45 1 Mathematics Term 1 STPM Chapter 1 Functions y x 0 1 1 y = ax y = x y = loga x Figure 1.19 Notice that the graph of y = loga x is the reflection of the graph y = ax about the line y = x. An important base of an exponential function is an irrational number symbolised by e whose value, correct to five decimal places, is 2.71828. The number e is called the natural base and the function f defined by f(x) = ex is called the natural exponential function. If y = ex , then x = loge y which can also be written as x = ln y. Therefore, if f : x ↦ ex , x R, then f –1 : x ↦ ln x, x R+ . The logarithmic function g defined by g(x) = loge x, usually written as g(x) = ln x, is called the natural logarithmic function. Example 39 Sketch the graph of y = 3x . Hence, sketch the graphs of (a) y = 3x + 1 (b) y = 3–x (c) y = –3x Solution: The graph of y = 3x is as shown on the right. (a) By adding 1 unit to the graph of y = 3x , the graph of y = 3x + 1 can be obtained, as shown on the right. The asymptote of the graph is y = 1. 0 y y = 3x x 1 0 y y = 3x + 1 y = 3x y = 1 x 1 2
46 1 Mathematics Term 1 STPM Chapter 1 Functions (b) By reflecting the graph of y = 3x about the y-axis, the graph of y = 3–x is obtained, as shown on the right. (c) By reflecting the graph of y = 3x about the x-axis, the graph of y = –3x is obtained, as shown on the right. Exercise 1.8 1. Sketch the graph of (a) y = 2x (b) y = 2–x (c) y = –2x (d) y = 32x 2. Sketch each of the following curves, showing the asymptotes. (a) y = 3x – 1 (b) y = 1 – 3x (c) y = 2(3x ) – 1 (d) y = 22x + 1 3. Sketch the graph of (a) y = log10 x (b) y = log10 (–x) (c) y = log8 (x – 1) 4. Sketch the graph of (a) y = 1 + log6 (1 – 2x) (b) y = log2 (x2 ) (c) y = log2 (x2 – 1) 5. Sketch the graph of (a) y = x2 e–x , x R (b) y = (1 – x)ex , x R (c) y = x ln x, x R+ (d) y = ln x x , x R+ Properties of exponents and logarithms Exponents If a is a real number and n a positive integer, then an = a × a × a × ……… × a (n factors) In the term an , a is known as the base and n the exponent, which is also known as the index. 0 y y = 3x y = 3–x x 1 0 1 –1 y x y = –3x y = 3x
47 1 Mathematics Term 1 STPM Chapter 1 Functions Properties of exponents If m, n Q and a R+ , then 1. am × an = am + n, 2. am ÷ an = am – n, 3. (am) n = amn. Zero, negative and fractional exponents (a) Consider am ÷ an = am – n, (a . 0). When m = n, an ÷ an = an – n 1 = a0 . When m = 0, a0 ÷ an = a0 – n a0 an = a–n 1 an = a–n (b) Consider am × an = am + n. When m = n = 1 2 , a —1 2 × a —1 2 = a —1 2 + —1 2 = a. But a × a = a. Hence, we can conclude that a —1 2 = a. Similarly, a —1 4 × a —1 4 × a —1 4 × a —1 4 = a. This gives the result that a —1 4 = 4 a. In general, if n is a positive integer, then a —1 n = n a a —m n = (a —1 n ) m = (n a ) m a —m n = (am) —1 n = n am Example 40 Evaluate. (a) 8 —2 3 (b) 16—3 4 (c) 1 8 27 2 – —2 3 Solution: (a) 8 —2 3 = (8—1 3 ) 2 (b) 16—3 4 = (24 ) —3 4 = ( 3 8 ) 2 = 2 4 × —3 4 = 22 = 23 = 4 = 8
48 1 Mathematics Term 1 STPM Chapter 1 Functions (c) 1 8 272 – —2 3 = 1 27 8 2 —2 3 = 1 3 27 3 8 2 2 = 1 3 2 2 2 = 9 4 Surds We learnt that irrational numbers are numbers which cannot be expressed in the form m n , where m, n Z. Some irrational numbers are in the form of surds such as 2 and 3 7 . Expessions which involve surds can usually be simplified. For example, ( 2 · 3 ) 2 = 2 · 3 · 2 · 3 = ( 2)2 · ( 3)2 = 2 · 3 = 6 Hence, 2 · 3 = 6 Properties of surds 1. ab = a · b for a, b R 2. a b = a b for a, b R, b ≠ 0 The expression 1 2 + 3 can be simplified as follows: 1 2 + 3 = 1 2 + 3 × 2 – 3 2 – 3 = 2 – 3 4 – 3 = 2 – 3 This way of simplifying an expression involving a surd is called rationalising the denominator. Notice that when a + b is multiplied by a – b , we get ( a + b )( a – b ) = ( a ) 2 – ( b ) 2 = a – b a – b is called the conjugate of a + b , and vice versa. Example 41 Rationalise the denominator in each of the following cases. (a) 2 3 3 5 (b) 1 2 + 1 (c) 3 – 5 1 + 3 5 Solution: (a) 2 3 3 5 = 2 3 3 5 × 5 5 = 2 15 15
49 1 Mathematics Term 1 STPM Chapter 1 Functions (b) 1 2 + 1 = 1 2 + 1 × 2 – 1 2 – 1 Multiply numerator and denominator with conjugate surd 2 – 1. = 2 – 1 ( 2)2 – 1 = 2 – 1 (c) 3 – 5 1 + 3 5 = 3 – 5 1 + 3 5 × 1 – 3 5 1 – 3 5 = 3 – 9 5 – 5 + 15 1 – (3 5 ) 2 = 18 – 10 5 –44 = 5 5 – 9 22 Logarithms We already know that the number 16 may be expressed in exponential form, i.e. 16 = 24 , where 4 is the exponent and 2 is the base. We may also express 16 = 24 in another form, i.e. log2 16 = 4 Read as ‘logarithm of 16 to base 2 is 4’. Similarly, 25 = 52 or log5 25 = 2 In general, if a and x are positive real numbers with x = an , then the logarithm of x with base a is n. Hence, x = an ⇔ loga x = n, where a, x R+ Note: Since 1 = a0 and a = a1 , loga 1 = 0 and loga a = 1 . Example 42 Evaluate (a) log10 1000 (b) log10 0.01 (c) log3 243 Solution: (a) Let x = log10 1000 Then 10x = 1000 = 103 x = 3 Hence, log10 1000 = 3 (b) Let y = log10 0.01 Then 10y = 0.01 = 10–2 y = –2 Hence, log10 0.01 = –2 (c) Let z = log3 243 Then 3z = 243 = 35 z = 5 Hence, log3 243 = 5
