka/Hence 1 + tan2 A = r2 x2 = sebelah kanan 1 + tan2 A = sek2 A Sebelah kanan: kosek2 A = 1 sin2 A = r2 y2 Right: cosec2A = 1 sin2A = r2 y2 Sebelah kiri: 1 + kot2 A = 1 + x2 y2 = (x2 + y2 ) y2 Left: 1 + cot2 A = 1 + x2 y2 = (x2 + y2) y2 Tetapi/But x2 + y2 = r2 Maka/Hence 1 + kot2 A = r2 y2 = sebelah kanan 1 + kot2 A = kosek2 A. 1 – tan2 A 1 + tan2 A = 1 − 2 sin2 A Sebelah kiri = 1 – tan2 A 1 + tan2 A Left = (1 − sin2 A kos2 A) ÷ (1 + sin2 A kos2 A) (1 − sin2 A cos2 A ) ÷ (1 + sin2 A cos2 A ) = (kos2 A kos2 A − sin2 A kos2 A) × kos2 A kos2 A + sin2 A ( cos2 A cos2 A − sin2 A cos2 A ) × cos2 A cos2 A + sin2 A = kos2 A − sin2 A cos2 A − sin2A = 1 − sin2 A − sin2 A = 1 − 2 sin2 A = Sebelah kanan/Right side Gunakan/Use kos2 A + sin2 A = 1 cos2 A + sin2 A = 1

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 141 6 (a) kos A 1 – sin2 A = sek A cos A 1 – sin2 A = sec A (b) 1 − 2 sin2 A = 2 kos2 A − 1 1 − 2 sin2 A = 2 cos2 A − 1 (c) tan A + kot A = sek A kosek A tan A + cot A = sec A cosec A (d) kot2 A + kos2 A = kosek2 A – sin2 A cot2 A + cos2 A = cosec2 A – sin2 A (e) 1 + kos A 1 – kos A = (kosek A + kot A)2 Sebelah kiri: kos A 1 – sin2 A = kos A kos2 A = 1 kosA Left: cos A 1 – sin2A = cos A cos2A = 1 cos A = sek/ sec A = sebelah kanan/right side Sebelah kiri / Left side = 1 − 2(1 − kos2 A) 1 − 2(1 − cos2 A) = 1 − 2 + 2 kos2 A 1 − 2 + 2 cos2 A = 2 kos2 A − 1 (sebelah kanan) 2 cos2 A − 1 (right side) Sebelah kiri / Left side = sin A kos/ cos A + kos/ cos A sinA = sin2 A + kos2 A kos A sin A = 1 kos A sin A = sin2 A + cos2 A cos A sin A = 1 cos A sin A = sek A kosek A (sebelah kanan) = sec A cosec A (right side) Sebelah kanan / Right side = (kosek A + kot A)2 (cosec A + cot A)2 = kosek2 A + 2 kosek A kot A + kot2 A cosec2 A + 2 cosec A cot A + cot2 A = 1 sin2 A + 2 kos A sin2 A + kos2 A sin2 A 1 sin2A + 2 cos A sin2A + cos2 A sin2A = 1 + 2 kos A + kos2 A sin2 A = (1 + kos A)2 sin2 A 1 + 2 cos A + cos2 A sin2A = (1 + cos A)2 sin2A Sebelah kiri / Left side kot2 A + kos2 A = kosek2 A − 1 + kos2 A cot2 A + cos2 A = cosec2 A − 1 + cos2 A = kosek2 A – (1 − kos2 A) cosec2 A – (1 − cos2 A) = kosek2 A – sin2 A (sebelah kanan) cosec2 A – sin2 A (right side) = (1 + kos A)2 1 – kos2 A = (1 + kos A)2 (1 – kos A)(1 + kos A) (1 + cos A)2 1 – cos2A = (1 + cos A)2 (1 – cos A)(1 + cos A) = 1 + kos A 1 – kosA (sebelah kiri) 1 + cos A 1 – cos A (left side)

