BAB 5 Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
(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
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.
(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 12 jam hingga 3 jam.
a piece of cake. The time taken is between 1 21 hours to
Y is a variable that represents the time taken to make
3 hours.
Y = {y: 1 12 ≤ y ≤ 3} adalah selanjar kerana sudah menentukan dengan tepat masa yang sewajar.
Y = {y: 1 12 ≤ 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.
98
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian BAB 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: P[X = 2] = P[3, 3, 3’] + P[3, 3’, 3] + P[3’, 3, 3]
X = {0, 1, 2, 3}
1 = 3 5 × 1 × 1
6 6 6
63
1 5
6 3 3’ = 72
5
3 6 P[X = 3] = P[3, 3, 3] = 1 × 1 × 1
1 6 6 6
5 63
1 3’ 1
6 6 5 3’ = 216
6 X=x 0 1 23
1 P[X = x]
125 25 51
63 216 72 72 216
5 1 35 3’
6
6 3’ 6
1
3
5 6 3’ Tip
6 3’
5
6 Kebarangkalian peristiwa A berlaku, P(A) = n(A) ,
n(S)
L1 L2 L3
dengan keadaan n(A) ialah bilangan kesudahan
5 5 5 bagi peristiwa A dan n(S) ialah bilangan
6 6 6
P[X = 0] = P[3’, 3’, 3’] = × × kesudahan dalam ruang sampel.
The probability of an event A happening, P(A) =
n(A)
= 125 n(S) , where n(A) is the number of outcomes of
216
event A and n(S) is the number of outcomes of
P[X = 1] = P[3, 3’, 3’] + P[3’, 3, 3’] + P[3’, 3’, 3]
the sample space.
= 3 5 × 5 × 1 = 75 = 25
6 6 6 216 72
99
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
BAB 5 (a) Sebuah beg mengandungi 2 biji guli merah dan (b) 30% pelajar menggunakan bas sekolah
4 biji guli hijau. Sebiji guli dicabut dan selepas ke sekolah dari suatu sekolah tertentu. X
mencatatkan warnanya, guli itu dikembalikan mewakili bilangan pelajar menggunakan bas
ke dalam beg sebelum sebiji guli dikeluarkan sekolah jika 3 pelajar dipilih secara rawak.
lagi. X mewakili bilangan kali mendapat guli 30% of the students take bus to the school from a
certain school. X represents the number of students
hijau dalam dua cabutan berturut-turut. who take school bus if 3 students are chosen at
random.
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.
X = {0, 1, 2} X = {0, 1, 2, 3}
2 3 3
10 10 B
2 3H B
3 37 B
H M 10 10 7 B’
1 10
73 3
3 10 10 10 B
2 B’ B’
1 3H 7 7 B’
M 10 10
3 1 M 3
10 B
3
B
P[X = 0] = P[M, M] 7 B’
10
= 1 × 1 = 1 3
3 3 9 10 B
P[X = 1] = P[H, M] + P[M, H] B’
7 B’
= 2 × 1 + 1 × 2 10
3 3 3 3
= 4
9
P[X = 2] = P[H, H]
= 2 × 2 = 4
3 3 9
1703 343
P[X = 0] = P[B’, B’, B’] = = 1000
X=x 0 1 2 P[X = 1] = 3C1 3 1702 = 441
10 1000
P[X = x] 1 4 4
9 9 9 1 3C23 2 170 189
P[X = 2] = 10 = 1000
P[X = 3] = 11303 = 27
1000
X=x 0 1 2 3
P[X = x]
343 441 189 27
1000 1000 1000 1000
100
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
4. Pemboleh ubah X mempunyai taburan kebarangkalian yang berikut. BAB 5
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
CONTOH (a) P(X = x) = x untuk/for x = {0, 1, 2, 3}
6
12 – x
P(X = x) = 36 untuk/for x = {0, 2, 4, 6}. (i) P(X = 0) = 0 =0
6
Penyelesaian:
P(X = 1) = 1
(i) P(X = 0) = 12 – 0 = 1 6
36 3
P(X = 2) = 2 = 1
P(X = 2) = 12 – 2 = 5 6 3
36 18
P(X = 3) = 3 = 1
P(X = 4) = 12 – 4 = 2 6 2
36 9
X=x 012 3
P(X = 6) = 12 – 6 = 1
36 6 P(X = x) 0 1 1 1
63 2
Jadual taburan kebarangkalian bagi X.
Probability distribution table for X.
X=x 0 2 4 6 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
=0+ 1 + 1
P(X = x) 1 5 21 6 1 + 2
3
3 18 9 6
= 1 (Terbukti)/(Proved)
Untuk menunjukkan X ialah satu pemboleh (ii) P(X = x)
ubah rawak diskret, kita mesti tunjukkan
∑Ton0 P[X = x] = 1. a random discrete variable, we
= = 1.
show that X is
P[X x]
must show ∑ n
0
P(X = 0) + P(X = 2) + P(X = 4) + P(X = 6) =
1 5 2 1
3 + 18 + 9 + 6 = 1 (Terbukti)/(Proved)
(ii) P(X = x) 1
2
12
36 1
3
10
36 1
6
8
36 x
0 0123
6
36
x
0 0246
101
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
(b) P(X = x) = (x + 1)2 untuk/for x = {1, 2, 3, 4} (ii)
54
P(X = x)
BAB 5 (i) P(X = 1) = (1 + 1)2 = 2
54 27 25
54
P(X = 2) = (2 + 1)2 = 1
54 6 16
54
P(X = 3) = (3 + 1)2 = 8
54 27 9
54
P(X = 4) = (4 + 1)2 = 25 4
54 54 54
X=x 1234 x
0 1234
P(X = x) 2 1 8 25
27 6 27 54
P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 2 + 1 + 8 + 25
27 6 27 54
= 1 (Terbukti)/(Proved)
5. Selesaikan setiap soalan yang berikut. (a) X ialah suatu pemboleh ubah rawak diskret
Solve each of the following questions. TP 4 dengan taburan kebarangkalian yang berikut
CONTOH X is a random discrete variable with the following
probability distribution.
X ialah suatu pemboleh ubah rawak diskret
dengan taburan kebarangkalian yang berikut. X 24 6 8
P(X = x) 0.14 0.21 0.38 0.27
X is a random discrete variable with the following
probability distribution. Cari/Find
(i) P(X ≤ 6)
X 0123 ii) P(2 ≤ X ≤ 6)
P(X = x) 0.24 0.31 0.28 0.17
Cari/Find (i) P(X ≤ 6) = 1 − P(X = 8)
(i) P(X > 1) = 1 − 0.27 = 0.73
(ii) P(0 < X < 3) (ii) P(2 ≤ X ≤ 6) = 1 − P(X = 8)
= 0.73
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
102
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
(b) Y ialah suatu pemboleh ubah rawak diskret (c) Z ialah suatu pemboleh ubah rawak diskret BAB 5
dengan taburan kebarangkalian yang berikut. dengan taburan kebarangkalian yang berikut.
Y is a random discrete variable with the following Z is a random discrete variable with the following
probability distribution. probability distribution.
{0.1 untuk y = 1,3,5,7 p untuk z = 0,1
P(Z = p + q untuk z =
P(Y = y) = 0.2 untuk y = 2,4,6 z) = 2
0 untuk y yang lain q untuk z = 3,4}
Tunjukkan bahawa Y ialah suatu pemboleh (i) Jika P(Z > 2) = 4 , cari nilai p dan q.
9
ubah rawak diskret. Seterusnya, cari 4
9
Show that Y is a random discrete variable. Then, find If P(Z > 2) = , find the values of p and q.
(i) P(Y < 4)
(ii) P(3 < Y < 5) (ii) Bina satu jadual taburan kebarangkalian Z.
Y 1234567 Construct a table of the probability distribution of Z.
P(Y = y) 0.1 0.2 0.1 0.2 0.1 0.2 0.1
(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
∑ 7 P(Y = y) = 0.1 × 4 + 0.2 × 3 4
0 q + q = 9
= 0.4 + 0.6 = 1
2
Maka, Y ialah suatu pemboleh ubah rawak q = 9
diskret. P(Z = 0) + P(Z = 1) + P(Z = 2) = 5
9
Hence, y is a random discrete variable.
(i) P(Y < 4) = P(Y = 1) + P(Y = 2) + P(Y = 3) 5
9
= 0.1 + 0.2 + 01 = 0.4 p + p + p + q =
(ii) P(3 < Y < 5) = P(Y = 4) 2 5
9 9
= 0.2 3p + =
3p = 3
9
= 1
9
(ii) z 0 1234
P(Z = z)
1 1122
9 9399
(iii) P(Z = z)
3
9
2
9
1
9
z
0 01234
103
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
NOTA IMBASAN
Taburan Binomial
5.2
Binomial Distribution
BAB 5 NOTA IMBASAN 4. Oleh itu, kebarangkalian mendapat x kejayaan ialah
Hence, the probability of getting x successes is given by
1. Suatu eksperimen dengan 2 kesudahan sahaja iaitu P(X = x) = nCr pr(1 − p)n − r, x = 0, 1, 2, 3 …. n.
kejayaan dan kegagalan dikenali sebagai eksperimen (Kadang-kala, kita menggunakan q sebagai tanda
Bernoulli.
kegagalan, iaitu q = 1 – p.)
An experiment with only two outcomes that is success and failure (Sometimes, we use q for the probability of failure, which is
is a Bernoulli experiment.
q = 1 – p.)
2. Jika eksperimen Bernoulli diulangi n kali secara tak 5. Katakan X adalah pemboleh ubah rawak diskret
bersandar, maka eksperimen tersebut dikenali sebagai
eksperimen Binomial. binomial dengan parameter n dan p, maka pengiraan
min, varians dan sisihan piawai ialah
When the Bernoulli experiment is repeated n times independently, When X is the binomial discrete random variable with parameters
then the experiment is called a Binomial experiment n and p, then to find the mean, the variance and the standard
deviation, we have
3. Katakan X adalah pemboleh ubah rawak diskret Min / mean, E(X) = np
yang menunjukkan bilangan kejayaan dalam suatu Varians / variance, Var(X) = np(1 − p)
eksperimen Binomial, maka X dikatakan bertaburan Sisihan piawai = np(1− p) , dengan n ialah bilangan
Binomial dengan simbol B(n, p), dengan n ialah percubaan dan p ialah kebarangkalian bagi kejayaan.
bilangan percubaan yang tak bersandar dan p ialah Standard deviation = np(1− p) , where n is the number of trials
kebarangkalian mendapat kejayaan. and p is the probability of success.
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.
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
CONTOH (a) Sebiji bola dikeluarkan daripada satu kotak
X ialah keputusan perlawanan suatu pertandingan yang mengandungi 5 bola merah dan 3 bola
bola sepak. Keputusan terdiri daripada menang,
tewas atau seri. biru. Selepas mencatatkan warna, bola itu
X is the result of a football game. the results can be win, dikembali ke dalam kotak sebelum sebiji bola
loss or draw.
dikeluarkan lagi. Proses ini berulang 4 kali dan
Penyelesaian:
Bukan, kerana terdapat tiga kesudahan, iaitu X ialah bilangan kali mendapat bola merah.
menang, tewas dan seri.
Binomial hanya mempunyai dua jenis kesudahan. A ball is drawn from a box which contains 5 red
and 3 blue balls. After noting the colour, the ball is
This is not because there are three outcomes, that are returned into the box before another ball is drawn.
win, lose and draw. This process is repeated 4 times and X is the number
Binomial has only two types of outcomes. of times of getting a red ball.
Ini ialah percubaan Bernoulli, kerana
kebarangkalian mendapat bola merah setiap
5
kali ialah tak bersandar, iaitu tetap = 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
5
fixed = 8 and the trial occurs 4 times, n = 4.
104
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
(b) 85% orang pelajar dari kelas A mempunyai (c) Satu huruf dipilih secara rawak daripada BAB 5
telefon pintar. X ialah bilangan kali mendapat
seorang pelajar mempunyai telefon pintar jika perkataan BACAAN. X ialah bilangan kali
proses diulangi 10 kali.
memilih huruf A jika semua huruf tidak boleh
85% of the students from class A has smartphones.