50 1 Mathematics Term 1 STPM Chapter 1 Functions Properties of logarithms Since logarithm is another term for exponent, therefore, the properties of logarithms are closely connected to the properties of exponents. Let x = am and y = an where x, y R+ Then, loga x = m and loga y = n. (a) xy = am × an = am + n Therefore, loga xy = m + n = loga x + loga y (b) x y = am an = am – n Therefore, loga x y = m – n = loga x – loga y (c) xp = (am) p = amp Therefore, loga x p = mp = p loga x For all x, y, a, p R+ 1. loga xy = loga x + loga y 2. loga x y = loga x – loga y 3. loga xp = p loga x Example 43 Express as a single logarithm (a) 1 2 log2 25 + log2 3 – 2 log2 15 (b) loga p2 + 2 loga q – 2 Solution: (a) 1 2 log2 25 + log2 3 – 2 log2 15 = log2 25—1 2 + log2 3 – log2 152 = log2 5 + log2 3 – log2 225 = log2 5 × 3 225 = log2 1 15 = log2 1 – log2 15 = –log2 15 (b) loga p2 + 2 loga q – 2 = 2 loga p + 2 loga q – 2 loga a = 2 (loga p + loga q – loga a) = 2 loga 1 pq a 2 loga a = 1
51 1 Mathematics Term 1 STPM Chapter 1 Functions Change of base of a logarithm In order to find the logarithm of a number not to base 10, we need to change it to logarithm base 10 first before we can find its value, using logarithmic tables or calculators. Example 44 Given that log10 2 = 0.3010 and log10 5 = 0.6990, find the value of log2 5. Solution: Let x = log2 5 Then 2x = 5 Taking logarithms to base 10 for both sides of equation, log10 2x = log10 5 x log10 2 = log10 5 x = log10 5 log10 2 = 0.6990 0.3010 = 2.322 (correct to 3 decimal places) In general, to change the base of a logarithm from a to b, we can use the following method. If loga c = x then c = ax Taking logarithms to base b for both sides of equation, logb c = logb ax logb c = x logb a x = logb c logb a = loga c In the special case when b = c, we get loga b = logb b logb a = 1 logb a Hence, the rules involved in changing the base are loga c = logb c logb a loga b = 1 logb a We often need to change the base when solving exponential equations. Note: The following results may be proven. 1. For a b and c 1, ca cb and logc a logc b. 2. For a b and 0 c 1, ca cb and logc a logc b
52 1 Mathematics Term 1 STPM Chapter 1 Functions Exercise 1.9 1. Evaluate. (a) 50 (b) 36—1 2 (c) 49–—1 2 (d) 81–—3 4 (e) 1 1 2 2 –3 (f) 100—3 2 (g) 1 1 4 2 –—3 2 (h) 12—1 2 · 3—1 2 (i) 32—1 2 · 2–—1 2 9 —1 2 · 8—1 2 (j) –––––––– (k) 27—1 4 · 3—1 4 (l) (0.0625)–—1 4 2 —1 2 2. Express in its simplest form (a) 18 + 8 (b) 50 + 32 – 72 (c) ( 2 – 3 ) 2 (d) (4 3 – 3 2 ) 2 (e) 12 × 27 (f) 32 × 15 ÷ 24 (g) (2 3 – 5 ) (2 3 + 5 ) (h) (3 7 + 5 2 )(3 7 – 5 2 ) 3. Rationalise the denominator for each of the following surds. (a) 10 5 (b) 2 3 (c) 2 2 + 2 (d) 5 3 + 2 (e) 3 2 5 – 3 (f) 1 + 2 3 – 1 (g) 3 + 2 2 5 – 3 (h) 3 3 – 4 6 – 2 3 (i) 1 2 + 1 + 1 2 – 1 (j) 1 5 – 2 – 1 5 + 2 4. Find the value of each of the following logarithms. (a) log3 81 (b) log27 3 (c) log4 0.5 (d) log100 10 (e) log8 0.25 (f) log0.5 8 (g) log5 5 5 (h) log5 0.04 5. Express as a single logarithm (a) log3 8 – log3 6 (b) 4 log2 3 – log2 9 (c) 2 loga 5 + loga 4 – 2 loga 10 (d) 3 2 log3 9 – 2 log3 6 (e) 2 log8 1 2 3 2 – log8 1 8 9 2 (f) log10 (x + 1) – log10 (x2 – 1) (g) 1 2 log10 (x – 1) + 2 log10 (x – 2) (h) log10 (p2 – q2 ) – log10 (p + q) + log10 p – 2 log10 q 6. Evaluate (a) loga 32 loga 2 (b) log3 x log9 x (c) (loga 27)(log3 a)
53 1 Mathematics Term 1 STPM Chapter 1 Functions Equations and inequalities involving exponential and logarithmic expressions An equation or an inequality may involve either an exponential expression or a logarithmic expression, or both, as shown in the following examples. Example 45 Solve the following exponential equations (a) 8x = 32, (b) 22x – 5 · 2x + 4 = 0. Solution: (a) 8x = 32 Expressing both sides of the equation with the same base 2, (23 ) x = 25 23x = 25 Equating the exponents, we get 3x = 5 x = 5 3 (b) 22x – 5 · 2x + 4 = 0 (2x ) 2 – 5 · 2x + 4 = 0 Let y = 2x and substituting into above equation, y2 – 5y + 4 = 0 Hence, (y – 1)(y – 4) = 0 i.e. y – 1 = 0 or y – 4 = 0 y = 1 or y = 4 2x = 1 or 2x = 4 2x = 20 or 2x = 22 x = 0 or x = 2 Example 46 Solve the equation 9x + 1 – 3x + 3 – 3x + 3 = 0. Solution: 9x + 1 – 3x + 3 – 3x + 3 = 0 9 · 9x – 33 · 3x – 3x + 3 = 0 9 · (32 ) x – 27 · 3x – 3x + 3 = 0 9 · (3x ) 2 – 28 · 3x + 3 = 0 (9 · 3x – 1)(3x – 3) = 0 Hence, 9 · 3x – 1 = 0 or 3x – 3 = 0 3x = 1 9 or 3x = 3 = 3–2 = 31 Equating the exponents, we have x = –2 or x = 1.
54 1 Mathematics Term 1 STPM Chapter 1 Functions Example 47 Solve the following equations. (a) 5x = 8 (b) log3 x – 4 logx 3 + 3 = 0 Solution: (a) 5x = 8 log10 5x = log10 8 x log10 5 = log10 8 x = log10 8 log10 5 = 0.9031 0.6990 = 1.292 (b) log3 x – 4 logx 3 – 3 = 0 log3 x – 4 log3 x – 3 = 0 1 logx 3 = ––––– log3 x (log3 x) 2 – 3 log3 x – 4 = 0 Multiply both sides of the equation with log3 x. Let y = log3 x y2 – 3y – 4 = 0 (y + 4)(y – 1) = 0 y = –4 or y = 1 Therefore, log3 x = –4 or log3 x = 1 x = 3–4 or x = 31 = 1 81 = 3 Example 48 Find the set of values of x which satisfy each of the following inequalities. (a) 5x 120 (b) x —7 2 , 49 (c) 1 2 3 2 x , 0.001 Solution: (a) 5x 120 log10 5x log10 120 x log10 5 log10 120 x(0.6990) 2.0792 x 2.0792 0.6990 x 2.975 (c) 1 2 3 2 x , 0.001 x log10 1 2 3 2 log10 (0.001) x (log10 2 – log10 3) log10 (0.001) x(0.3010 – 0.4771) –3 x(–0.1761) –3 x . –3 –0.1761 x . 17.036 (b) x —7 2 , 49 log10 x —7 2 , log10 49 7 2 log10 x , log10 49 log10 x , 2 7 log10 49 log10 x , 2 7 (1.6902) log10 x , 0.4829 log10 x , log10 (3.0402) x , 3.0402 Inequality sign reversed when divided by a negative number.