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 142 BAB 6 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 6.5 Rumus Sudut Majmuk dan Rumus Sudut Berganda Addition Formulae and Double Angle Formulae 1. Rumus majmuk Multiple formulae (a) sin (A ± B) = sin A kos B ± kos A sin B sin (A ± B) = sin A cos B ± cos A sin B (b) kos (A ± B) = kos A kos B ± sin A sin B cos (A ± B) = cos A cos B ± sin A sin B (c) tan (A ± B) = tan A ± tan B 1 ± tan A tan B 2. Rumus sudut berganda Double angle formulae (a) sin 2A = 2 sin A kos A sin 2A = 2 sin A cos A (b) kos 2A= kos2 A – sin2 A = 1 − 2 sin2 A = 2 kos2 A − 1 (c) tan 2A = 2 tan A 1 – tan2 A 3. Rumus sudut separuh Half angle formulae (a) sin A = 2 sin A 2 kos A 2 sin A = 2 sin A 2 cos A 2 (b) kos A = kos2 A 2 − sin2 A 2 cos A = cos2 A 2 − sin2 A 2 = 1 − 2 sin2 A 2 = 2 kos2 A 2 − 1 2 cos2 A 2 − 1 (c) tan A = 2 tan A 2 1 – tan2 A 2 15. Buktikan identiti trigonometri yang berikut dengan menggunakan rumus sudut majmuk bagi sin (A ± B), kos (A ± B) atau tan (A ± B). Prove the following trigonometric identities by using the multiple angle formulae of sin (A ± B), cos (A ± B) or tan (A ± B). TP 4 CONTOH Buktikan kot (A − B) = kot A kot B + 1 kot B − kot A . Prove cot (A − B) = cot A cot B + 1 cot B − cot A . Sebelah kanan / Right side = kot A kot B + 1 kot B − kot A / cot A cot B + 1 cot B − cot A = (kos A kos B sin A sin B + 1) ÷ (kos B sin B − kos A sin A ) ( cos A cos B sin A sin B + 1) ÷ ( cos B sin B − cos A sin A ) = ( kos A kos B + sin A sin B sin A sin B ) × ( sin B sin A kos B sin A – kos A sin B ) ( cos A cos B + sin A sin B sin A sin B ) × ( sin B sin A cos B sin A – cos A sin B) = kos (A – B) sin (A – B) = kot (A − B) / cos (A – B) sin (A – B) = cot (A − B) Tip Mula dengan sebelah yang lebih kompleks. Teknik ialah perlu lihat bentuk sebutan dalam sebelah kiri. Di sini, sebelah kiri terdiri daripada sebutan dalam trigo kosinus, maka perlu tukar sebelah kanan kepada sebutan dalam kosinus juga. Always starts with the more complex side. The technique is to see the end results on the other side. Here, the left side consists of the terms with cosine, hence need to convert the terms on the right into the terms containing cosine. Tip kos A kos B + sin A sin B = kos (A − B) cos A cos B + sin A sin B = cos (A − B) kos B sin A – kos A sin B = sin (A − B) cos B sin A − cos A sin B = sin (A − B)

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 143 6 (a) Buktikan kos(A + B) sin(A – B) = 1 – tan A + tan B tan A – tan B . Prove cos(A + B) sin(A – B) = 1 – tan A + tan B tan A – tan B . (b) Buktikan sin(A – B) kos A kos B = tan A – tan B. Prove sin(A – B) cos A cos B = tan A – tan B. (c) Buktikan sin (A + B) + sin (A – B) sin (A + B) – sin (A – B) = tan A tan B . Prove sin (A + B) + sin (A – B) sin (A + B) – sin (A – B) = tan A tan B . kos (A + B) sin (A – B) = 1 – tan A + tan B tan A – tan B sebelah kanan/Right side = 1 – sin A sin B cos A cos B sin A cos A – sin B cos B = cos A cos B – sin A sin B cos A cos B sin A cos B – cos A sin B cos A cos B = cos (A + B) sin (A – B) = Sebelah kiri/Left side sin (A – B) kos A kos B = tan A – tan B Sebelah kanan/Right side = sin A cos A – sin B cos B = sin A cos B – cos A sin B cos A cos B = sin (A – B) cos A cos B = Sebelah kiri/Left side sin (A + B) + sin (A – B) sin (A + B) – sin (A – B) = tan A tan B Sebelah kiri/Left side = sin A cos B + cos A sin B + sin A cos B – cos A sin B sin A cos B + cos A sin B – sin A cos B + cos A sin B = 2 sin A cos B 2 cos A sin B = sin A cos A × cos B sin B = tan A × cot B = tan A tan B = Sebelah kanan/Right side