X is the number of times of choosing a student with a diulangi dan proses ini dibuat 3 kali.
smartphone if this process is repeated 10 times
A letter is chosen at random from the word BACAAN.
Ini ialah percubaan Bernoulli, kerana terdapat X is the number of times of choosing A if all the letters
dua kesudahan sahaja, dengan p = 0.85 dan cannot be repeated and this process is done three
percubaan ini berlaku 10 kali, n = 10. times.
This is a Bernoulli trial because there are two Ini bukan percubaan Bernoulli, kerana
outcomes, where p = 0.85 and this trial occurs 10
times, n = 10. kebarangkalian mendapat A pertama kali, p =
3 2 1
6 , kedua kali p = 5 dan ketiga kali p = 4
dengan setiap kali p berubah, iaitu bersandar.
This is not Bernoulli trial, because the probability to
3 2
get A the first time, p = 6 , the second time p = 5 and
the third time p = 1 , where each time p changes,
4
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: 1 . Katakan X ialah bilangan kali
(i) Kebarangkalian mendapat gambar dalam satu lambungan, p = 2
mendapat gambar, maka X = {0, 1, 2, 3, 4, 5}.
1
The probability of getting a head in one toss, p = 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
2 5 P(X = 2)
2 2 =Th5eCp2ro12ba2bi1lit–yof getting heads in tosses =
1 3 = 5
2 16
(ii) P(X < 2) = P(X = 0) + P(X = 1)
= 5C0 1 20 1 − 1 25 + 5C1 1 211 − 1 24
2 2 2 2
=
3
16 Sudut Kalkulator
Tekan 5 SHIFT nCr 2 × ( 1 a b/c 2 ) x2
Press
( 1 – 1 a b/c 2 ) SHIFT x2 = 5
16
105
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
BAB 5 (a) Di sepanjang suatu jalan raya, kebarangkalian (b) Dalam suatu kaji selidik, 90 daripada 100
kemalangan berlaku pada suatu hari ialah
5%. Cari kebarangkalian dalam dua minggu orang yang didedahkan kepada orang yang
berlakunya kemalangan jalan raya
mempunyai penyakit Covid 19 akan diuji
Along a certain highway, the probability of road
accident in a certain day is 5%. Find the probability positif. Jika 8 orang daripada kumpulan tersebut
in two weeks of a road accident
dipilih secara rawak, cari peratus bahawa
(i) tepat 4 kali./exactly 4 times.
(ii) kurang daripada 12 kali. In a research, 90 out of 100 people exposed to the
Covid-19 sickness are tested positive. If 8 people are
less than 12 times. chosen at random from this group of people, find the
percentage that
(iii) 5 hingga 7 kali./5 to 7 times.
(i) semua dijangkiti Covid-19.
n = 14, p = 0.05, q = 0.95
(i) P(X = 4) all contracted Covid-19.
= 14C4(0.05)4(0.95)10 = 0.0037
(ii) P(X < 12) = 1 − P[X = 12] − P(X = 13) + (ii) lebih daripada 6 orang dijangkiti covid-19.
P(X = 14)
= 1 − (14C12(0.05)12(0.95)2 + more than 6 of them contracted covid-19.
14C13(0.05)13(0.95)1 + 14C14(0.05)14)
=1 n = 8, p = 0.9, q = 0.1
(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 + (i) P(X = 8) = 8C8(0.9)8 = 0.4305
14C7( 0.57)(0.95)7] = 0.00043 Maka, 43.05% dijangkiti Covid-19.
Hence, 43.05% contracted Covid-19.
(c) 95% daripada murid di sebuah sekolah
mematuhi undang-undang sekolah. Jika 10 (ii) P(X > 6) = P(X = 7) + P(X = 8)
orang murid dipilih secara rawak dari sekolah
itu, cari kebarangkalian bahawa = 8C7(0.9)7(0.1)1 + 8C8(0.9)8
= 0.8131
95% of the students in a school comply with school
rules. If 10 students are chosen at random from the Maka, 81.31% dijangkiti Covid-19.
school, find the probability that Hence, 81.31% contracted Covid-19.
(i) lebih daripada 8 orang murid mematuhi (d) Kebarangkalian Karol akan menang dalam
undang-undang sekolah.
pertandingan tenis ialah 2 . Jika dia mengambil
more than 8 students comply with school rules. 3
(ii) 2 hingga 8 orang murid mematuhi undang- bahagian dalam n pertandingan, kebarangkalian
undang sekolah.
dia tidak menang semua pertandingan ialah
2 to 8 of them comply with school rules.
0.0014. Cari
The probability that Karol will win in the tennis
n = 10, p = 0.95, q = 0.05 competition is 2 . If he takes part in n competitions,
3
(i) P(X > 8) = P(X = 9] + P(X = 10)
the probability that he does not win all the
= 10C9(0.95)9(0.05)1 + 10C10(0.95)10 competitions is 0.0014. Find
= 0.9139
(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.
(ii) P(2 ≤ X ≤ 8) p= 2 , q = 1
3 3
= 1 − P(X = 0) − P(X = 1) − P(X = 9) − (i) P(X = 0) = 0.0014 1
2 3
P(X = 10) = nC0 3 20 2n = 0.0014
= 0.0861 n log 1 2 = log 0.0014
3
log0.0014
n = =6
1
log 3
(ii) P(X > 4) = P(X = 5) + P(X = 6)
= 6C5 2 25 1 2 + 6C6 2 26
3 3 3
= 0.3512
106
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
8. Selesaikan masalah yang berikut. BAB 5
Solve the following problems. TP 4
CONTOH
Diketahui bahawa 20% daripada murid di P (X = x)
sebauh sekolah mempunyai iPad. Jika 4 0.5
orang murid dipilih secara rawak dari sekolah
itu, tentukan taburan binomial dengan 0.4
menggunakan graf.
0.3
It is known that 20% of the students in a school have
iPad. If 4 students are chosen at random from the 0.2 Tip
school, determine the binomial distribution using 0.1
graph. Hasil tambah semua
0 01234 kesudahan yang
Penyelesaian: mungkin ialah 1.
Katakan X mewakili bilangan murid x The sum of all possible
mempunyai iPad. outcomes is 1.
Maka, X ialah suatu pemboleh ubah rawak P(X = 0) + P(X = 1) +
binomial dengan n = 4, p = 0.2, q = 1 – 0.2 = 0.8 P(X = 2) +
P(X = 3) +
Let X represents the number of students have iPad P(X = 4) = 1
Hence, X is binomial random variable with n = 4,
p = 0.2.
P(X = 0) = 4C0(0.2)0(0.8)4 = 0.4096
P(X = 1) = 4C1(0.2)1(0.8)3 = 0.4096
P(X = 2) = 4C2(0.2)2(0.8)2 = 0.1536
P(X = 3) = 4C3(0.2)3(0.8)1 = 0.0256
P(X = 4) = 4C4(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 P (X = x)
0.65 + 2p = 1 0.30
2p = 0.35
p = 0.175
0.25
0.20
p
0.15
0.10
0.05 x
0
01 2345
107
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
BAB 5 (b) Sekeping duit syiling dilambangkan sebanyak P (X = x)
0.4
4 kali. Jika X mewakili bilangan kali mendapat 0.3
0.2
kepala, cari taburan binomial bagi X dan plot 0.1
grafnya. 0x
01 234
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 , q = 1
2 2
P(X = 0) = 4C0 1 20 1 24 = 0.0625
2 2
P(X = 1) = 4C1 1 21 1 23 = 0.2500
2 2
P(X = 2) = 4C2 1 22 1 22 = 0.3750
2 2
P(X = 3) = 4C3 1 23 1 21 = 0.2500
2 2
P(X = 4) = 4C4 1 24 1 20 = 0.0625
2 2
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
CONTOH (a) 3 biji dadu dilambungkan serentak.
1 daripada ayam yang dijual dalam pasar 3 dices are thrown simultaneously.
20
mempunyai berat yang kurang memuaskan (i) Hitung kebarangkalian untuk mendapat
dua ‘5’.
untuk dijual. Cari min dan sisihan piawai bagi
Calculate the probability of getting two ‘5’.
bilangan ayam yang kurang memuaskan untuk
(ii) Jika 3 biji dadu itu dilambungkan 216 kali,
dijual daripada 160 ekor ayam pada suatu hari berapa kalikah dijangka untuk mendapat
dua ‘5’ ?
tertentu.
If the 3 dices are thrown 216 times, estimate the
1 of the chickens sold in the market have the masses number of times of getting two ‘5’?
20
which are not suitable to be sold. Find the mean and 1
(i) n = 3, p = 6
standard deviation of the number of chickens which are
P(mendapat dua ‘5’) = 3C2 1 22 5 2
not suitable to be sold from 160 chickens on a particular 6 6
day. 5
72
Penyelesaian: =
Katakan X mewakili bilangan ayam yang kurang (ii) E(X) = np = 216 × 5 = 15 kali
72
memuaskan untuk dijual.
Let X be the number of chickens which are not suitable
to be sold.
n = 160, p = 1 , q = 19
20 20
Min/Mean = E(X) = np = 160 × 1 = 8
20
2 2Sisihan piawai = 160 1 19 = 2.76
20 20
Standard deviation
108
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
(b) Satu pemboleh ubah rawak diskret X (c) X bertaburan binomial dengan B(n, p). Jika q
ialah 0.35 dan sisihan piawai ialah 1.5, cari
bertaburan binomial Bn, 3 2. Jika min ialah
14 X is a binomial distribution with B(n, p). If q is 0.35
3, cari and the standard deviation is 1.5, find
BAB 5
A random discrete variable X has a binomial (i) nilai n.
2 n, 3 . If the mean is 3, find the value of n.
distribution B 14
(ii) min / mean
(i) nilai n.
the value of n.
(i) q = 0.35, p = 0.65
(ii) varians dan sisihan piawai.
Sisihan piawai = 1.5 = n(0.65)(0.35)
the variance and the standard deviation.
Standard deviation
(i) E(X) = np = 3 n = 1.52
(0.65)(0.35)
n 134 2 = 3
= 10
n = 14 (ii) Min/Mean = np
(ii) Varians = npq =
14 × 3 × 11 = 2 5 = 10(0.65)
14 14 14
= 6.5
Sisihan piawai = 1.54
Standard deviation
10. Selesaikan setiap yang berikut.
Solve each of the following. TP 5
CONTOH (a) 2% daripada lampu yang dihasilkan
Dalam suatu temasya sukan sekolah, oleh sebuah kilang adalah rosak. Cari
kebarangkalian seorang murid mendapat markah bilangan lampu yang perlu disemak supaya
untuk rumah sukannya ialah 3 . Cari bilangan kebarangkalian bahawa sekurang-kurangnya
8
satu lampu yang rosak adalah lebih daripada
minimum murid yang mendapat sekurang-
0.95.
kurangnya satu markah jika kebarangkalian
2% of the bulbs produced by a factory is faulty. Find
adalah lebih daripada 0.65. the number of bulbs which must be inspected so
that the probability that at least one bulb is faulty is
In a school sports meet, the probability of a student 0.95.
getting a mark for the sports house is 3 . Find the p = 0.02, q = 0.98
8 P(X ≥ 1) > 0.95
1 – P(X = 0) > 0.95
minimum number of students getting at least one mark 1 − nC0(0.02)0(0.98)n > 0.95
if the probability is more than 0.65. 0.98n< 0.05
n log 0.98 < log 0.05
Katakan X mewakili bilangan murid yang
mendapat markah.