55 1 Mathematics Term 1 STPM Chapter 1 Functions Exercise 1.10 1. In the following equations, find the values of x. (a) 2x = 8 (b) 3x = 27—2 3 (c) 3x = 27—1 4 (d) 10x = 0.01 (e) (2x ) 3 = 16 (f) (2x ) 2 = 25 2. Solve the following exponential equations. (a) 2x – 3 = 4x + 1 (b) 32x · 3x – 1 = 9 (c) 2x · 2x + 1 = 1 2 (d) 3x · 22x – 3 = 18 3. Solve the following exponential equations. (a) 22x – 9 · 2x + 8 = 0 (b) 32x – 10 · 3x + 9 = 0 (c) 4x – 3 · 2x + 1 + 8 = 0 (d) 22x + 1 + 4 = 2x + 3 + 2x (e) 32x – 3 – 4 · 3x – 2 + 1 = 0 (f) 16x – 5 · 22x – 1 + 1 = 0 4. Given that log3 2 = a and log3 5 = b, express each of the following in terms of a and b. (a) log3 60 (b) log3 6.4 (c) log10 2 5. Given that log5 x = p, express each of the following in terms of p. (a) log5 5x2 (b) logx 5 (c) log25 x (d) logx 0.2 6. Find the value of x in each of the following equations. (a) log8 x = 1.5 (b) log3 x = –0.7 (c) log3 x = 2.2 (d) log2 x = 0.05 (e) logx 65 = –6 (f) logx 5 = 1.2 7. Solve each of the following equations. (a) 5x = 15 (b) 22x + 1 = 0.01 (c) 10x = 0.3 (d) 8(22x ) = 5 (e) 3x – 2 = 52x + 1 8. Solve each of the following equations. (a) log2 x4 + log2 4x = 12 (b) log3 x + log3 (x + 6) = 3 (c) log3 x = 4 logx 3 (d) 2 log4 x + 3 logx 4 = 7 (e) 3 log8 x = 2 logx 8 + 5 (f) log5 x + logx 25 = 3 9. Solve each of the following inequalities. (a) 2x 9 (b) (4.8)x + 1 3.6 (c) 3x + 1 4x – 1 (d) 12x2 102x (e) 1 3 5 2 2x + 1 0.001 10. Find the set of values of x for the following inequalities. (a) 22x – 5(2x ) + 6 0 (Hint: let 2x = y) (b) ex – 3e–x 2 (Hint: multiply both sides by ex )
56 1 Mathematics Term 1 STPM Chapter 1 Functions 1.4 Trigonometric Functions Trigonometric ratios of angles of any magnitude Consider any point P(x, y) on the coordinate plane, with OP = r. When the line OP is rotated in an anti-clockwise direction from the positive x-axis, the angle formed, q, is considered positive. However, if the line OP is rotated in a clockwise direction from the positive x-axis, the angle formed is considered negative. P(x, y) y x r x θ y O P(x, y) –y x r x –θ y O Figure 1.20(a) Figure 1.20(b) For any angle q, we define the trigonometric ratios as x y y cos q = —r sin q = —r tan q = —x r r x sec q = —x csc q = —y cot q = —y The x-axis and y-axis divide the coordinate plane into four regions, or quadrants. The signs of the x and y coordinates in each quadrant are as shown in Figure 1.38(a). If P is in the first quadrant, q is acute and all the trigonometric ratios are positive. y x O First quadrant Second quadrant Fourth quadrant Third quadrant x fi 0 y ff 0 x ff 0 y ff 0 x fi 0 y fi 0 x ff 0 y fi 0 y x O sin θ fi 0 sin θ fi 0 cos θ ff 0 cos θ fi 0 tan θ ff 0 tan θ fi 0 sin θ ff 0 sin θ ff 0 cos θ ff 0 cos θ fi 0 tan θ fi 0 tan θ ff 0 Figure 1.21(a) Figure 1.21(b) By taking note of the signs of the x and y coordinates of P as P moves from quadrant to quadrant, the signs for sin q, cos q and tan q can be determined. Figure 1.21(b) shows the signs of the trigonometric ratios sin q, cos q and tan q in each quadrant. The numerical value of a trigonometric ratio for an angle q of any magnitude is the same as the trigonometric ratio of the acute angle between OP and the x-axis. The position of the quadrant in which OP lies will determine the sign of the trigonometric ratio, either positive or negative.
57 1 Mathematics Term 1 STPM Chapter 1 Functions The diagrams below show the position of the point P in each quadrant and the value of the acute angle a. (i) First quadrant (ii) Second quadrant (0 q 90°) (90° q 180°) x P y θ α = O x P y θ α = 180° – θ O Figure 1.22(a) Figure 1.22(b) a = q a = 180° – q sin q = sin a sin q = sin a cos q = cos a cos q = –cos a tan q = tan a tan q = –tan a (iii) Third quadrant (iv) Fourth quadrant (180° q 270°) (270° q 360°) x P y θ α = – 180° θ O x P O y θ α = 360° – θ Figure 1.22(c) Figure 1.22(d) a = q – 180° a = 360° – q sin q = –sin a sin q = –sin a cos q = –cos a cos q = cos a tan q = tan a tan q = –tan a Note: 1. Angles are measured in terms of degrees or radians, where 360° = 2π radians. 2. For any angle q, sin (–q) = –sin q, cos (–q) = cos q, tan (–q) = –tan q, 3. For any angle q, we define csc q = 1 sin q , sec q = 1 cos q , cot q = 1 tan q .
58 1 Mathematics Term 1 STPM Chapter 1 Functions Periodic and symmetric properties of trigonometric functions For each of the six trigonometric ratios, we know that the value of the ratio repeats itself after a fixed period. π π π 1 For example, sin — = sin 12π + —2 = sin 14π + —2 = … = — 6 6 6 2 π π π 3 cos — = cos 12π + —2 = cos 14π + — 2 = … = — 6 6 6 2 π π π 1 and tan — = tan 1π + —2 = tan 12π + —2 = … = — 6 6 6 3 We say that the trigonometric functions are periodic. The period for sin x and cos x are both 2π, and for tan x, the period is π. The shape of the graph for each trigonometric function repeats itself after each period. The graphs for sin x and cos x are symmetrical about the line parallel to the y-axis and passing through either a maximum or a minimum point of the function. The graph for tan x does not have an axis of symmetry. Graphs of trigonometric functions Graph of y = sin x Period –1 1 0 –2π –π π 2π 3π 4π y x 3_π 2 3π – _ 2 5_π 2 7_π 2 _π 2 π – _ 2 y = sin x Figure 1.23 The graph of y = sin x displays the following properties: (a) The graph is continuous for all x R. (b) The range of the function is –1 sin x 1. (c) The graph repeats itself after each period, for example, from 0 to 2π, from 2π to 4π and so on. (d) The graph is symmetrical about the line passing through a maximum or minimum point and parallel to the y-axis (x = 0). For example, the graph is symmetrical about the line x = 3π 2 .
59 1 Mathematics Term 1 STPM Chapter 1 Functions Graph of y = cos x Period y = cos x –1 1 0 –2π –π π 2π 3π 4π y x Figrue 1.24 The graph of y = cos x displays the following properties: (a) The graph is continuous for all x R. (b) The range of the function is –1 cos x 1. (c) The graph is periodic with period 2π. (d) The graph is symmetrical about the line x = 0, or any line passing through a maximum or minimum point and parallel to the line x = 0. Graph of y = tan x Period –1 1 –2π –π 0 π 2π 3π 4π y x 3_π 2 3π – _ 2 5_π 2 7_π 2 _π 2 π – _ 2 y = tan x Figure 1.25 The graph of y = tan x is different from the graphs of y = sin x or y = cos x. The graph of y = tan x displays the following properties: (a) The graph is discontinuous (i.e. not continuous). nπ (b) The graph is undefined when x = —, where n is an odd integer. 2 (c) The range of the function is –∞ tan x ∞. (d) The graph is periodic with period π.