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 144 BAB 6 16. Terbitkan rumus sudut berganda bagi sin 2A, kos 2A dan tan 2A dan juga rumus sudut separuh. Derive the double angle formulae of sin 2A, cos 2A and tan 2A and also the half angle formulae. TP 4 (a) Diberi kos (A + B) = kos A kos B – sin A sin B dan jika A = B, buktikan Given that cos (A + B) = cos A cos B – sin A sin B and if A = B, prove that kos 2A = kos2 A − sin2 A = 2 kos2 A − 1 = 1 − 2 sin2 A Seterusnya nyatakan kos θ jika A = θ 2 . Then, state that cos θ if A = θ 2 . (b) Dengan menggunakan sin (A + B) = sin A kos B + kos A sin B dan kos (A + B) = kos A kos B – sin A sin B, jika A = B, bukitkan By using sin (A + B) = sin A cos B + cos A sin B and cos (A + B) = cos A cos B – sin A sin B, if A = B, prove that tan (2A) = 2 tan A 1 − tan2 A Seterusnya, nyatakan tan θ jika A = θ 2 . Then, state tan θ if A = θ 2 . CONTOH Jika/If A = B Maka, kos (A + A) = kos A kos A − sin A sin A Hence, cos (A + A) = cos A cos A − sin A sin A kos 2A = kos2 A – sin2 A cos 2A = cos2 A – sin2 A Oleh kerana kos2 A = 1 − sin2 A Since cos2 A = 1 − sin2 A Maka/Hence kos/ cos 2A = 1 − sin2 A – sin2 A = 1 − 2 sin2 A Jika/If sin2 A = 1 − kos2 A sin2 A = 1 − cos2 A Maka/Hence kos 2A = kos2 A – (1 − kos2 A) cos 2A = cos2 A – (1 − cos2 A) = 2 kos2 A − 1 /2 cos2 A − 1 Jika/If A = θ 2 , kos θ = kos2 θ 2 – sin2 θ 2 cos θ = cos2 θ 2 – sin2 θ 2 = 2 kos2 θ 2 − 1 2 cos2 θ 2 − 1 = 1 − 2 sin2 θ 2 Jika/If A = B tan (A + A) = sin (A + A) kos (A + A) = 2 sin A kos A kos2 A – sin2 A Perhatikan: sin 2A ≠ 2 sin A 2A bererti sudut dua kali ganda, misalnya A = 302 . Maka, 2(300 ) = 600 , menjadi sin 600 , manakala 2 sin 300 = 2 × sin 300 . Jika A = θ 4 , maka sin θ 2 = 2 sin θ 4 kos θ 4 . 2A means twice the angle, for example A = 302 . Then, 2(300 ) = 600 , becomes sin 600 , however 2 sin 300 = 2 × sin 300 . If A = θ 4 , then sin θ 2 = 2 sin θ 4 cos θ 4 . tan 2A = 2 sin A kos A kos2 A kos2 A – sin2 A kos2 A = 2 tan A 1 – tan2 A Jika A = θ 2 , maka tan θ = 2 tan θ 2 1 – tan2 θ 2 Gunakan sin (A + B) = sin A kos B + kos A sin B dan jika A = B, buktikan sin 2A = 2 sin A kos A. Seterusnya, nyatakan sin θ jika A = θ 2 . Using sin (A + B) = sin A cos B + cos A sin B and if A = B, prove that sin 2A = 2 sin A cos A. Hence, state sin θ if A = θ 2 . Penyelesaian: Jika/If A = B Maka, sin (A + A) = sin A kos A + kos A sin A Hence, sin (A + A) = sin A cos A + cos A sin A sin 2A = 2 sin A kos A sin 2A = 2 sin A cos A Jika/If A = θ 2 , maka/hence sin 2( θ 2 ) = 2 sin θ 2 kos θ 2 sin 2( θ 2 ) = 2 sin θ 2 cos θ 2 sin θ = 2 sin θ 2 kos θ 2 sin θ = 2 sin θ 2 cos θ 2

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 145 6 17. Dengan menggunakan sin 2A, kos 2A atau tan 2A, dan rumus-rumus yang sesuai, buktikan yang berikut. By using sin 2A, cos 2A or tan 2A, and other suitable formulae, prove the following. TP 4 (a) Buktikan 1 + sin 2A = (kos A + sin A)2 . Prove that 1 + sin 2A = (cos A + sin A)2. (b) Buktikan 1 + tan2 A 1 – tan2 A = sek 2A. Prove that 1 + tan2 A 1 – tan2 A = sec 2A. (c) Buktikan 1 – kos 2A 1 + kos 2A = tan2 A. Prove that 1 – cos 2A 1 + cos 2A = tan2 A. (d) Buktikan tan θ 2 = sin θ 1 + kos θ Prove that tan θ 2 = sin θ 1 + cos θ (e) Buktikan tan A + kot A = 2 kosek 2A Prove that tan A + cot A = 2 cosec 2A CONTOH Buktikan kosek A = kos 2A sin A + sin 2A kos A . Prove that cosec A = cos 2A sin A + sin 2A cos A . Penyelesaian: Sebelah kanan / Right side kos 2A sin A + sin 2A kos A = kos 2A kos A + sin 2A sin A sin A kos A = kos (2A – A) sin A kos A = kos A sin A kos A = 1 sin A = kosek A Sebelah kanan / Right side (kos A + sin A)2 = kos2 A + 2 kos A sin A + sin2 A = 1 + 2 sin A kos A = 1 + sin 2A (sebelah kiri) Sebelah kiri / Left side 1 – kos 2A 1 + kos 2A = 1 – (1 – 2 sin2 A) 1 + (2 kos2 A – 1) = 2 sin2 A 2 kos2 A = tan 2 A Sebelah kiri / Left side sin A kos A + kos A sin A = sin2 A + kos2 A sinA kos A = 1 sin A kos A = 2 2 sin A kosA = 2 sin 2A = 2 kosek 2A 2 cosec 2A Sebelah kiri / Left side kos2 A + sin2 A kos2 A kos2 A – sin2 A kos2 A = 1 kos 2A = sek 2A Sebelah kanan / Right side sin θ 1 + kos θ = 2 sin θ 2 kos θ 2 1 + (2 kos2 θ 2 – 1) = 2 sin θ 2 2 kos θ 2 = tan θ 2

Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 146 BAB 6 18. Tanpa menggunakan kalkulator, cari nilai yang berikut. Without using a calculator, find the value of the following. TP 3 (a) sin 132° kos 72° − kos 132° sin 72° sin 132° cos 72° − cos 132° sin 72° = sin (132° − 72°) = sin 60° = 3 2 (b) tan 165° (c) kos/ cos 15° 19. Selesaikan yang berikut dengan 0° ≤ x ≤ 360°. Solve the following where 0° ≤ x ≤ 360°. TP 3 CONTOH kos/ cos 75° Penyelesaian: Menggunakan rumus sudut majmuk dan sudutsudut khas Use the multiple angle formulae and the special angles kos/ cos 75° = kos (45° + 30°) = kos 45° kos 30° – sin 45° sin 30° = ( 1 2 )( 3 2 )−( 1 2 )( 1 2 ) = 3 – 1 2 2 = tan (120° + 45°) = (tan 120° + tan 45°) 1 – tan 120° tan 45° = (– 3 + 1) 1 – (– 3 ) = 1 – 3 1 + 3 = 4 – 2 3 –2 = 3 – 2 = kos (45° − 30°) = kos 45° kos 30° + sin 45° sin 30° = ( 1 2 )( 3 2 ) + ( 1 2 )( 1 2 ) = ( 3 + 1) 2 2 6.6 Aplikasi Fungsi Trigonometri Application of Trigonometric Functions CONTOH (i) sin x = −0.3516 (ii) kos/ cos 2x = 0.4517 Penyelesaian: (i) Sinus ialah negatif pada sukuan III dan IV Sine is negative in quadrant III and quadrant IV. Sudut rujukan x = sin–1 (0.3516) = 20° 35’ Sudut x sebenar Actual angle x = 180° + 20° 35’ dan 360° − 20° 35’ = 200° 35’ dan 339° 25’ (ii) 0° ≤ x ≤ 360°, maka 0° ≤ 2x ≤ 720° Sudut rujukan, 2x = kos–1 (0.4517) Reference angle = 63°9’ kosinus ialah positif pada sukuan I dan IV cosine is positive in quadrant I and quadrant IV 2x = 63° 9’, 360° − 63° 9’, 360° + 63° 9’, 720° − 63° 9’ 2x = 63° 9’, 296° 51’, 423° 9’, 656° 51’ x = 31° 35’, 148° 26’, 211° 35’, 328° 26’ x y 20fi35ff 20fi35ff

BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri 147 6 Perhatian Jangan bahagi sudut rujukan dengan 2 dahulu, iaitu 63° 9’ 2 = 31° 35’. Do not divide the reference angle by 2, that is 63° 9’ 2 = 31° 35'. Tip Langkah-langkah penyelesaian. Steps to solve. 1. Cari sudut rujukan tanpa tanda negatif. Find the reference angle without the negative sign. 2. Letak sudut rujukan pada sukuan yang betul. Place the reference angle at the correct quadrant. 3. Sudut sebenar mesti bermula dari paksi-x yang positif dalam arah lawan jam seperti ditunjukkan The real angle must starts from the positive x-axis in the anticlockwise direction as shown in the example. Sudut Kalkulator SHIFT sin 0.3516 = 20.59° (a) kos/ cos x = −0.6283 (b) 2 tan x = 2.52 (c) sin 2x = −0.357 (d) kos/ cos 2x = −0.712 x y 51fi5ff 51fi5ff x y 51fi34ff 51fi34ff x y 20fi55ff 20fi55ff x y 44fi36ff 44fi36ff x = 180° – 51° 5’, 180° + 51.5’ = 128° 55’, 231° 5’ sin 2x = –0.357 2x = 200° 55’, 339° 5’ 560° 55’, 699° 5’ x = 100° 28’, 169° 33’, 280° 28’, 349° 33’ 2 tan x = 2.52 tan x = 1.26 x = 51° 34’, 180° + 51.34’ = 51° 34’, 231° 34’ kos 2x = 0.712 2x = 135° 24’, 224° 36’ 495° 24’, 584° 36 x = 67°42’. 112° 18’ 247° 42’, 292° 18’