Let X represents the number of students that get marks
P(X ≥ 1) > 0.65 n > log0.05
log0.98
1 – P(X = 0) > 0.65 Tip
P(X = 0) < 0.35 n > 148.28
n = 149
2 2 nC0305n < 0.35 log0.35
8 8 2n <
5 log 5
8 8
n log < log 0.35 –0.4559
n< –0.2041
log 0.35
n >
5 n > 2.23
log 8 n=3
n > 2.23
∴n = 3
109
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
BAB 5 (b) Lima dadu dilambungkan serentak. (c) Kebarangkalian pelanggan yang membuat
Five dice are thrown simultaneously. tempahan meja di sebuah restoran tetapi tidak
(i) Apakah kebarangkalian untuk mendapat hadir ialah 0.2. Restoran itu cuma boleh muat
tiga ‘2’? 50 meja. Jika terdapat tempahan sebanyak
What is the probability of getting three ‘2’s? 5 meja pada suatu hari, cari kebarangkalian
(ii) Jika lima dadu itu dilambungkan 550 kali, bahawa
berapa kali mendapat tiga ‘2’? The probability of a customer who booked the table
reservation at a restaurant but not present is 0.2. The
If the five dice are thrown 550 times, how many restaurant can only accommodate 50 tables. If there
times can get three ‘2’s? are 5 tables booked on a day, find the probability
that
(i) P(X = 3) = 5C3 1 23 5 22
6 6 (i) tepat 3 pelanggan hadir.
= 0.0322 exactly 3 customers present.
(ii) E(X) = np
= 550 × 0.0322 (ii) kurang daripada 2 tempahan meja hadir.
= 17.68
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) = 5C3(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
5.3 Taburan Normal
Normal Distribution
NOTA IMBASAN Kekerapan Lengkung ini
Frequency simetri pada x = .
1. Suatu pemboleh ubah rawak selanjar, X adalah The curve is
bertaburan normal jika graf fungsi kebarangkaliannya symmetry at x = .
mempunyai ciri-ciri berikut:
P(a < X < b)
A continuous random variable, X is normally distributed if the
probability of function graph has the following characteristics: 0 ax = b x
(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). 2. Jika suatu pemboleh ubah rawak normal, X mempunyai
(ii) Jumlah luas di antara lengkung dan paksi-x ialah 1. min, µ = 0 dan sisihan piawai, σ = 1, X dikatakan
The total area between the curve and the x-axis is 1. bertaburan normal piawai, iaitu X ~ N(0, 1).
(iii) Kebarangkalian bagi X yang mempunyai nilai di
antara a dan b, ditulis sebagai P(a < X < b) = luas If the normal random variable, X has a mean, µ = 0 and standard
rantau berlorek dalam rajah. deviation, σ = 1, X is said to be the standard normal distribution,
The probability of X that have values between a and b, that is X ~ N(0, 1).
is written as P(a < X < b) = area of shaded region in the
diagram 3. Tpskaiaobwruparaiianmwneaoliaralmutaiaulrusbmkooluer-shz.Zdi=tuxka–σrµk,edpeandgaatnabZudrainnanmoarmkaanl
110
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
nThoermnaolrdmisatlribduisttiroibnubtyiotnhecafonrmbeulaconxv–eµrte, dwhinetroe the standard f(z)
Z is called the
σ
standard score or z-score.
4. Kebarangkalian untuk suatu pemboleh ubah rawak BAB 5
normal X boleh ditentukan daripada sifir taburan
normal melalui langkah-langkah berikut: Z
The probability of the random normal variable X can be 0z
determined from the normal distribution table by using the steps
below: (ii) Dapatkan nilai P(Z > z) daripada jadual
kebarangkalian hujung atas bagi taburan normal
(i) f(x) Proses pemiawaian dalam buku sifir.
Standardising process
Find the value of P(Z > z) from the probability table of the
Z = x – upper end of the normal distribution table.
0 x
11. Selesaikan yang berikut. (a) Rajah menunjukkan graf taburan normal bagi
suatu pemboleh ubah rawak selanjar X.
Solve the following. TP 2
The diagram shows the normal distribution of a
CONTOH continuous random variable X.
Rajah menunjukkan graf taburan normal bagi f(x)
suatu pemboleh ubah rawak selanjar X.
The diagram shows the normal distribution of a
continuous random variable X.
f(x)
(ii)
x 0 23 32 38 x
0 35 8
(i) Nyatakan nilai µ. dalam (i) Nyatakan nilai µ.
State the value of µ. State the value of µ.
(ii) Lorek rantau P(3 < x < µ). (ii) Nyatakan nilai P(X > µ).
Shade the region of P(3 < x < µ). State the value of P(X > µ).
(iii) Ungkapkan rantau berlorek (iii) Cari P(23 < X < 32).
tatatanda kebarangkalian. Find P(23 < X < 32).
Express the shaded region in probability (i) µ = 32
(ii) P(X > 32) = 0.5
notation. (iii) P(23 < X < 32) = 0.5 – 0.16
= 0.34
Penyelesaian: (iii) P(5 < X < 8)
(i) µ = 5
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)
111
BAB 5 Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian (c) Suatu pemboleh ubah rawak selanjar X
bertaburan normal. Lakarkan satu graf taburan
(b) Rajah menunjukkan graf taburan normal bagi normal untuk menunjukkan yang berikut.
suatu pemboleh ubah rawak selanjar X.
A continuous random variable X has a normal
The diagram shows the normal distribution of a distribution. Sketch the normal graph to show the
continuous random variable X. following.
f(x) (i) min, µ = 2.5.
0 54 65 76 x mean, µ = 2.5.
Jika P(X < 76) = 0.26, cari (ii) Lorekkan P(2.8 < X < 3.0).
If P(X < 76) = 0.26, find Shade P(2.8 < X < 3.0).
(i) P(X > 54). (iii) Lorekan secara anggaran P(X < x) = 0.6.
(ii) P(54 ≤ X ≤ 76).
Shade the estimate region of P(X < x) = 0.6.
(i) P(X > 54) = 0.26
(ii) P(54 ≤ X ≤ 76) = 1 − 2(0.26) f(x)
= 0.48
x
0 2.5 2.8 3.0
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 Min, µ Sisihan piawai, σ Skor-z
Variable X Mean, µ Standard deviation, σ z-score
CONTOH 40 10 Z = 35− 40 = −0.5
10
35
(a) 0.8 = X − 67 67 15 0.8
15 −2.1
1.85
X = 15(0.8) + 67
= 79
(b) 4.6 −2.1 = 4.6 – µ 0.9
0.9
µ = 4.6 + 2.1(0.9)
= 6.49
(c) 130 143 1.85 = 130 − 143
σ
–13
σ = 1.85
= 7.03
112
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
13. Dengan menggunakan jadual taburan normal piawai, cari nilai bagi setiap yang berikut diberi bahawa Z ialah BAB 5
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
CONTOH (a) P(Z < –0.4)
P(Z > 0.3) f(z) P(Z < –0.4) = P(Z > 0.4)
P (Z > 0.3) = 0.3446
Penyelesaian:
P(Z > 0.3) = 0.3821 f(z)
0 0.3 z
–0.4 0 z
Sudut Kalkulator Graf ialah simetri pada f(z) = 0.
P(Z > 0.3) Graph is symmetrical about f(z) = 0.
Tekan: MODE MODE 1 SHIFT 3 3 0 . 3 =
0.38209
(b) P(Z > –1.2) (c) P(–1.5 < Z < 2.1)
1 – P(Z > 1.5) – P(Z > 2.1)
P(Z > –1.2) = 1 – 0.0668 – 0.0179
= 1 – P(Z ≥ 1.2) = 0.9153
= 1 – 0.1151
= 0.8849 f(z)
Jumlah luas di bawah graf ialah 1. –1.5 0 2.1 z
Total area under the graph is 1.
f(z)
–1.2 0 z
14. Cari skor-z bagi setiap yang berikut. (a) P(Z < z) = 0.0375
skor-z = –1.78
Find the z-score for each of the following. TP 3
f(z)
CONTOH
P(Z > z) = 0.2451
Penyelesaian:
skor-z = 0.69
f(z)
Luas = 0.2451
Area
z
z0
0z z
113
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
(b) P(Z > z) = 0.9126 (c) P(Z ≤ z) = 0.8251
BAB 5 1 – P(Z ≤ z) = 0.9126 1 – P(Z > z) = 0.8251
P(Z ≤ z) = 0.0874 P(Z > z) = 0.1749
skor-z = –1.357 skor-z = 0.935
Luas adalah lebih daripada 0.5, maka skor-z Luas lebih daripada 0.5, maka skor-z ialah
ialah negatif. positif.
Area is more than 0.5, so z-score is negative. Area is more than 0.5, so z-score is positive.
f(z) f(z)
z0 z 0z z
15. Cari nilai k bagi setiap yang berikut.
Find the value of k in each of the following. TP 3
CONTOH (a) µ = 15, σ = 22, P(X > k) = 0.4152
µ = 350, σ = 38, P(X > k) = 0.65 P(Z > k − 15 ) = 0.4152
22
Penyelesaian:
k − 15 = 0.214
P(X > k) = 0.65 22
P X − µ > k –338502 = 0.65 k = 19.71
σ
PZ > k –338502 = 0.65 f(z)
1 – PZ > k –338502 = 0.65 0.4152
P(Z < k –338502 = 0.35 0 0.214 z
skor-z = –0.385
k –33850 = –0.385
∴k = 335.37 (b) µ = 250, σ = 16, P(X < k) = 0.32
f(z) PZ < k − 250 2 = 0.32
16
0.35 0.65
z k − 250 = –0.468
16
k–350 0
38 k = 242.51
Tip f(z) z
0.32
Luas lebih daripada 0.5, maka skor-z ialah
negatif. –0.468 0
Area is more than 0.5, then z-score is negative
114
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
(c) µ = 40.2, σ = 12.5, P(35 < X < k) = 0.458 (d) µ = 35, σ = k, P(30 < X < 40) = 0.578
P 35 – 40.2 < Z< k – 40.2 = 0.458 P 30 – 35 < Z < 40 – 35 = 0.578
12.5 12.5 k k
BAB 5
P–0.416 < Z< k – 40.2 2 = 0.458 P– 5 < Z < 5 2 = 0.578
12.5 k k
1 – P(Z < –0.416) – P1Z > k – 40.2 2 = 0.458 1 – 2P(Z ≥ 5 ) = 0.578
12.5 k
P1Z < k – 40.2 2 = 0.2033 1 2 PZ≥ 5 = 0.211
12.5 k
f(z) 0 .578
f(z)
0.3387 ∴ k – 40.2 = 0.83 0.211 5 = 0.803
12.5 k
0.2033
z k = 50.58 z k = 6.23
–0.416 0 0.83 – 5 0 5
k k
16. Selesaikan yang berikut.
Solve the following. TP 5
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) f(z) Tip
–0.9655 0 0.7586 z
= P130 − 135.6 < X–µ < 140 −5.8135.62 Lakar graf untuk menentukan rantau yang
5.8 σ betul dahulu.
Sketch the graph to determine the correct
= P(−0.9655 < Z < 0.7586) region first.
= 1 − P(Z > 0.7586) – P(Z > 0.9655)
= 1 – [P(Z > 0.7586) + P(Z > 0.9655)]
= 1 – 0.3912 Sudut Kalkulator
= 0.6088
(ii) P(X ≥ x) = 0.8 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).
P X – µ ≥ x − 135.6 2 = 0.8
σ 5.8
Tekan MODE MODE 1 SHIFT DISTR 3 0.7586 ) + SHIFT DISTR 3
x − 135.6 Press
PZ 5.8 2 = 0.8
≥ 0.9655 ) = 0.3912
1 − PZ ≤ x − 135.6 2 = 0.8
≤ 5.8
x − 135.6 2 = 0.2
PZ 5.8
x–135.6 0 z
x − 135.6 5.8
5.8 = − 0.842
115
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
x = 135.6 – 5.8(0.842) P(X > x2) = P(X < x1) = 0.25
= 130.72 P3 X – µ > x2 −51.835.64 = 0.25
σ
(iii) Kuartil bawah ialah 25% dan kuartil atas
BAB 5 iTahleahlow75e%r q,uparetrilleuisca2r5i%x1adnadnthxe2.upper quartile is P1Z > x2 −51.835.62 = 0.25
75%, need to find x1and x2. 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 0 x2 x1 = 135.6 – (139.51 − 135.6)
= 131.69
(a) Skor suatu ujian bulanan bagi sebuah kelas (b) Suatu hutan simpan yang terdiri daripada
yang mengandungi 155 orang bertaburan 10 000 batang pokok perlu ditebang. Bilangan
normal dengan min 53 dan sisihan piawai 9. pokok yang perlu ditebang adalah bertaburan
The scores of the monthly test of a class with 155 normal dengan min μ dan sisihan piawai σ.
students are normally distributed with mean 53 and
standard deviation 9. Kebarangkalian kurang daripada 60 batang
(i) Cari kebarangkalian skor calon yang pokok ditebang adalah 0.8 dan kebarangkalian
mendapat antara 42 dengan 58. kurang daripada 70 batang pokok ditebang
Find the probability of the score of a candidate adalah 0.9. Hitung nilai min dan varians bagi
to get between 42 dengan 58.
taburan ini.