60 1 Mathematics Term 1 STPM Chapter 1 Functions Graph of y = a sin (bx + c) The graph of y = a sin x, (a 0) is similar to the graph y = sin x and intersects the x-axis at the same points. The maximum value of a sin x is a and the minimum value is –a. 1 Figure 1.25 shows the graphs of y = sin x, y = — sin x and y = 2 sin x. 2 0 0.5 y x 3_π 2 _π 2π 2 π –2 –1 2 –0.5 1 y = 1 sin x _ 2 y = sin x y = 2sin x —– 5π 2 Figure 1.26 The graph of y = sin (x + c) can be obtained by shifting the graph of y = sin x c units to the left if c 0 and c units to the right if c 0. The graph of y = cos x can be obtained by shifting the graph of y = sin x translationally π 2 units to the left. Hence, cos x = sin (x + π 2 ). 0 y x π 2π 3π –1 1 y = sin(x + π ) _ 2 y = sin(x – π ) _ y = sinx 2 Figure 1.27 The function f(x) = sin x, where x R, is periodic with period 2π. Hence the graph y = sin x completes a cycle in the period 2π. The function g(x) = sin 2x, where x R, is periodic with period π, i.e. 2π 2 . Hence, the graph y = sin 2x completes a cycle in the period π, for two cycles for the domain [0, 2π]. In general, for the function f(x) = sin kx, where x R and k 0, the period of the function is 2π k and the graph y = sin kx completes k cycles for the domain [0, 2π].
61 1 Mathematics Term 1 STPM Chapter 1 Functions The graph of y = a sin (bx + c) can be sketched by following the steps below: 1. Sketch the graph of y = sin bx. 2. Sketch the graph of y = sin (bx + c) by shifting the graph y = sin bx translationally c b units to the left (if c 0), or c b units to the right (if c 0). 3. Sketch the graph of y = a sin (bx + c) by multiplying the maximum value and minimum value (+1 and –1) of y = sin (bx + c) by a. Example 49 Sketch, on the same diagram, the graphs of y = cos x and y = cos 1 2 x for 0 x 2π. Solution: The shape of the graph of y = cos 1 2 x is similar to that of the graph of y = cos x. 2π The period of the graph y = cos 1 2 x is ––––, i.e. 4π. 1 2 Hence, the graph of y = cos 1 2 x completes a cycle in the domain [0, 4π], or half a cycle in the domain [0, 2π]. The graph of y = cos 1 2 x is as below: π 2π y x 0 1 –1 y = cos 1 x _ 2 y = cos x Example 50 Sketch the graph of y = 2 sin 1 3 2 x + 1 4 π2 for 0 x 2π. Solution: By comparing the graph of y = 2 sin 1 3 2 x + 1 4 π2 with that of y = sin x, we deduce, 2π (a) the period of the function is ––––, i.e. 4π 3 , 3 2 (b) it completes 3 2 cycles in the domain [0, 2π], 1 4 π (c) it is shifted translationally ––––, i.e. π 6 units to the left, 3 2 (d) in the domain [0, 2π], –2 2 sin 1 3 2 x + 1 4 π2 2.
62 1 Mathematics Term 1 STPM Chapter 1 Functions The graph of each function is as follows. 0 y x _ 3 – _ 6 π π 2 –1 1 y = sin 3 x _ 2 0 y x _ 6 –1 1 – _ 6 0 y x _ 6 –2 2 5 — 3 7 — 3 — 2 7 — 6 3 — 2 11 — 6 11 — 6 13 — 6 15 — 6 15 — 6 5 — 6 — 2 7 — 6 3 y = sin ( —x + — ) 2 4 3 y = 2 sin ( — x + — ) 2 4 8 — 3 2 — 3 4 — 3 Period π π π π π π π π π π π π π π π π π π π π π π π Graphs of inverse trigonometric functions The function f : x ↦ sin x is a many-to-one mapping for the domain x R. Hence, it does not have an inverse function. However, if we restrict the domain of the function f to 3– π 2 , π 2 4, then the function is a oneto-one function. In this case, the function f : x → sin x can have an inverse. This inverse function of sin x is represented by sin–1 x. Hence, for f : x ↦ sin x, – π 2 x π 2 , f –1 : x ↦ sin–1 x, –1 x 1. By similarly restricting the domains of cos x and tan x to [0, π] and 1– π 2 , π 2 2 respectively, their inverses also exist. For f : x ↦ cos x, 0 x π, f –1 : x ↦ cos–1 x, –1 x 1; and f : x ↦ tan x, – π 2 x π 2 , f –1 : x ↦ tan–1 x, –∞ x ∞.
63 1 Mathematics Term 1 STPM Chapter 1 Functions The graphs of y = sin x, y = sin–1 x, y = cos x, y = cos–1 x, y = tan x and y = tan–1 x are as shown below. y x 0 1 –1 y = sin x π – _ 2 _π 2 y x –1 0 1 y = sin–1x π – _ 2 _π 2 Figure 1.28(a) y x 0 1 –1 y = cos x _π 2 π y x 0 –1 1 y = cos–1x π _π 2 Figure 1.28(b) y x 0 y = tan x π – _ 2 _π 2 y x 0 π – _ 2 _π 2 y = tan–1x Figure 1.28(c) Exercise 1.11 1. Sketch, on separate diagrams, each of the following curves for 0° x 360°. (a) y = 3 cos x (b) y = cos 2x (c) y = cos (x – 30°) 2. Sketch each of the following curves for 0 x 2π. (a) y = sin 1x + π 4 2 (b) y = sin 2x (c) y = 2 sin 12x + π 4 2 (d) y = cos 3x (e) y = 3 cos 13x + π 3 2 (f) y = sin 1 2 x (g) y = 2 sin 1 3 x (h) y = tan 2x 3. Sketch, on separate diagrams, each of the following curves for 0 x π. (a) y = tan 2x (b) y = tan 21x – π 4 2 (c) y = tan1 π 2 – 2x2
64 1 Mathematics Term 1 STPM Chapter 1 Functions Pythagorean identities Let P(x, y) be any point on the coordinate plane with the line OP inclined at an angle q with the positive x-axis, as shown in Figure 1.29. In the right-angled triangle OPN, x2 + y = r 2 By Pythagoras' Theorem. Dividing both sides of the above equation by r 2 , x2 r 2 + y2 r 2 = 1 Since x r = cos q and y r = sin q, hence cos2 q + sin2 q = 1 ………… 1 Dividing equation 1 with cos2 q, we have cos2 q cos2 q + sin2 q cos2 q = 1 cos2 q 1 + tan2 q = sec2 q ………… 2 Dividing equation 1 with sin2 q, we have cos2 q sin2 q + sin2 q sin2 q = 1 sin2 q cot2 q + 1 = csc2 q …………3 Equation 1, 2 and 3 are true for all values of q, and form the basic identities in trigonometry. We may use the symbol '≡' for identities. Hence, for any angle q, cos2 q + sin2 q ≡ 1 1 + tan2 q ≡ sec2 q cot2 q + 1 ≡ csc2 q These are called The Pythagorean identities. Example 51 If sin q = 5 13 and q is an obtuse angle, find cos q and tan q. Solution: Using cos2 q + sin2 q = 1 cos2 q = 1 – sin2 q = 1 – 1 5 132 2 = 1 – 25 169 = 144 169 \ cos q = ± 12 13 Since q is an obtuse angle, cos q = – 12 13. P(x, y) y r N x θ O y x Figure 1.29
65 1 Mathematics Term 1 STPM Chapter 1 Functions tan q = sin q cos q = 5 13 ÷ (– 12 13) = – 5 13 × 13 12 = – 5 12 Therefore, if sin q = 5 13 and q is obtuse, then cos q = – 12 13 and tan q = – 5 12. Example 52 Simplify. (a) sin2 q 1 – cos q (b) sin q cos q + cos q sin q Solution: (a) sin2 q 1 – cos q (b) sin q cos q + cos q sin q = 1 – cos2 q 1 – cos q = sin2 q + cos2 q cos q sin q = (1 – cos q)(1 + cos q) 1 – cos q = 1 cos q sin q = 1 + cos q = sec q csc q Example 53 Prove that (csc A – sin A)(sec A – cos A) ≡ 1 tan A + cot A . Solution: (csc A – sin A)(sec A – cos A) = 1 1 sin A – sin A21 1 cos A – cos A2 = 1 1 – sin2 A sin A 21 1 – cos2 A cos A 2 = 1 cos2 A sin A 21 sin2 A cos A 2 = cos A sin A = cos A sin A sin2 A + cos2 A Since sin2 A + cos2 A = 1 1 = sin2 A + cos2 A cos A sin A 1 = sin A cos A + cos A sin A = 1 tan A + cot A Hence, (csc A – sin A)(sec A – cos A) ≡ 1 tan A + cot A .
66 1 Mathematics Term 1 STPM Chapter 1 Functions Exercise 1.12 1. For this question, the use of a calculator is not permitted. (a) If cos θ = 4 5 and θ is an acute angle, find the values of sin θ, tan θ and sec θ. (b) If tan θ = – 8 15 and θ is an obtuse angle, find the values of sin θ, sec θ and cot θ. (c) If cot x = 7 24 and x is non-acute, find the values of sin x, cos x and csc x. (d) If sin x = 12 13 and x is an acute angle, find the values of cos x, cos (x + 90°) and tan (x + 180°). 2. Simplify each of the following expressions. (a) 1 – cos2 q tan q (b) sin q 1 – sin2 q (c) tan q 1 + tan2 q (d) 1 – sec2 q 1 – csc2 q (e) 1 + tan2 q 1 – sin2 q — (f) 1 cos q 1 + cot2 q 3. Simplify each of the following expressions. (a) (sin2 θ – 2)2 – 4 cos2 θ (b) cos4 q – sin4 q cos q – sin θ (c) sin q csc q – cot θ (d) tan2 q + cos2 q sin q + sec θ 4. Prove each of the following identities. (a) sec A – cos A ≡ sin A tan A (b) sec2 A + csc2 A ≡ sec2 A csc2 A (c) 1 – sin q 1 + sin q ≡ (sec θ – tan θ) 2 (d) cot2 q – 1 cot2 q + 1 ≡ 1 – 2 sin2 θ (e) cos x 1 – tan x + sin x 1 – cot x ≡ sin x + cos x (f) tan2 x + cot2 x ≡ sec2 x + csc2 x – 2 (g) sin A 1 + cos A + 1 + cos A sin A ≡ 2 sin A (h) (1 + sin θ + cos θ) 2 ≡ 2(1 + sin θ)(1 + cos θ) The compound angle formulae Let A and B be any two angles. Then A + B or A – B is known as a compound angle. To find sin (A + B), sin (A – B), cos (A + B) and cos (A – B), we need to find the formula for each of them in terms of sin A, cos A, sin B and cos B.
67 1 Mathematics Term 1 STPM Chapter 1 Functions The diagram below may be used to obtain the formulae for the compound angles A + B and A – B. K L A B O M P Q R A Figure 1.30 Let OPQ and OQR be two right-angled triangles. Notice that RM is perpendicular to OP, and QK is perpendicular to RM. Then, ∠LRQ = A. From ∆OMR, sin (A + B) = MR OR = MK + KR OR = PQ + KR OR = PQ OR + KR OR = PQ OQ × OQ OR + KR QR × QR OR = sin A cos B + cos A sin B Hence, sin (A + B) ≡ sin A cos B + cos A sin B …………… By substituting B with (–B), sin (A – B) ≡ sin A cos (–B) + cos A sin (–B) But cos (–B) = cos (360° – B) = cos B and sin (–B) = sin (360° – B) = –sin B Hence, sin (A – B) ≡ sin A cos B – cos A sin B …………… b Similarly, cos (A + B) = OM OR = OP – MP OR = OP – KQ OR = OP OR – KQ OR = OP OQ × OQ OR – KQ RQ × RQ OR = cos A cos B – sin A sin B Hence, cos (A + B) ≡ cos A cos B – sin A sin B …………… c By substituting B with (–B), cos (A – B) ≡ cos A cos (–B) – sin A sin (–B) ≡ cos A cos B + sin A sin B Hence, cos (A – B) ≡ cos A cos B + sin A sin B …………… d
68 1 Mathematics Term 1 STPM Chapter 1 Functions Based on the formulae for sin (A ± B) and cos (A ± B) obtained above, the formulae for tan (A ± B) can be obtained as follows: tan (A + B) ≡ sin (A + B) cos (A + B) ≡ sin A cos B + cos A sin B cos A cos B – sin A sin B Dividing each term on the right hand side (RHS) by cos A cos B, sin A cos A + sin B cos B tan (A + B) ≡ 1 – sin A cos A . sin B cos B ≡ tan A + tan B 1 – tan A tan B Hence, tan (A + B) ≡ tan A + tan B 1 – tan A tan B …………… e In the same way, or substituting B with (–B), Hence, tan (A – B) ≡ tan A – tan B 1 + tan A tan B …………… f For identities , c and e, when A = B, we will obtain the multiple angle formulae, i.e. sin 2A ≡ 2 sin A cos A cos 2A ≡ cos2 A – sin2 A ≡ 2 cos2 A – 1 ≡ 1 – 2 sin2 A tan 2A ≡ 2 tan A 1 – tan2 A If the angle A in the above formulae for a multiple angle is replaced by A 2 , the half-angle formulae can be obtained, i.e. sin A ≡ 2 sin A 2 cos A 2 cos A ≡ cos2 A 2 – sin2 A 2 ≡ 2 cos2 A 2 – 1 ≡ 1 – 2 sin2 A 2 2 tan A 2 tan A ≡ 1 – tan2 A 2
69 1 Mathematics Term 1 STPM Chapter 1 Functions Example 54 Without using tables or a calculator, find each of the following in surd form. (a) sin 15° (b) tan 105° (c) cos (–165°) Solution: (a) sin 15° = sin (60° – 45°) = sin 60° cos 45° – cos 60° sin 45° = 3 2 · 1 2 – 1 2 · 1 2 = 3 – 1 2 2 × 2 2 = 1 4 ( 6 – 2) (b) tan 105° = tan (60° + 45°) = tan 60° + tan 45° 1 – tan 60° tan 45° = 3 + 1 1 – 3 = ( 3 + 1)2 (1 – 3 )(1 + 3 ) = 3 + 2 3 + 1 1 – 3 = 4 + 2 3 –2 = –(2 + 3) (c) cos (–165°) = cos (360° – 165°) = cos 195° = –cos 15° = –cos (60° – 45°) = –(cos 60° cos 45° + sin 60° sin 45°) = –1 1 2 · 1 2 + 3 2 · 1 2 2 = – (1 + 3 ) 2 2 × 2 2 = – 1 4 ( 2 + 6 ) Example 55 Without using tables or a calculator, find the values of (a) 1 2 (cos 75° – sin 75°), (b) 1 + tan 15° 1 – tan 15° Solution: (a) 1 2 (cos 75° – sin 75°) = 1 2 cos 75° – 1 2 sin 75° = cos 45° cos 75° – sin 45° sin 75° = cos (45° + 75°) = cos 120° = –cos 60° = – 1 2 From the formulae sin (A – B) = sin A cos B – cos A sin B From the formulae tan (A + B) tan A + tan B = ———————– 1 – tan A tan B From the formulae cos (A – B) = cos A cos B + sin A sin B 1 Replace —– with 2 cos 45° and sin 45°.