(ii) Sekiranya calon gagal jika mendapat
A reserve forest has 10 000 trees need to be cut
kurang daripada 40, berapa peratus pelajar down. The number of trees that need to be cut down
are normally distributed with mean μ and standard
gagal? deviation σ.
A candidate fails if he gets less than 40, what is The probability of less than 60 trees to be cut down
the percentage of the students fail? is 0.8 and the probability of less than 70 trees to be
cut down is 0.9. Calculate the value of mean and
(iii) Guru yang menyemak ingin meluluskan variance.
74% pelajarnya, cari skor lulus paling P(X < 60) = 0.8
rendah yang dapat mencapai matlamat P1Z < 60 – µ 2 = 0.8
σ
guru.
60 – µ = 0.842
The reviewing teacher wants to approve 74% of σ
his students, find the lowest passing score that
can achieve the teacher’s goal.
(i) µ = 53, σ = 9 60 – µ = 0.842σ ...➀
P[42 < X < 58] P(X < 70) = 0.9
= P 42 – 53 < Z< 58 – 53 4 70 –
9 9 σ
P1Z < µ 2 = 0.9
= P– 11 < Z < 5 4 70 – µ = 1.282
9 9 σ
= 0.5999 PZ < 40 – 53 4 70 – µ = 1.282σ ...➁
(ii) P(X < 40] = 9
➁–➀
= 0.0743
10 = 0.44 σ
0.0743 × 100% = 7.43%
σ = 22.73
(iii) P[X > m] = 0.74
varians/variance = 516.53
P3Z m – 53 4 =
> 9 0.74 µ = 60 – 0.842(22.73)
m –9 5m3 == –0.643 = 40.86
53 – 9(0.643)
= 47.21
116
PRAKTIS Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
SPM 15
Kertas 1 3. Rajah menunjukkan graf taburan binomial BAB 5
X ~ B(4, p).
1. Rajah menunjukkan graf taburan normal piawai
dengan min μ dan sisihan piawai σ. 2017 the diagram shows the binomial distribution graph
X ~ B(4, p).
2014 The diagram shows a standard normal distribution
graph with mean µ and the standard deviation σ. P[X = x]
f(z) k
h
0 z 5
27
(a) Nyatakan nilai min dan sisihan piawai 4
27
State the value of the mean and the standard
deviation. 1
9
(b) Cari luas rantau berlorek.
0 0 12 34 x
Find the area of the shaded region.
(a) min = 0 dan sisihan piawai = 1 (a) Ungkapkan P(X = 0) + P(X > 2) dalam sebutan
h dan k.
Mean = 0 and standard deviation = 1
Express P(X = 0) + P(X > 2) in terms of h and k.
(b) P(0 < Z < 1) = 0.3413
(b) Cari nilai p.
Find the value of p.
(a) P(X = 0) + P(X > 2) = 1 – h – k
(b) P(X = 0) = 4C0 p0 (1 − p)4 = 4
27
1 − p = 0.037
2. Pemboleh ubah X bertaburan normal dengan min
p = 0.963
32 dan sisihan piawai σ. Diberi skor-z ialah 1.8
2015 apabila X = 38. Cari Kertas 2
The variable X is normally distributed with mean 32 and 1. (a) Didapati bahawa 22% daripada murid
the standard deviation σ. Given the z-score is 1.8 when membawa makanan tengah hari ke sekolah.
X = 38. Find
2016 Jika 7 orang murid itu dipilih secara rawak,
(a) nilai σ. cari kebarangkalian tepat 4 orang murid
membawa bekal makanan tengah hari ke
the value of σ. sekolah.
(b) nilai k jika P(X < k) = 0.4213. It is found that 22% of the students bring packed
lunch to school. If 7 students are chosen at random,
the value of k if P(X < k) = 0.4213. find the probability that exactly 4 students bring
packed lunch to school.
(a) 1.8 = 38 −32 , σ = 3.33
(b) P(X < σ < (b) Jisim satu bungkus mi goreng dari sebuah
k −32 kedai makanan bertaburan normal dengan
k) = P(Z 3.33 ) = 0.4213 min 0.46 kg dan sisihan piawai m kg. Diberi
bahawa 18% daripada bungkusan mi goreng
k3−.3332 = −0.199 mempunyai jisim lebih daripada 0.52 kg.
k = 31.34
117
BAB 5 Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian (i) Jisim minimum gred A ialah 1.2 kg. Jika
The mass of a packed fried mee from a restaurant sebuah nanas dipilih secara rawak, cari
is normally distributed with the mean of 0.46 kg
and the standard deviation of m kg. Given that 18% kebarangkalian bahawa nanas itu adalah
of the packed fried mee have a mass of more than
0.52 kg. gred A.
(i) Hitung nilai m. The minimum mass of grade A is 1.2 kg. If
a pineapple is chosen at random, find the
Calculate the value of m. probability that it is of grade A.
(ii) Diberi bahawa 375 bungkus mi goreng (ii) Cari jisim minimum gred B jika 15%
dijual pada suatu hari, cari bilangan
bungkus mi goreng yang mempunyai daripada nanas adalah gred C.
jisim antara 0.4 kg dengan 0.5 kg.
Find the minimum mass of grade B if 15% of the
Given that 375 packed fried mee are sold on pineapples are grade C.
a certain day, find the number of the packed
fried mee that have the masses between 0.4 kg (b) Dalam perlawanan bola keranjang antara
and 0.5 kg.
sekolah, kebarangkalian sekolah W menang
(a) n = 7, p = 0.22
ialah 40%. Pasukan sekolah W menyertai
P(X = 4) = 7C4(0.22)4(0.78)3
= 0.0389 dalam n perlawanan. Kebarangkalian untuk
(b) (i) Min 0.46 kg dan sisihan piawai m kg. menang sekali adalah 8 kali kebarangkalian
Mean 0.46 kg and the standard deviation of kalah dalam semua perlawanan.
m kg.
In an inter-school basketball tournament, the
P(X > 0.52) = 0.18 probability that school W wins is 40%. The team
of school W participated in n tournaments. The
PZ > 0.52 – 0.462 = 0.18 probability of winning once is 8 times the probability
m of losing all the games.
0.52 – 0.46 = 0.916 (i) Cari nilai n.
m m =
0.52 – 0.46 Find the value of n.
0.916
(ii) Hitung sisihan piawai bagi bilangan
kemenangan.
Calculate the standard deviation for the number
of wins.
= 0.066 kg (a) (i) P(X > 1.2) =
(ii) P=(P01.40.<40X.–0<606.04.65)< Z < PZ > 1.2 – 0.75 2 = 0.0668
0.3
0.5 – 0.46 2
0.066 (ii) P(X < m) =
= 0.5461 PZ < m – 0.75 2 = 0.15
0.3
0.5461 × 375 = 205 bungkus
m – 0.75 = −1.036
0.3
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 × nC0(0.6)n
n × 00..64 = 8
2. (a) Jisim bagi nanas yang dihasilkan di
sebuah ladang bertaburan normal dengan n = 8 × 0.6 = 12
2017 min 0.75 kg dan sisihan piawai 0.3 kg. 0.4
Nanas itu dikelaskan kepada tiga gred, A, B (ii) Var(X) = npq = 12(0.4)(0.6)
dan C mengikut jisimnya, gred A > gred B > Sisihan piawai = 1.7
gred C. Standard deviation
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.
118
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
3. (a) Satu kajian menunjukkan bahawa Gerai itu menawarkan 3 lontaran bagi setiap BAB 5
perbelanjaan pelanggan di sebuah pasar raya
2018 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.
f(x)
permainan. Pelanggan perlu membayar RM4 bagi
50% 12% setiap permainan. Patung beruang akan diberi jika
pelanggan dapat menjatuhkan semua tin bagi tiga
90 140 (RM) lontaran itu dalam satu permainan. Awang ingin
(i) Cari sisihan piawai bermain, kebarangkalian dia dapat menjatuhkan
Find the standard deviation. semua tin bagi setiap lontaran ialah 0.7.
The stall offers 3 throws for each game. A customer needs
(ii) Jika 45 orang pelanggan dipilih secara to pay RM4 for each game. A teddy bear will be given if
rawak, cari bilangan pelanggan yang
membelanja antara RM50 dengan the player can topple all the tins in all the three throws in
RM100.
a game. Awang wants to play the game, the probability
If 45 customers are chosen at random, find that he can topple all the tins for each throw is 0.7.
the number of customers who spend between
RM50 and RM100. (a) Awang akan bermain jika dia mempunyai
(b) Didapati bahawa 23% pelanggan membelanja sekurang-kurangnya 91% peluang
kurang daripada RMy. Cari nilai y.
memenangi sekurang-kurangnya satu
Given that 23% of the customers spend less than
RMy. Find the value of y. 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
(a) (i) Min/Mean = 90 winning at least a teddy bear with the spending of
RM24. From your calculation, suggest to Awang
P(X > 140) = 0.12 whether he should play or not.
P(X > 140) = PZ > 140 – 90 2 (b) Apakah bilangan minimum permainan yang
σ
perlu jika dia ingin memenangi tiga patung
= 0.12 beruang?
140σ– 90 = 1.175 What is the minimum number of games needed if he
wants to win three teddy bears?
σ = 42.55 (a) RM24 boleh bermain 6 set permainan
(ii) P(50 < X < 100) = RM24 can play 6 sets of game.
P 5402–.5950 < Z < 10402.–55902 = 0.4193 n=6
P(menjatuhkan semua tin dalam tiga lontaran
0.4193 × 45 = 18.87 berturut-turut)
( b) P (X < y) = PZ < y42–.5950 2 = 19 P(topple all the tins in three consecutive throws)
= (0.7)3 = 0.343
= 0.23 P(X ≥ 1) = 1 − P(X = 0)
y – 90 = −0.739 = 1 − 6C0(0.343)0(0.657)6
42.55 = 0.9196
y = RM58.56 Peratus menang ialah 91.96%. Awang patut
bermain.
Percentage to win is 91.96%. Awang should play
(b) E(x) = np
4. Rajah menunjukkan susunan tin bagi suatu 3 = n(0.3430)
permainan di pesta ria sekolah. n = 8.75 Praktis
Dia perlu bermain 9 set. SPM
2019 The diagram shows the arrangement of the tins for a Ekstra
game in a school bazaar. He needs to play 9 sets.
119
Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian
Sudut KBAT KBAT
Ekstra
BAB 5 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.
f(x)
34% 34%
13.5% 13.5%
2.5% 2.5% xm
14.0 15.2 15.8 17.0 17.6 18.2
16.4
(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?
(a) min/mean = 16.4
P[X > 17] = 0.16
P3Z > 17 – 16.4 4 = 0.16
σ
17 – 16.4 = 0.994
σ
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 4 = 0.01563
0.6036
0.01563 × 1000 = 15.63 pokok
Quiz 5
120
BAB Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri BAB 6
6 Fungsi Trigonometri
Trigonometric Functions
6.1 Sudut Positif dan Sudut Negatif
Positive Angles and Negative Angles
NOTA IMBASAN
1. Rajah di bawah menunjukkan garis lurus OP berputar 4. Jika garis OP berputar lebih daripada 360° atau 2π,
pada pusat O di atas satah Cartes dan membentuk sudut yang dibentuk akan melebihi 360° atau 2π.
sudut θ1 dan θ2 bermula dengan paksi-x yang positif.