70 1 Mathematics Term 1 STPM Chapter 1 Functions (b) 1 + tan 15° 1 – tan 15° = tan 45° + tan 15° 1 – tan 45° tan 15° = tan (45° + 15°) = tan 60° = 3 Example 56 The angle A is obtuse with sin A = 3 5 , while the angle B is acute with sin B = 12 13. Without finding the values of A and B, determine the values of (a) cos (A – B), (b) tan (A + B). Solution: In order to be able to use the compound angle formulae, we need to know the values of cos A, cos B, tan A and tan B. Based on the identity sin2 A + cos2 A ≡ 1, we have cos2 A ≡ 1 – sin2 A = 1 – 1 3 5 2 2 = 1 – 9 25 = 16 25 cos A = ± 4 5 Since A is obtuse, cos A = – 4 5 . tan A ≡ sin A cos A 3 5 = – 4 5 = – 3 4 Similarly, cos2 B ≡ 1 – sin2 B = 1 – 1 12 132 2 = 1 – 144 169 = 25 169 cos B = ± 5 13 Since B is acute, cos B = 5 13 . tan B ≡ sin B cos B 12 13 = 5 13 = 12 5 Replace 1 with tan 45° to obtain the form tan A + tan B ———————– = tan(A + B) 1 – tan A tan B
71 1 Mathematics Term 1 STPM Chapter 1 Functions (a) cos (A – B) ≡ cos A cos B + sin A sin B = 1– 4 5 21 5 132 + 1 3 5 2112 132 = – 4 13 + 36 65 = 16 65 (b) tan (A + B) ≡ tan A + tan B 1 – tan A tan B 1– 3 4 2 + 1 12 5 2 = 1 – 1– 3 4 2112 5 2 33 20 = 1 + 9 5 = 33 56 Sums and differences of sines and of cosines Consider the compound angle formulae again, i.e. sin (A + B) ≡ sin A cos B + cos A sin B ............................... sin (A – B) ≡ sin A cos B – cos A sin B ..............................b cos (A + B) ≡ cos A cos B – sin A sin B ..............................c cos (A – B) ≡ cos A cos B + sin A sin B ...............................d + b : sin (A + B) + sin (A – B) ≡ 2 sin A cos B .....................................................e – b : sin (A + B) – sin (A – B) ≡ 2 cos A sin B .....................................................f c + d : cos (A + B) + cos (A – B) ≡ 2 cos A cos B .................................................... c – d : cos (A + B) – cos (A – B) ≡ –2 sin A sin B ................................................... Let A + B = P and A – B = Q. Then, 2A= P + Q ⇒ A = 1 2 (P + Q) and 2B= P – Q ⇒ B = 1 2 (P – Q). Substituting these into e, f, and , we get sin P + sin Q ≡ 2 sin 1 2 (P + Q) cos 1 2 (P – Q) sin P – sin Q ≡ 2 cos 1 2 (P + Q) sin 1 2 (P – Q) cos P + cos Q ≡ 2 cos 1 2 (P + Q) cos 1 2 (P – Q) cos P – cos Q ≡ –2 sin 1 2 (P + Q) sin 1 2 (P – Q)
72 1 Mathematics Term 1 STPM Chapter 1 Functions Example 57 Express each of the following as a single product. (a) sin 6θ + sin 4θ (b) cos 4θ – cos 2θ Solution: (a) By using sin P + sin Q ≡ 2 sin 1 2 (P + Q) cos 1 2 (P – Q) and letting P = 6θ and Q = 4θ, sin 6θ + sin 4θ ≡ 2 sin 1 2 (6θ + 4θ) cos 1 2 (6θ – 4θ) i.e. sin 6θ + sin 4θ ≡ 2 sin 5θ cos θ (b) By using cos P – cos Q ≡ –2 sin 1 2 (P + Q) sin 1 2 (P – Q) and letting P = 4θ and Q = 2θ, cos 4θ – cos 2θ ≡ –2 sin 1 2 (4θ + 2θ) sin 1 2 (4θ – 2θ) cos 4θ – cos 2θ ≡ –2 sin 3θ sin θ Example 58 Factorise cos θ – cos 3θ + cos 5θ – cos 7θ. Solution: cos θ – cos 3θ + cos 5θ – cos 7θ ≡ (cos θ – cos 7θ) – (cos 3θ – cos 5θ) By using cos P – cos Q ≡ –2 sin 1 2 (P + Q) sin 1 2 (P – Q), we have cos θ – cos 3θ + cos 5θ – cos 7θ ≡ {–2 sin 1 2 (θ + 7θ) sin 1 2 (θ – 7θ)} – {–2 sin 1 2 (3θ + 5θ) sin 1 2 (3θ – 5θ)} ≡ –2 sin 4θ sin (–3θ) + 2 sin 4θ sin (–θ) ≡ 2 sin 4θ sin 3θ – 2 sin 4θ sin θ ≡ 2 sin 4θ (sin 3θ – sin θ) ≡ 2 sin 4θ . 2 cos 1 2 (3θ + θ) sin 1 2 (3θ – θ) ≡ 4 sin 4θ cos 2θ sin θ Example 59 Simplify sin (A + B) – sin A cos (A + B) + cos A . Solution: sin (A + B) – sin A ≡ 2 cos 1 2 (A + B + A) sin 1 2 (A + B – A) ≡ 2 cos (A + 1 2 B) sin 1 2 B cos (A + B) + cos A ≡ 2 cos 1 2 (A + B + A) cos 1 2 (A + B – A) ≡ 2 cos (A + 1 2 B) cos 1 2 B
73 1 Mathematics Term 1 STPM Chapter 1 Functions 2 cos (A + 1 2 B) sin 1 2 B ∴ sin (A + B) – sin A cos (A + B) + cos A = 2 cos (A + 1 2 B) cos 1 2 B sin 1 2 B = cos 1 2 B = tan 1 2 B Exercise 1.13 1. Express in surd form (a) sin 75° (b) cos 105° (c) tan (–15°) (d) cot 75° 2. Using the compound angle formulae, find the value of (a) sin 70° cos 80° + cos 70° sin 80° (b) 1 2 cos 15° – 1 2 sin 15° (c) cos 105° cos 15° + sin 105° sin 15° (d) 3 2 sin 60° + 1 2 cos 60° 3. Find the value of (a) tan 40° + tan 20° 1 – tan 40° tan 20° (b) tan 75° – 1 tan 75° + 1 (c) sin 75° + cos 75° (d) 3 cos 15° – sin 15° 4. A and B are acute angles with sin A = 12 13 and sin B = 4 5 . Find the values of sin (A + B) and cos (A + B). 5. If A is an acute angle with cos A = 5 7 and B is an obtuse angle with sin B = 1 5 , find the values of sin (A – B) and cos (A – B). 6. If 180° A 360° with tan A = 1 2 , and –90° B 90° with tan B = – 1 3 , find tan (A – B) and the value of the angle (A – B). 7. If tan (A – B) = 2 and tan B = 1 4 , find tan A. 8. Prove each of the following identities. (a) cot (X + Y) ≡ cot X cot Y – 1 cot X + cot Y (b) sin (45° + θ) + sin (45° – θ) ≡ 2 cos θ (c) sin (A + B) + cos (A – B) ≡ (sin A + cos A)(sin B + cos B) (d) sin (a + b) cos a cos b ≡ tan a + tan b (e) tan (A + B) – tan A ≡ sin B cos A cos (A + B)
74 1 Mathematics Term 1 STPM Chapter 1 Functions 9. Express as the sum or difference of two trigonometric ratios (a) 2 sin 2x cos x (b) 2 cos 3θ cos 5θ (c) 6 sin 3x sin 2x (d) sin 2θ cos 4θ (e) 2 cos (A + B) cos (A – B) (f) sin (θ + 75°) sin (θ + 15°) 10. Express as a product (a) sin x + sin 5x (b) cos 3A + cos 7A (c) sin 3x – sin 5x (d) cos A – cos 5A (e) sin 30° + sin 60° (f) cos 70° + cos 50° (g) sin (x + y) – sin (x – y) (h) cos 3A + sin (A – 90°) 11. Simplify (a) cos (x – 60°) + cos (x + 60°) (b) 3 cos θ – sin (θ + 60°) – sin (θ + 120°) 12. Simplify each of the following expressions. (a) cos 7θ + cos 3θ sin 7θ + sin 3θ (b) cos 5θ – cos θ sin 5θ – sin θ (c) cos x – cos 2x + cos 3x sin x – sin 2x + sin 3x (d) cos x – 2 cos 3x + cos 5x cos x + 2 cos 3x + cos 5x 13. Prove each of the following identities. (a) sin 2x + sin 2y sin 2x – sin 2y ≡ tan (x + y) tan (x – y) (b) cos 2x – cos 2y cos 2x + cos 2y ≡ tan (x + y) tan (y – x) (c) sin θ sin 2θ + sin 3θ sin 6θ sin θ cos 2θ + sin 3θ cos 6θ ≡ tan 5θ (d) sin x + sin 2x + sin 3x ≡ sin 2x (1 + 2 cos x) (e) cos 2θ + cos 4θ + cos 6θ + cos 12θ ≡ 4 cos 3θ cos 4θ cos 5θ Trigonometric equations Consider the equation cos θ = – 3 2 . Since the value of cos θ is negative, the angle θ must lie either in the second or third quadrant. Let α be the acute angle such that cos a = 3 2 . Then, α = 30°. Hence, the possible solutions for θ are 150°, 210° or –150°, –210°. For θ 360°, the possible values of θ are θ = 360° + 150° = 510° or θ = 360° + 210° = 570°. Since a trigonometric function is periodic, therefore a trigonometric equation will have an infinite number of solutions. Usually, a range of values is given to determine the solutions which satisfy a given trigonometric equation. Hence, for the equation cos θ = – 3 2 , the solutions are θ = 150° or 210°, for 0° θ 360°, or θ = –150° or 150°, for –180° θ 180°. α = 30° α = 30° θ y x Figure 1.31
75 1 Mathematics Term 1 STPM Chapter 1 Functions Example 60 Given that tan θ = 0.5744, find all the values of θ in the range 0° θ 360°, giving your answers correct to the nearest 0.1°. Solution: Given that tan θ = 0.5744. Since tan θ is positive, θ must lie in the first or the third quadrant. For tan a = 0.5744, a = 29.9° Hence, θ = 29.9° or 209.9°. If the trigonometric equations are in terms of fractions or multiples of θ, for example, sin 1 2 θ or cos 2θ, then the number of solutions in a given range is decreased or increased according to the fraction or multiple, as shown in the following examples. Example 61 Find the values of θ in the range 0° θ 360° for each of the following equations, giving your answers correct to the nearest 0.1°. (a) sin 1 2 θ = 0.5870 (b) cos 1 2 θ = –0.7384 Solution: (a) The range is 0° θ 360°. Hence, 0° 1 2 θ 180°. This means that for the equation sin 1 2 θ = 0.5870, we need to solve for θ such that 0° 1 2 θ 180°. Since sin 1 2 θ is positive, the value of 1 2 θ must be either in the first or the second quadrant. For sin a = 0.5870 a = 35.94° Hence, 1 2 θ = 35.94°, 144.06° i.e. θ = 71.88°, 288.12° = 71.9°, 288.1° (to nearest 0.1°) (b) For cos 1 2 θ = –0.7384 (negative), the value of 1 2 θ must be either in the second or third quadrant. For cos a = 0.7384 a = 42.4° Hence, 1 2 θ = 180° – 42.4°, 180° + 42.4° = 137.6°, 222.4° ∴ θ = 275.2° since 0° θ 360°. y x = 29.9° θ α α = 35.94° θ y x 1 – 2 α= 42.4° α θ y x 1 – 2
76 1 Mathematics Term 1 STPM Chapter 1 Functions Example 62 Solve the equation cos 2θ = 0.4756 for 0° θ 360°, giving your answers correct to the nearest 0.1°. Solution: The range is 0° θ 360°, hence, 0° 2θ 720°. This means that for the equation cos 2θ = 0.4756, we need to solve for θ such that 0° 2θ 720°. Since cos 2θ is positive, the value of 2θ must be either in the first or the fourth quadrant. For cos a = 0.4756, a = 61.6° Hence, 2θ = 61.6°, 298.4°, 421.6°, 658.4° i.e. θ = 30.8°, 149.2°, 210.8°, 329.2° Example 63 Solve the equation 2 cos2 θ – 3 cos θ + 1 = 0 for the values of θ in the range –π θ π. Solution: Given that 2 cos2 θ – 3 cos θ + 1 = 0 (2 cos θ – 1)(cos θ – 1) = 0 ∴ cos θ = 1 2 or cos θ = 1 For cos θ = 1 2 , θ = – π 3 or π 3 For cos θ = 1, θ = 0 Hence, the required solutions are θ = – π 3 , 0 and π 3 . Notice that Example 63 involves a quadratic equation in cos θ. Sometimes a quadratic equation involving more than one trigonometric ratio may be given. In this case, the basic identities and multiple angle formulae can be used to express it in terms of one trigonometric ratio only. Example 64 Solve the equation 1 – 2 sin θ – 4 cos 2θ = 0 for values of θ in the range 0 θ 2π in terms of π. Solution: Given that 1 – 2 sin θ – 4 cos 2θ = 0. Substituting cos 2θ ≡ 1 – 2 sin2 θ, we have 1 – 2 sin θ – 4(1 – 2 sin2 θ) = 0 1 – 2 sin θ – 4 + 8 sin2 θ = 0 8 sin2 θ – 2 sin θ – 3 = 0 (4 sin θ – 3)(2 sin θ + 1) = 0 ∴ 4 sin θ – 3 = 0 or 2 sin θ + 1 = 0 sin θ = 3 4 or sin θ = – 1 2 sin θ = 0.75 or sin θ = –0.5 For sin θ = 0.75, θ = 0.848 or 2.294 = 0.270π or 0.730π For sin θ = –0.5, θ = 7 6 π or 11 6 π Hence, the required solutions are θ = 0.270π, 0.730π, 7 6 π and 11 6 π. y x = 61.6° 2 θ α
77 1 Mathematics Term 1 STPM Chapter 1 Functions Example 65 Solve the equation 4 sin2 x = 3 sin 2x for values of x in the range 0° x 360° to the nearest 0.1°. Solution: Given that 4 sin2 x = 3 sin 2x. 4 sin2 x = 3(2 sin x cos x) 2 sin2 x = 3 sin x cos x sin x(2 sin x – 3 cos x) = 0 Hence, sin x = 0 or 2 sin x – 3 cos x = 0 i.e. sin x = 0 or tan x = 1.5 For sin x = 0, x = 0°, 180°, 360°. For tan x = 1.5, x = 56.3° or (180° + 56.3°) = 56.3° or 236.3° Hence, the solutions in the range 0° x 360° are 0°, 56.3°, 180°, 236.3° and 360°. Exercise 1.14 1. Find all values of θ between 0° and 360° such that (a) sin θ = 0.4, (b) tan θ = 1.8, (c) cos θ = –0.75, (d) sin θ = –0.72. 2. Find all values of θ between –180° and 180° such that (a) cos θ = 0.65, (b) sin θ = 0.83, (c) tan θ = –0.77, (d) cos θ = –0.368. 3. Solve each of the following equations for 0° θ 360°. (a) 4 sin θ = 3 cos θ (b) sin θ + cos θ = 0 (c) 4 tan θ = 3 sec θ (d) 2 tan θ cosec θ = 3 4. Solve each of the following equations for 0° x 360°. (a) 4 sin2 x = 3 (b) 2 cos2 x = cos x (c) 5 sin x cos x = sin x (d) sin2 x – sin x – 2 = 0 (e) 2 tan2 x = sec x tan x (f) 3 cos2 x = 2 sin x cos x 5. Solve each of the following equations for –π x π, in terms of π. (a) 2 tan x – 1 = cot x (b) 4 sin x – 7 = 2 cosec x (c) 3 – sec x = 2 cos x (d) 2 sin x tan x – 5 cos x = 3 sin x 6. Solve each of the following equations for values of θ in the range 0° θ 360°. (a) 8 sin2 θ + 2 cos θ – 5 = 0 (b) 2 sin θ = 3 cot θ (c) tan2 θ = sec θ + 5 (d) 7 sin2 θ + cos2 θ = 5 sin θ (e) 4 cot2 θ – 2 cot θ = 3 csc2 θ (f) 4 sec2 θ – 3 tan2 θ = 5 tan θ 7. Solve each of the following equations for values of θ in the range 0 x 2π, in terms of π. (a) cos 2x = 5 cos x + 2 (b) 2 sin x – 1 = 3 cos 2x (c) 4 sin2 x = tan2 x + 1 (d) 7 tan 2x = 4 sin x Do not cancel sin x from both sides since sin x = 0 is possible. tan x . 0 ⇒ x is in the first or the third quadrant.