If the line OP rotates more than 360° or 2π, the angle formed is
The diagram shows the line OP is rotated about the centre O on more than 360° or 2π.
a cartesian plane and forms the angles of θ1 and θ2 starting with
the positive x-axis. 5. Sudut rujukan ditandakan sebagai α ialah sudut
y tirus positif yang dibentuk oleh jejari pemutaran OP
P
terhadap paksi-x sahaja.
Sudut positif The reference angle marked as α is the positive acute angle
Positive angle
1 formed by the rotating radius of the OP against the x-axis only.
x
NOTA IMBA0SAN2 Sudut negatif (a) y (b) y
Negative angle
IP
II
2. Sudut θ1 diukur daripada paksi-x positif dalam arah ␣ ␣ x
lawan jam dikenali sebagai sudut positif manakala θ2 0 x0
bermula dengan paksi-x yang positif pada arah ikut jam
dikenali sebagai sudut negatif. ␣= ␣ = 180Њ –
The angle θ1 is measured from the positive x-axis in the (c) y (d) y
anticlockwise direction is defined as positive angle while the
angle θ2 is measured from the positive x-axis in the clockwise x x
direction is defined as a negative angle. ␣0 0 ␣
3. Satah Cartes ini dibahagikan kepada empat sukuan. III
The Cartesian plane is divided into four quadrants. ␣ = – 180Њ
90°
Sukuan Kedua Sukuan Pertama IV
Second Quadrant First Quadrant ␣ = 360Њ –
90° < θ < 180° 0° < θ < 90° π
π 2
atau/or 2 < θ < π atau/or 0° < θ <
180° II I
0° 360°
Sukuan Ketiga Sukuan Keempat 6. Sudut tirus adalah digunakan untuk mencari nisbah
Third Quadrant Fourth Quadrant trigonometri asas.
270°< θ < 360°
180° < θ < 270° The acute angle is used to find the basic trigonometric ratios.
atau/or π < θ < 3 π 3π < θ < 2π
2 2
III IV
121
BAB 6 Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
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) 35π
(iii)
Penyelesaian:
(i) y (ii) y y
II ␣ 108Њ
0
0 x ␣ 0 x x
295Њ ␣ = 65Њ III –139Њ
IV
Sukuan keempat Sudut rujukan Sudut rujukan
4th quadrant Reference angle Reference angle
Sudut rujukan/Reference angle = 180° − 139° = 180° − 108°
α = 360° − 295° = 41°
= 65° (dengan paksi-x) =72°
Tip
(with x-axis)
–139° ialah sudut negatif.
Perhatian: − 139° is a negative angle
y
295Њ x
␣ = 25Њ
α = 295° − 270°
= 25°
SALAH! WRONG!
(a) 121° (b) –316° (c) 520°
y y y
II
II x I x
121Њ ␣ ␣
0 0 520Њ
␣ 316Њ x
0
α = 180° − 121° α = 360° − 316° α = 180° − 160°
= 59° = 44° = 20°
122
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(d) –412° (e) – 32 π (f) 9 π
8
y
– 32 –2 × 180° 9 9 × 180°
π = 3 8 p = 8 BAB 6
0 x = –120° = 202.5°
–412Њ ␣
IV y y
α = 412° − 360° 60Њ 0x 9
= 52° III –120Њ 8x
␣0
α = 60° α = 9 p – p = π
8 8
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
CONTOH (a) (b)
Py y
y 42Њ 0
0
x 6 x
0 x
3
P
Penyelesaian: θ = 360° + (180° – 42°) θ = –3π – π
= 498° 6
θ = – 32 π kerana mengikut
arah jam. =– 17 π
6
θ = – 32 π because follows
clockwise direction. kerana arah ikut jam.
because follows clockwise
direction.
123
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
6.2 Nisbah Trigonometri bagi Sebarang Sudut
Trigonometric Ratios of any Angle
BAB 6 NOTA IMBASAN
1. Apabila sudut θ diletak dalam sukuan pertama seperti 4. Pertimbangkan sudut θ dengan pelengkapnya (90° − θ)
ditunjukkan dalam rajah, OQ = x, OY = y dan OP = r = dalam rajah di bawah, didapati bahawa
x2 + y2 , maka rajah Consider the angle θ with its complement (90° − θ) in the diagram
below, it is found that
When the angle θ is at the first quadrant as shown in the diagram,
y P
OQ = x, OY = y and OP = r = x2 + y2 , then the diagram y
(90Њ–)
y r
P(x , y)
ry x
0x
0x x
(a) sin θ = kos (90°− θ) = y
r
sin θ = y kos/ cos θ = x
x r (b) kos/ cos θ = sin (90° − θ) = x
r
tan θ = y = sin θ
x kos/cos θ (c) tan θ = kot/ cot (90° − θ) = y
x
2. Rajah yang berikut menunjukkan tanda-tanda nisbah
trigonometri mengikut sukuan. (d) kosek/ cosec θ = sek/ sec (90°− θ) = r
y
The following diagram shows the signs for the trigonometric
(e) sek/ sec θ = kosek/ cosec (90°− θ) = r
ratiNosOacTcoArdIinMgBtoAthSe AquNadrants. x
y (f ) kot/ cot θ = tan (90° − θ) = x
y
Sukuan II Sukuan I
Quadrant II Quadrant I 5. Sudut istimewa 30°, 45°, 60°, 90°, 180°, 270° dan 360°.
Special angles 30°, 45°, 60°, 90°, 180°, 270° and 360°.
sinus (+) All (+) Merujuk kepada rajah, jadual di bawah didapati.
sin θ = sin (180° – θ) Semua (+) Referring to the diagrams, the table below is found.
kos θ = –kos (180° – θ)
tan θ = –tan (180° – θ) x 60Њ √2 45Њ
Sukuan IV 2 1
Sukuan III Quadrant IV
Quadrant III sin θ = –sin(360° – θ) 1
sin θ = –sin(θ – 180°) kos θ = kos(360° – θ)
kos θ = –kos(θ – 180°) tan θ = –tan(360° – θ) 30Њ 45Њ
tan θ = tan(θ – 180°) kosinus (+) √3 1
tangen (+)
90° 180°
3. Salingan bagi nisbah trigonometri asas adalah seperti θ 30° 45° 60° 270° 360°
berikut.
sin θ 1 13 1 0 −1 0
The inverse of the basic trigonometric ratios are as follows. 2 22
kos/
kosekan/ cosecan θ = 1 = r , cos θ 3 11
sin θ y 2 22
tan θ
sekan/ secan θ = 1 = r 0 −1 0 1
kos / cos θ x
kotangen/ cotangen θ = 1 = x = kos θ 1 3 ∞0∞0
tan θ y sin θ 31
124
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
3. Ungkapkan setiap fungsi trigonometri berikut dalam sudut rujukan yang sepadan. BAB 6
Express each of the following trigonometric function in the corresponding reference angle. TP 2
CONTOH
(i) tan 315° (ii) kosek (–15°) y
cosec (–15°)
Penyelesaian: x
(i) y 0 15Њ
315Њ tan 315° = –tan 45°
0
x kosek (−15°)
45Њ = ␣ =cosseinc ((−−11155°°))
Tip Tip
– Tentukan sudut dalam sukuan yang betul dahulu. = 1 Tukar kosek θ = 1
Determine the angle in the right quadrant first. –sin15° sin θ
– Cari sudut rujukan.
Find the reference angle. = −kosek 15° Change cosec θ = 1
– Hanya kosinus positif dalam sukuan IV sin θ
Only positive cosine in quadrant IV. −cosec 15°
(a) sin 249° sin 249° = –sin 69° (b) sek (−200°) sek (–200°) = 1
kos(–200°)
y x sec (−200°)
x 1 = –sek 20°
69Њ 0 y –kos 20°
20Њ 0
(c) kot 219° kot/ cot 219° = 1 (d) kos (−152°) kos/ cos (−152°) = kos/ cos 28°
tan 219°
cot 219° cos (−152°)
y y
39Њ 0 x = kot/ cot 39° 0x
28Њ
152Њ
(e) −kosek (−32°) (f) –tan 3 π2
4
−cosec (−32°)
y –kosek(–32°) = – 1 y –tan 3 π2 = – tan 3 × 4180°2
sin(–32°) 4
135Њ
1 45Њ = –tan 135°
0 –32Њ x = – –sin 32° 0 x = −(−tan 45°)
= kosek/ cosec 32° = tan 45°
125
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(g) –sin – 83 π2 (h) –kot 7 π2
4
BAB 6 y 2 –cot 7
4π
0
60Њ –sin – 83 π2 = –sin(–480°) y –kot( 7 π) = – 1
4 tan 315°
x
= –(–sin 60°) =– 1
–tan 45°
= sin 60° 0 45Њ x
= kot/ cot 45°
4. Tentukan nilai nisbah trigonometri bagi sebarang sudut yang diberi.
Determine the value of the trigonometric ratio for any given angles. TP 3
CONTOH (a) Diberi/Given tan θ = 5 , 180° ≤ θ ≤ 270°.
Cari / Find 12
3
Diberi/Given sin θ= 5 , 90° ≤θ ≤ 180°. (i) sin θ
Cari / Find
(ii) sek/ sec θ
(i) tan θ
(iii) kot/ cot θ
(ii) kosek/ cosec θ
(iii) sek/ sec θ
Penyelesaian: y
(i) tan θ = 3 –12 x
–4 θ0
(ii) kosek/ cosec θ –5
13
1 1
= sin θ = 3
5 5 –5
= 3 13
(i) sin θ =
(iii) sek/ sec θ = 1 θ = 1 (ii) sek/ sec θ = 1 = 1 = – 1132
kos – 54 kos/ cos θ –12
13
= – 54 (iii) kot/ cot θ = 1 = 12
tan 5
θ
Tip
y
35 x
–4 0
– Tempatkan sudut dalam sukuan betul.
Place the angle in the correct quadrant.
– Lengkapkan sisi dengan teoram Pythagoras.
Complete the sides with Pythagoras theorem.
126
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(b) Diberi sek θ = − 2 , 90° ≤ θ ≤ 180°. (c) Diberi kot θ = − 3 , 270° ≤ θ ≤ 360°. BAB 6
Given sec θ = − 2 , 90° ≤ θ ≤ 180°. Given cot θ = − 3 , 270° ≤ θ ≤ 360°.
Cari / Find Cari / Find
(i) kot/ cot θ (i) kos/ cos θ
(ii) kosek/ cosec θ (ii) kosek/ cosec θ
(iii) sin θ (iii) sek/ sec θ
sek/sec θ = – 2 = 1 y kot/cot θ = – 3 y
cos θ
kos/cos θ = –1 √2 tan θ = –1 √3
2 1 x 3 0 x
3 –1
(i) kot/cot θ = –1 (i) kos/cos θ = 2 2
–1 0
(ii) kosek/cosec θ = 1 = 2 (ii) kosek/cosec θ = 1
sin θ sin θ
1
(iii) sin θ = 1 = – 12
2
= –2
(iii) sek/sec θ = 1
kos/cos θ
=1=2
33
2
5. Ungkapkan yang berikut dalam sebutan q. (a) Diberi/Given tan θ = −p, 90° ≤ θ ≤ 180°.
Cari/Find
Express the following in terms of q. TP 3
(i) kos/ cos θ
CONTOH (ii) kosek/ cosec θ
(iii) sek/ sec θ
Diberi kos θ = p, 0° ≤ θ ≤ 90°.
(i) kos/cos θ = –1
Given cos θ = p, 0° ≤ θ ≤ 90°. 1 + p2
Cari/Find (ii) kosek/cosec θ = 1 y
(i) tan θ sin θ √1+p2
(ii) sek/ sec θ
(iii) sin θ
Penyelesaian:
y
1 1 + p2 p
p
√1–p2 = x
0p x –1 0
(iii) sek/sec θ = 1
kos/cos θ
1 – p2
(i) tan θ = p = – 1 + p2
(ii) sek/ sec θ= 1 θ = 1
kos/cos p
(iii) sin θ = 1 – p2
127
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(b) Diberi/Given sin θ = – 2p , 180° ≤ θ ≤ 270°. (c) Diberi kot θ = 1 + p, 0° ≤ θ ≤ 90°.
Cari/Find
Given cot θ = 1 + p, 0° ≤ θ ≤ 90°.
(i) kot/ cot θ
Cari/Find
(i) sin θ
(ii) kosek/ cosec θ
(iii) kos/ cos θ
BAB 6(ii) sek/ sec θ
√2+2p+p 2(iii) kos/ cos θ
p2 – 4 y (i) sin θ = 1 y
2
(i) kot/cot θ = 2 + 2p + p2
1 –√p2–4 0 x (ii) kosek/cosec θ = 1
kos θ sin θ
(ii) sek/sec θ =
–2 p 1
= –p = 2 + 2p + p2 x
p2 – 4 (iii) kos/cos θ = (1+p)
1+p 0
(iii) kos/cos θ = – p2 – 4 2 + 2p + p2
2
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
CONTOH (a) kos θ = sin 13° 42’
(i) sin θ = kos 63° (ii) sek θ = kosek 54° cos θ = sin 13° 42'
sin θ = cos 63° sec θ = cosec 54°
y
Penyelesaian: 76Њ18Ј
(i)
(ii) sek θ = kosek 54° 13Њ42Ј x
90Њ– 0
r sec θ = cosec 54°
kos/cos θ = sin 13° 42’
y = kos 76° 18’
θ = 76° 18’
r 36Њ
0x y
r 27Њ y 54Њ
0x
63Њ 1 θ = 1 (b) kot θ = tan 82° 15’
0x kos sin 54°
y cot θ = tan 82° 15'
sin θ = kos 63° kos θ = sin 54° = r
= sin(90° – 63°) = kos 36° y
= sin 27°
∴θ = 36°
∴θ = 27°
7Њ45Ј
Tip 82Њ15Ј x
0
Tempatkan sudut dalam rajah.
Gunakan hubungan sudut pelengkap. k ot/cot θ = tan 82° 15’
sin θ = kos (90° – θ) = kot 7° 45’
Place the angle in the right quadrant. ∴θ = 7° 45’
Use the relation of complementary angle.
sin θ = cos (90° – θ)
128
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(c) kosek θ = sek p° (d) sin θ = kos 34° 54’ (e) tan (90° − θ) = kot 68° 34’
cosec θ = sec p° sin θ = cos 34° 54' tan (90° − θ) = cot 68° 34'
y y y BAB 6
90Њ–pЊ 55Њ6Ј 21Њ26Ј
pЊ x 34Њ54Ј x 68Њ34Ј x
0 0 0
kosek/ cosec θ = sek/ sec p° sin θ = kos/ cos 34° 54’ tan (90° − θ) = kot/ cot 68° 34’
= sin 55° 6’ = tan 21° 26’
1 θ = 1 θ = 55° 6’ 90° − θ = 21° 26’
sin kos/ cos p° 9 0° − 21° 26’ = θ
θ = 68° 34’
sin θ = kos/ cos p°
= sin 90° − p°
θ = 90° − p
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
CONTOH (a) kos/ cos 23°
sin 23° kos/ cos 23° = sin(90° − 23°)
= sin 67°
Penyelesaian: = 0.9205
sin 23° = kos/ cos (90° − 23°)
= kos/ cos 67°
= 0.3907
Tip
Gunakan hubungan sudut pelengkapnya
Use the complementary angle relationship.
(b) tan 23° (c) sek 23° = kosek (90° – 23°)
sec 23° = cosec (90° – 23°)
tan 23° = kot/ cot = (90 – 23°) sek/ sec 23° = kosek/ kosec 67°
= kot/ cot 67° = 1 = 1
tan 67° sin 67°
= 1 = 1
2.3559 0.9205
= 0.4245 = 1.0864
129
BAB 6 Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
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
CONTOH
(i) kos/ cos 210° (ii) sek/ sec (−135°) = kos/ 1
(ii) sek/ sec (−135°) cos (–135°)
Penyelesaian: y = 1 45°
–kos/ cos
(i) kos/ cos 210° = −kos/ cos 30° x
= – 23 –415Њ 0 1
– 1 √2 = – 2
y Tip
–√3 210Њ x Lukis dan lengkapkan sisi segi tiga yang mengandungi sudut
30Њ 0 khas yang tertentu dalam sukuan yang dikehendaki.
Draw and complete the sides of the triangle which contains
–1 2
certain special angle in the correct quadrant.
(a) tan 225° (b) sek/ sec (−240°)
tan 225° = tan 45° sek/ sec (−240°) = kos/ 1
= 1 cos(–240°)
= 1 60°
–kos/ cos
y = 1 = –2
– 12
–1 225Њ y
x
45Њ
–1 0
√2 2
√3 60Њ 0 x
–1
–240Њ
(c) – kosek/ cosec (330°) (d) – sin (−300°)
– kosek/ cosec (330°) = – 1 – sin (−300°)
sin 330° = – sin 60°
= – 23
= – 1
–sin 30°
= 1 =2 y y
1
√3 x
2 0 30Њ –1
2 2 √3
x
60Њ
01
130
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(e) sin 315° + kos/ cos 135° (f) sek 420° + kosek (−120°)
sec 420° + cosec (−120°)
sin 315° + kos/ cos 135° sek/sec 420° + kosek/cosec (−120°)
= – 12 + – 12 y = kos/ 1 420° + sin 1 BAB 6
(–120°)
= – 22 cos
√2 1x = 1 + 1 y
1 45Њ kos/ cos 60° –sin 60°
45Њ –1 = 1 –2
–1 13
2
√2 2
2= 2 – 2 = 2 3 – 1
33 60Њ √3
x
60Њ 1
6.3 Graf Fungsi Sinus, Kosinus dan Tangen
Graphs of Sine, Cosine and Tangent Functions
NOTA IMBASAN
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
NveOrtTicAal dIiMsplBacAemSeAntNup or down of the graph
2. Graf/Graph y = a sin bx + c y b berubah y y = │sin x │
y = sin x b changes 1
y a berubah 1 y = sin 2x
y = sin x a changes
1
Amplitud/Amplitude
0 x 0 x 0 x
2 2 2
–1 –1 –1
1 kala/cycle
y 1 y c berubah
y = 2 sin x 2 c changes
1 y
2 y = sin x 1 1
1 y = sin x + 2
0
1 1 –1
x 2 x 2
2 y = –│sin x │
0 10 x
–1 2
–
–2 –1 y = sin x – 1
2
y = – 3 sin x –1 1
2 2
131
MatemNaOtiTkATaIMmbBaAhSanA NTingkatan 5 Bab 6 Fungsi Trigonometri
BAB 6 3. Graf y = a kos bx + c y b berubah y y = │kos x│
Graph y = a cos bx + c y = kos 2x b changes 1 y = │cos x│
y = cos 2x
y y = kos x x
y = –kos x y = cos x 1 2
y = –cos x
x x 0 y = tan x +1
1 2 0 2
0 –1 –1
–1 y = kos x
y = cos x
1 kala
1 cycle
y
N1OTA IMBASAN x
2
0 y = kos x – 1
–1 y = cos x – 1
NOTA IMBASAN
4. Graf/ Graph y = a tan bx + c
y y = tan x y y = tan 2x y
0 2 x 2 x 1 2 x
0
0
y
y = │tan x│
x
0 2
y = –│tan x│
132
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
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
CONTOH BAB 6
y = sin x
x 0 π π 3 π π 5 π 3 π 7 π 2π
4 2 4 4 2 4
360°
0° 45° 90° 135° 180° 225° 270° 315° 0
sin x 0 0.71 1 0.71 0 –0.71 –1 –0.71
sin x
1
0 x
90Њ 180Њ 270Њ 360Њ
–1
(a) y = kos/ cos x
x 0 π π 3 π π 5 π 3 π 7 π 2π
kos/cos x 4 2 4 4 2 4
0° 90° 180° 360°
1 45° 0 135° –1 225° 270° 315° 1
0.71 kos/ cos x –0.71 –0.71 0 0.71
1
0 x
90Њ 180Њ 270Њ 360Њ
–1
133
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(b) y = tan x
0 π π 3 π π 5 π 3 π 7 π 2π
4 2 4 4 2 4
BAB 6 x 0° 180° 360°
tan x 0 45° 90° 135° 0 225° 270° 315° 0
1 ∞ –1 1 ∞ –1
tan x
4
2
0 x
90Њ 180Њ 270Њ 360Њ
–2
–4
10. Lakar graf fungsi trigonometri yang berikut.
Sketch the following trigonometric function graph. TP 3
CONTOH
(i) y = – 12 sin 2x untuk/for 0 ≤ x ≤ 2π. Tip
Penyelesaian:
– Tentukan nilai a = amplitud (di sini ialah a = – 12 )
– 21
y y = – 1 sin 2x Determine the value of a = amplitude (a = )
1 2 – Tentukan berapa kala dalam 2π.
2 (b = 2 kala)
Determine the number of cycles in 2π. (b = 2 cycles)
– Ada c untuk anjakan atas atau bawah (c = 0)
x Is there any c of vertical up or down movement (c = 0)
2
0
– 1
2
134
(a) y = 3 kos 2x untuk 0≤ x≤ 2π. Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
4
(b) y = 1 + kos 2x untuk 0 ≤ x ≤ π.
y = 1 + cos 2x for 0 ≤ x ≤ π.
y= 3 cos 2x for 0 ≤ x ≤ 2π.
4
BAB 6
y y = 3 kos 2x y y = 1+ kos 2x
4 y = 1+ cos 2x
3 3
4 y = 4 cos 2x 2
1
0 x x
2 0
–1
– 3 –2
4
(c) y = tan 2x − 1 untuk/for 0 ≤ x ≤ π. (d) y = 3 kos 1 x untuk 0 ≤ x ≤ 2π.
2
1
y=3 cos 2 x for 0 ≤ x ≤ 2π.
y y = tan 2x–1 y y = 3│kos 1 x│
2
0 3 1
–1 x 2 y = 3│cos 2 x│
1
x
0 2
–1
–2
–3
(e) y = −sin 3 x − 1 untuk/for 0≤ x ≤ 2π. (f) y = 2 − |tan x| untuk/for 0 ≤ x ≤ 2π.
4
y
y 2
0 2 x
1
y = 2 – │tan x│
x
0 2
–1
–2 3
4
y = –│sin x│–1
135
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
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
BAB 6 CONTOH (a) y = a kos bx + c
y = a sin bx + c
y = a cos bx + c
yy
12
1 0 x
2 –2 2
x
0 2
Penyelesaian: y = a kos bx + c
1 a = −2
2
Amplitud/ Amplitude, a= 1 1 kala dalam 2π
2
1 1
b= 2 kala dalam 2π / 2 cycle in 2π 1 1 cycle in 2π
2
1 3
b= 2 b= 2 ; c = 0
Anjakan ke atas dari 0
Upward shift from 0
c = 1
2a
= −2 ;b= 12 kala/cycle ; c= 0
(b) y = a | kos bx | + c 2 (c) y = |a sin bx + c |
x
y = a | cos bx | + c y
3
y 2
1
0 0
–2 x
2
a = −2 ; b = 1 kala/cycle ; c = 0 a = 2 ; b = 1 kala/cycle ; c = 1
2
136
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
12. Selesaikan persamaan trigonometri dengan menggunakan graf.
Solve the trigonometric equations by using graphs. TP 4
CONTOH BAB 6
Dengan menggunakan skala dan paksi yang Apabila/When x = 0, y = 3
x = π, y = – 34 4
sama, lakarkan graf lengkung y = kos 2x dan
garis lurus y= 3 1 − 2x untuk 0 ≤x ≤ 3 π.
4 π 2
6x + 4 kos 2x = 3 / 6x + 4 cos 2x = 3
Nyatakan bilangan penyelesaian yang ππ
6x
memuaskan π + 4 kos 2x = 3. kos 2x = 3 − 6x
4 4π
With the same scale and on the same axes, sketch the
2curve y = cos 2x and the straight line y = 3 1 − 2x y = 3 1 − 2x 2
4 π 4 π
3
for 0 ≤ x ≤ 2 π. State the number of solutions which Maka, ada dua penyelesaian.
6x Hence, there is two solutions.
π
satisfies + 4 cos 2x = 3.
Penyelesaian Tip
y y = kos 2x Titik persilangan ialah penyelesaian.
1 y = cos 2x The intersection point is a solution.
0 x
3 2
2
–1 y = 3 (1– 2x )
4
(a) (i) Lakar graf lengkung y = tan x untuk y y = tan x
0 ≤ x ≤ 2π.
1
Sketch the curve of y = tan x for 0 ≤ x ≤ 2π. x
(ii) Pada paksi yang sama, lakarkan satu 0 2
graf yang sesuai untuk menyelesaikan
persamaan π(1 – tan x) = 2x. Nyatakan π(1 – tan x) = 2x
bilangan penyelesaian yang memuaskan
π(1 – tan x) = 2x. 1 – tan x = 2x
π
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.
1 – 2x = tan x
π
2x
Maka/Hence y = 1– π
Apabila/When x = 0 y = 1
x = π y = –1
Terdapat 3 penyelesaian
There are 3 solution
137
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(b) (i) Lakar graf y = |2 sin 2x| untuk 0 ≤ x ≤ 2π. (c) (i) Lakar graf y = −2 sin 3 x untuk 0 ≤ x ≤ 2π.
2
Sketch the graph of y = |2 sin 2x| for 0 ≤ x ≤ 2π.
(ii) Pada paksi yang sama, lakar satu graf yang Sketch the curve y = −2 sin 3 x for 0 ≤ x ≤ 2π.
2
BAB 6 sesuai untuk menyelesaikan persamaan
2|2 sin 2x| − 3 = − 2x . (ii) Pada paksi yang sama, lakar satu graf
π
Nyatakan bilangan penyelesaian yang yang sesuai, y = k supaya terdapat satu
2x
memuaskan 2|2 sin 2x| − 3 = − π . penyelesaian sahaja. Cari nilai yang
On the same axes, sketch a suitable graph to mungkin bagi k.
solve the equation 2|2 sin 2x| − 3 = − 2x . State On the same axes, sketch a suitable graph,
π y = k so that there is only one solution. Find the
possible values of k
the number of solutions which satisfies
2|2 sin 2x| − 3 = − 2x . (i) y = –2 sin 3 x 0 ≤ x≤ 2π
π 2
(i) y y = │2 sin 2x│ y y = –2 sin 3 x
2
2 2 y=k
x x
0 2 0 2
–2 –2 y = k
(ii) 2|2 sin 2x| – 3 = –2x
π
(ii) y = k = −2 atau/or 0 < k ≤ 2
3 x
|2 sin 2x| = 2 – π
∴ y= 3 – x
2 π
Apabila/When x = 0, y= 3
2
x = 2π y= 3 –2=– 1
2 2
5 penyelesaian/solutions
(d) (i) Lakar graf y = −2 kos 3 x untuk 0 ≤ x ≤ 2π. (i) 3
4 4
3 y y = –2 kos x
4 y = 2x
Sketch the curve y = −2 cos x for 0 ≤ x ≤ 2π. 2 y cos 3 x
= –2 4
(ii) Pada paksi yang sama, lakar satu graf yang x
2
sesuai untuk menyelesaikan persamaan 0
3 –2
π + 2 kos 4 x = 0 untuk 0 ≤ x ≤ 2π.
2x
Nyatakan bilangan penyelesaian itu.
On the same axes, sketch a suitable graph to
π 3
solve the equation 2x + 2 cos 4 x = 0 for (ii) π = –2 kos 3 x
2x 4
0 ≤ x ≤ 2π. State the number of solutions.
π π 1
∴y = 2x = 2 x
2 penyelesaian/solutions
138
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(e) (i) Lakar graf y = 2 sin 2x + 1 untuk (i) y y = 2 sin 2x+1
NO TA0SkI≤MetxBch≤AthS32eAπcN.urve y = 2 sin 2x + 1 for 0 ≤ x ≤ 3 3 BAB 6
2 π. 2
(ii) Seterusnya, pada paksi yang sama, lakar 1
satu graf yang sesuai untuk menyelesaikan 0 x
–1 2 3
persamaan 2 sin 2x + 2 kos x + 1 = 0. 2
–2
Nyatakan bilangan penyelesaian.
(ii) 2 sin 2x + 2 kos x + 1 = 0
On the same axes, sketch a suitable graph to 2 sin 2x + 1 = –2 kos x
solve the equation 2 sin 2x + 2 cos x + 1 = 0. ∴ y = –2 kos x
2 penyelesaian/solutions
State the number of solutions.
6.4 Identiti Asas
Basic Identities
NOTA IMBASAN (b) 1 + tan2 A = sek2 A
1 + tan2 A = sec2 A
1. Identiti asas / Basic identities:
(a) sin2 A + kos2 A = 1 (c) 1 + kot2 A = kosek2 A
sin2 A + cos2 A = 1 1 + cot2 A = cosec2 A
13. Dengan rajah yang diberi, buktikan rumus yang berikut.
With the diagram given, prove the following formula. TP 3
CONTOH Penyelesaian:
yrA = xr ......
Diberi satu titik P(x, y) dan OP membuat sudut sin A = ➀
A dengan paksi-x, buktikan sin2 A + kos2 A = 1. kos/ cos ➁
Given a point P(x, y) and OP makes an angle A with Menurut teorem Pythagoras,
the x-axis, prove that sin2 A + cos2 A = 1.
According to Pythagoras’ theorem,
y
P(x, y) x2 + y2 = r2
ry x2 + y2 =1
r2 r2
y2 x2
Dari ➀ dan ➁, sin2A = r2 , kos2A = r2 .
A x y2 x2
0x From ➀ and ➁, sin2A = r2 , cos2A = r2 .
Maka, sin2A + kos2A = 1 (Terbukti)
Hence, sin2A + cos2A = 1 (Proved)
139
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri (b) Dengan kaedah yang serupa, buktikan 1 +
(a) Dengan kaedah yang serupa, buktikan kot2 A = kosek2 A
1 + tan2 A = sek2 A.
With the similar method, prove that 1 + cot2 A =
With the similar method, prove that 1 + tan2 A = sec2 A. cosec2 A.
BAB 6 Sebelah kanan: sek2A = 1 = r2 Sebelah kanan: kosek2A = 1 = r2
kos2A x2 sin2A y2
1
Right: sec2A = 1 = r2 Right: cosec2A = sin2A = r2
cos2A x2 y2
Sebelah kiri: 1 + tan2 A = 1+ y2 = (x2 + y2) Sebelah kiri: 1 + kot2 A = 1 + x2 = (x2 + y2)
x2 x2 y2 y2
(x2 + y2) (x2 + y2)
Left: 1 + tan2 A = 1 + y2 = x2 Left: 1 + cot2 A = 1 + x2 = y2
x2 y2
Tetapi/But x2 + y2 = r2 Tetapi/But x2 + y2 = r2 r2 = sebelah kanan
Maka/Hence 1 + tan2A = r2 = sebelah kanan Maka/Hence 1 + kot2 A = y2
x2 1 + kot2 A = kosek2 A.
1 + tan2A = sek2A
14. Buktikan identiti trigonometri yang berikut dengan menggunakan identiti asas.
Prove the following trigonometric identities with the basic identities. TP 4
CONTOH
1 – tan2 A = 1− 2 sin2 A
1 + tan2 A
Sebelah kiri = 1 – tan2 A
1 + tan2 A
Left = (1 − sin2 AA) ÷ (1 + sin2 A )
kos2 kos2 A
(1 − sin2 A )÷ (1 +
sin2 A )
cos2 A cos2 A
= (kkooss22 A − sin2 AA) × kos2 A Gunakan/Use
A kos2 kos2 A + sin2 A kos2 A + sin2 A = 1
cos2 A cos2 A + sin2 A = 1
( cos2 A − sin2 A ) × cos2 A + sin2 A
cos2 A cos2 A Tip
= kos2 A − sin2A Sentiasa bermula dengan sebelah yang
lebih kompleks.
cos2 A − sin2A Always starts with a more complex side.
= 1 − sin2 A − sin2 A
= 1 − 2 sin2 A
= Sebelah kanan/Right side
140
(a) kos A = sek A Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
1 – sin2 A
cos A (b) 1 − 2 sin2 A = 2 kos2 A − 1
1 – sin2 A = sec A BAB 6
1 − 2 sin2 A = 2 cos2 A − 1
Sebelah kiri: kos A = kos A = 1
1 – sin2A kos2A kos A Sebelah kiri / Left side
= 1 − 2(1 − kos2 A)
Left: cos A = cos A = 1
1 – sin2A cos2A cos A 1 − 2(1 − cos2 A)
= sek/ sec A = 1 − 2 + 2 kos2 A
= sebelah kanan/right side
1 − 2 + 2 cos2 A
= 2 kos2 A − 1 (sebelah kanan)
2 cos2 A − 1 (right side)
(c) tan A + kot A = sek A kosek A (d) kot2 A + kos2 A = kosek2 A – sin2 A
tan A + cot A = sec A cosec A cot2 A + cos2 A = cosec2 A – sin2 A
Sebelah kiri / Left side = sin A + kos/ cos A Sebelah kiri / Left side
kos/ cos sin A kot2 A + kos2 A = kosek2 A − 1 + kos2 A
A
cot2 A + cos2 A = cosec2 A − 1 + cos2 A
= sin2 A + kos2 A = 1
kos A sin A kos A sin A = kosek2 A – (1 − kos2 A)
= sin2 A + cos2 A = 1 cosec2 A – (1 − cos2 A)
cos A sin A cos A sin A
= kosek2 A – sin2 A (sebelah kanan)
= sek A kosek A (sebelah kanan)
cosec2 A – sin2 A (right side)
= sec A cosec A (right side)
(e) 1 + kos A = (kosek A + kot A)2
1 – kos A
Sebelah kanan / Right side = (1 + kos A)2 = (1 – (1 + kos A)2 A)
= (kosek A + kot A)2 1 – kos2A kos A)(1 + kos
(cosec A + cot A)2 (1 + cos A)2 = (1 + cos A)2
1 – cos2A (1 – cos A)(1 + cos A)
= kosek2 A + 2 kosek A kot A + kot2 A
1+ kos A
cosec2 A + 2 cosec A cot A + cot2 A 1– kos A
= 1 + 2 kos A + kos2 A = (sebelah kiri)
sin2A sin2A sin2A
1 + cos A (left side)
1 2 cos A cos2 A 1 – cos A
sin2A + sin2A + sin2A
= 1 + 2 kos A + kos2 A = (1 + kos A)2
sin2A sin2A
1 + 2 cos A + cos2 A (1 + cos A)2
sin2A = sin2A
141
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
6.5 NOTRAuImMuBsASSuAdNut Majmuk dan Rumus Sudut Berganda
Addition Formulae and Double Angle Formulae
BAB 6 NOTA IMBASAN
1. Rumus majmuk 3. Rumus sudut separuh
Half angle formulae
Multiple formulae
(a) sin (A ± B) = sin A kos B ± kos A sin B (a) sin A = 2 sin A kos A
2 2
kscioonss((AA(A±±±BB))B==)s=cionksAoAcscooAssBkB±o±scosBisn±AAsssiininnBBA sin B
(b) sin A = 2 sin A cos A
2 2
(c) tan (A ± B) = 1ta±ntaAn±AtatannBB (b) kos A = kos2 A − sin2 A
2 2
2. Rumus sudut berganda cos A = cos2 A − sin2 A
2 2
Double angle formulae
(a) sin 2A = 2 sin A kos A = 1 − 2 sin2 A
2
sin 2A = 2 sin A cos A
(b) kos 2A = kos2A – sin2A = 2 kos2 A − 1
2
= 1 − 2 sin2 A
= 2 kos2 A − 1 2 cos2 A − 1
2
(c) tan 2A = 2 tan A
1 – tan2 A A
2 tan 2
(c) tan A = A
2
1 – tan2
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 . Tip
kot B − kot A
Mula dengan sebelah yang lebih kompleks.
Prove cot (A − B) = cot A cot B + 1 . Teknik ialah perlu lihat bentuk sebutan dalam sebelah kiri. Di
cot B − cot A sini, sebelah kiri terdiri daripada sebutan dalam trigo kosinus,
Sebelah kanan / Right side maka perlu tukar sebelah kanan kepada sebutan dalam
kosinus juga.
= kot A kot B + 1 / cot A cot B + 1 Always starts with the more complex side.
kot B − kot A cot B − cot A
The technique is to see the end results on the other side. Here,
kos A kos B kos B kos A
= ( sin A sin B + 1) ÷ ( sin B − sin A ) the left side consists of the terms with cosine, hence need to
convert the terms on the right into the terms containing cosine.
( cos A cos B + 1) ÷ ( cos B − cos A ) Tip
sin A sin B sin B sin A
kos A kos B + sin A sin B = kos (A − B)
= ( kos A kos B + sin A sin B ) × ( sin B sin A )
sin A sin B sin A – kos A
kos B sin B
( cos A cos B + sin A sin B ) × ( cos B sin B sin A sin B ) cos A cos B + sin A sin B = cos (A − B)
sin A sin B sin A – cos A
kos B sin A – kos A sin B = sin (A − B)
= kos (A – B) = kot (A − B) / cos (A – B) = cot (A − B) cos B sin A − cos A sin B = sin (A − B)
sin (A – B) sin (A – B)
142
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
(a) Buktikan kos(A + B) = 1 – tan A + tan B . (b) Buktikan sin(A – B) = tan A – tan B.
sin(A – B) tan A – tan B kos A kos B
Prove cos(A + B) = 1 – tan A + tan B . Prove sin(A – B) = tan A – tan B.
sin(A – B) tan A – tan B cos A cos B
BAB 6
kos (A + B) = 1 – tan A + tan B sin (A – B) = tan A – tan B
sin (A – B) tan A – tan B kos A kos B
sebelah kanan/Right side Sebelah kanan/Right side
1– sin A sin B = sin A – sin B
cos A cos B cos A cos B
=
sin A sin B sin A cos B – cos A sin B
cos A – cos B = cos A cos B
cos A cos B – sin A sin B = sin (A – B) = Sebelah kiri/Left side
cos A cos B
= cos A cos B
sin A cos B – cos A sin B
cos A cos B
= cos (A + B) = Sebelah kiri/Left side
sin (A – B)
(c) Buktikan sin (A + B) + sin (A – B) = tan A .
sin (A + B) – sin (A – B) tan B
Prove sin (A + B) + sin (A – B) = tan A .
sin (A + B) – sin (A – B) tan B
sin (A + B) + sin (A – B) = tan A
sin (A + B) – sin (A – B) 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 B
cos A sin B
= tan A × cot B
= tan A = Sebelah kanan/Right side
tan B
143
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
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
BAB 6 CONTOH (a) Diberi kos (A + B) = kos A kos B – sin A sin B
Gunakan sin (A + B) = sin A kos B + kos A sin B dan jika A = B, buktikan
dan jika A = B, buktikan sin 2A = 2 sin A kos A. Given that cos (A + B) = cos A cos B – sin A sin B and
if A = B, prove that
Seterusnya, nyatakan sin θ jika A = θ . kos 2A = kos2A − sin2A
2
Using sin (A + B) = sin A cos B + cos A sin B and if = 2 kos2A − 1
A = B, prove that sin 2A = 2 sin A cos A. = 1 − 2 sin2 A θ
2
Hence, state sin θ if A= θ . Seterusnya nyatakan kos θ jika A = .
2
Penyelesaian: Then, state that cos θ if A = θ .
2
Jika/If A = B
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 Maka, kos (A + A) = kos A kos A − sin A sin A
sin 2A = 2 sin A kos A Hence, cos (A + A) = cos A cos A − sin A sin A
sin 2A = 2 sin A cos A kos 2A = kos2 A – sin2 A
θ cos 2A = cos2 A – sin2 A
2
Jika/If A = , maka/hence Oleh kerana kos2 A = 1 − sin2 A
θ θ θ Since cos2 A = 1 − sin2 A
2 2 2
sin 2( ) = 2 sin kos Maka/Hence kos/ cos 2A = 1 − sin2 A – sin2 A
sin 2( θ ) = 2 sin θ cos θ = 1 − 2 sin2 A
2 2 2
Jika/If s in2 A = 1 − kos2 A
sin θ =2 sin θ kos θ
2 2 sin2 A = 1 − cos2 A
Maka/Hence kos 2A = kos2 A – (1 − kos2 A)
cos 2A = cos2 A – (1 − cos2 A)
sin θ = 2 sin θ cos θ
2 2 = 2 kos2 A − 1 /2 cos2 A − 1
Perhatikan: Jika/If A = θ , kos θ = kos2 θ – sin2 θ
sin 2A ≠ 2 sin A 2 2 2
2A bererti sudut dua kali ganda, misalnya A = 302. Maka,
cos θ = cos2 θ – sin2 θ
2(300) = 600, menjadi sin 600, 2 2
θ
manakala 2 sin 300 =2 × sin 300. Jika A = 4, maka = 2 kos2 θ −1
θ θ θ 2
sin 2 = 2 sin 4 kos 4 .
2A means twice the angle, for example A = 302. Then, θ
2(300) = 600, becomes sin 600, 2 cos2 2 −1
however 2 sin 300 = 2 × sin 300. θ = 1−2 sin2 θ
θ θ θ 4 2
If A = 4 , then sin 2 = 2 sin 4 cos .
(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 − tan2A
θ
Seterusnya, nyatakan tan θ jika A = 2 .
Then, state tan θ if A = θ .
2
Jika/If A = B sin (A + A) 2 sin A kos A 2 sin A kos A 2 tan A
tan (A + A) = kos (A + A) kos2 A – sin2 A 1 – tan2A
= tan 2A = kos2 A =
kos2 A – sin2 A θ
2
Jika A= θ , kos2 A θ= 2 tan
2 maka tan
θ
1 – tan2 2
144
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
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
CONTOH (a) Buktikan 1 + sin 2A = (kos A + sin A)2. BAB 6
Buktikan kosek A = kos 2A + sin 2A . Prove that 1 + sin 2A = (cos A + sin A)2.
sin A kos A
Sebelah kanan / Right side
Prove that cosec A = cos 2A + sin 2A . (kos A + sin A)2 = kos2 A + 2 kos A sin A + sin2 A
sin A cos A = 1 + 2 sin A kos A
= 1 + sin 2A (sebelah kiri)
Penyelesaian:
Sebelah kanan / Right side
kos 2A + sin 2A = kos 2A kos A + sin 2A sin A
sin A kos A sin A kos A
= kos (2A – A)
sin A kos A
= kos A = 1
sin A kos A sin A
= kosek A
(b) Buktikan 1 + tan2 A = sek 2A. (c) Buktikan 1– kos 2A = tan2 A.
1 – tan2 A 1+ kos 2A
Prove that 1 + tan2 A = sec 2A. 1 – cos 2A
1 – tan2 A Prove that 1 + cos 2A = tan2 A.
Sebelah kiri / Left side Sebelah kiri / Left side
kos2 A + sin2 A 1 – kos 2A = 1 – (1 – 2 sin2A)
kos2 A 1 + kos 2A 1 + (2 kos2A – 1)
kos2 A – sin2 A = 2 sin2A = tan 2A
2 kos2A
kos2 A
= 1 = sek 2A
kos 2A
(d) Buktikan tan θ = sin θ (e) Buktikan tan A + kot A = 2 kosek 2A
2 1 + kos θ
Prove that tan A + cot A = 2 cosec 2A
Prove that tan θ = sin θ
2 1 + cos θ
Sebelah kanan / Right side Sebelah kiri / Left side
2 sin θ kos θ sin A + kos A = sin2A + kos2 A = 1
2 2 kos A sin A sin A kos A sin A kos A
sin θ
1 + kos θ = θ – 1) = 2 = 2
2 2 sin A kos A sin 2A
1 + (2 kos2
2 sin θ = 2 kosek 2A
2
= = tan θ 2 cosec 2A
2
2 kos θ
2
145
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
18. Tanpa menggunakan kalkulator, cari nilai yang berikut.
Without using a calculator, find the value of the following. TP 3
BAB 6 CONTOH (a) sin 132° kos 72° − kos 132° sin 72°
sin 132° cos 72° − cos 132° sin 72°
kos/ cos 75°
Penyelesaian: = sin (132° − 72°)
= sin 60°
Menggunakan rumus sudut majmuk dan sudut-
3
sudut khas = 2
Use the multiple angle formulae and the special angles
kos/ cos 75° = kos (45° + 30°)
= kos 45° kos 30° – sin 45° sin 30°
=( 1 )( 23)−( 12)( 1 )
2 2
= 3– 1
22
(b) tan 165° (c) kos/ cos 15°
= tan (120° + 45°) = kos (45° − 30°)
= kos 45° kos 30° + sin 45° sin 30°
= (tan 120° + tan 45°)
1 – tan 120° tan 45°
1 )( 3) 1 1
= (– 3+ 1) = ( 2 2 +( 2 )( 2 )
1– (– 3)
( 3 + 1)
1– 3 = 2 2
= 1+ 3
= 4–2 3
–2
= 3 –2
6.6 Aplikasi Fungsi Trigonometri
Application of Trigonometric Functions
19. Selesaikan yang berikut dengan 0° ≤ x ≤ 360°.
Solve the following where 0° ≤ x ≤ 360°. TP 3
CONTOH
(i) sin x = −0.3516 = 180° + 20° 35’ dan 360° − 20° 35’
(ii) kos/ cos 2x = 0.4517 = 200° 35’ dan 339° 25’
(ii) 0° ≤ x ≤ 360°, maka 0° ≤ 2x ≤ 720°
Penyelesaian: y Sudut rujukan, 2x = kos–1 (0.4517)
Reference angle = 63°9’
(i) Sinus ialah negatif pada kosinus ialah positif pada sukuan I dan IV
sukuan III dan IV cosine is positive in quadrant I and quadrant IV
Sine is negative in quadrant x 2x = 63° 9’, 360° − 63° 9’, 360° + 63° 9’,
20Њ35Ј 720° − 63° 9’
III and quadrant IV. 20Њ35Ј
2x = 63° 9’, 296° 51’, 423° 9’, 656° 51’
Sudut rujukan x x = 31° 35’, 148° 26’, 211° 35’, 328° 26’
= sin–1 (0.3516)
= 20° 35’
Sudut x sebenar
Actual angle x
146
Matematik Tambahan Tingkatan 5 Bab 6 Fungsi Trigonometri
Perhatian Sudut Kalkulator
Jangan bahagi sudut rujukan dengan 2 dahulu, SHIFT sin 0.3516 = 20.59°
iaitu 63° 9’ = 31° 35’.
2
63° 9’ BAB 6
Do not divide the reference angle by 2, that is 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.
(a) kos/ cos x = −0.6283 (b) 2 tan x = 2.52
y y
51Њ5Ј x 51Њ34Ј x
51Њ5Ј
51Њ34Ј
x = 180° – 51° 5’, 180° + 51.5’ 2 tan x = 2.52
= 128° 55’, 231° 5’ tan x = 1.26
x = 51° 34’, 180° + 51.34’
= 51° 34’, 231° 34’
(c) sin 2x = −0.357 (d) kos/ cos 2x = −0.712
y y
20Њ55Ј x 44Њ36Ј x
20Њ55Ј 44Њ36Ј
sin 2x = –0.357 kos 2x = 0.712
2x = 200° 55’, 339° 5’ 2x = 135° 24’, 224° 36’
560° 55’, 699° 5’ 495° 24’, 584° 36
x = 100° 28’, 169° 33’, 280° 28’, 349° 33’ x = 67°42’. 112° 18’
247° 42’, 292° 18’
147
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
eBook M&M 2021 Matematik Tam Tg5
Discover the best professional documents and content resources in AnyFlip Document Base.
Search