78 1 Mathematics Term 1 STPM Chapter 1 Functions Equations in the form a cos θ + b sin θ = c The equation such as 3 cos q + sin q = 1 can be solved if the expression 3 cos q + sin q is expressed in the form r cos (q – a), where r ( 0) and a (0° a 90°) are values to be determined. Let r cos (q – a) ≡ a cos q + b sin q Then, r (cos q cos a + sin q sin a) ≡ a cos q + b sin q i.e. (r cos a) cos q + (r sin a) sin q ≡ a cos q + b sin q Equating the coefficient of cos q: r cos a = a …………… Equating the coefficient of sin q : r sin a = b ……………b Squaring and b and adding : r 2 cos2 a + r 2 sin2 a = a2 + b2 r 2 (cos2 a + sin2 a) = a2 + b2 r 2 = a2 + b2 r = a2 + b2 b ÷ : r sin a r cos a = b a tan a = b a Hence, the expression a cos q + b sin q can be expressed in the form r cos (q – a), where r = a2 + b2 and tan a = b a . Example 66 By expressing 3 cos q + sin q in the form r cos (q – a), where r and α are values to be found, solve the equation 3 cos q + sin q = 1 for 0° q 360°. Solution: Let r cos (q – a) ≡ 3 cos q + sin q ∴ r cos q cos a + r sin q sin a ≡ 3 cos q + sin q (r cos α) cos q + (r sin a) sin q ≡ 3 cos q + sin q Equating the coefficient of cos q : r cos a = 3 ……… Equating the coefficient of sin q : r sin a = 1 ………b Squaring and b and adding : r 2 cos2 a + r 2 sin2 a = 9 + 1 r 2 (cos2 a + sin2 a) = 10 r 2 = 10 r = 10 b ÷ : tan a = 1 3 Since a is acute. a = 18°26' Hence, the equation 3 cos q + sin q = 1 can be written as 10 cos (q – 18°26') = 1 cos (q – 18°26') = 1 10 = 0.3162 q – 18°26' = 71°34', 288°26' q = 90°, 306°52' Hence, the values of q that satisfy the equation 3 cos q + sin q = 1 for 0° q 360° are 90° and 306°52'. Positive value to be taken.
79 1 Mathematics Term 1 STPM Chapter 1 Functions In the above example, we used the expression r cos (q – a). In other cases, the expressions r cos (q + a), r sin (q + a) and r sin (q – a) are also used. The following illustrate the different substitutions used: a cos q + b sin q ≡ r cos (q – a) a cos q – b sin q ≡ r cos (q + a) a sin q + b cos q ≡ r sin (q + a) a sin q – b cos q ≡ r sin (q – a) By choosing the appropriate substitution, the values of r ( 0) and α (0° α 90° or 0 α π 2 ) may be determined. Example 67 Express 5 sin q – 8 cos q in the form r sin (q – a), where r 0 and 0 a π 2 . State the maximum and minimum values of 5 sin q – 8 cos q. Hence, solve the equation 5 sin q – 8 cos q = 6 for 0 q 2π in terms of π. Solution: Let r sin (q – a) ≡ 5 sin q – 8 cos q ∴ r sin q cos a – r cos q sin a ≡ 5 sin q – 8 cos q (r cos a) sin q – (r sin a) cos q ≡ 5 sin q – 8 cos q Equating coefficient of sin q : r cos a = 5 …………… Equating coefficient of cos q : r sin a = 8 ……………b 2 + b2 : r 2 (cos2 a + sin2 a) = 25 + 64 r 2 = 89 r = 89 b ÷ : tan a = 8 5 = 1.6 a = 0.322π Hence, 5 sin q – 8 cos q = 89 sin (q – 0.322π) Maximum value of sin (q – 0.322π) is 1. Minimum value of sin (q – 0.322π) is –1. Hence, the maximum value of 5 sin q – 8 cos q is 89 , and the minimum value of 5 sin q – 8 cos q is – 89 . For the equation 5 sin q – 8 cos q = 6, we write 89 sin (q – 0.322π) = 6 sin (q – 0.322π) = 6 89 = 0.6360 ∴ q – 0.322π = 0.219π or 0.781π q = 0.219π + 0.322π or 0.781π + 0.322π = 0.541π or 1.103π Hence, the solutions for 5 sin q – 8 cos q = 6 are 0.541π and 1.103π.
80 1 Mathematics Term 1 STPM Chapter 1 Functions Exercise 1.15 1. By expressing cos 2θ and sin 2θ in terms of tan θ, solve the following equations in the range 0° θ 360°. (a) cos 2θ – 2 sin 2θ = 2 (b) 5 cos 2θ + 2 sin 2θ = 3 2. Express each of the following functions in the form r cos (θ ± a), where r 0 and 0° a 90°. Hence, find the maximum and minimum values of the function, stating the values of θ, between 0° and 360°, where the maximum and minimum values occur. (a) cos θ – sin θ (b) 3 cos θ – sin θ (c) 5 cos θ + 12 sin θ (d) 2 cos θ + sin θ 3. Express each of the following functions in the form r cos (θ ± a) or r sin (θ ± a), where r 0 and 0 a π 2 . (a) 5 sin θ – 7 cos θ (b) 6 cos θ + 3 sin θ (c) 10 cos θ – 5 sin θ (d) 7 sin θ + 4 cos θ 4. By using the expressions in the form r cos (θ ± a), solve each of the following equations for θ in the range 0° θ 360°. (a) 2 cos θ + 5 sin θ = 4 (b) 4 cos θ – 3 sin θ = 1 (c) 3 cos θ + 7 sin θ + 3 = 0 (d) 15 cos θ – 8 sin θ + 10 = 0 5. Solve the equations in question 4 by expressing in the form r cos (2θ + a). 6. Solve the following equations for 0° x 360°. (a) 3 cos x – 5 sin x = 2 (b) 4 cos x – 6 sin x = 5 7. Show that the expression 8 cos2 x – 6 sin x cos x + 2 may be expressed in the form r cos (2x + a) + s, and determine the values of the constants r, s and a. Hence, or otherwise, find the greatest and the least values of the expression as x varies. Trigonometric inequalities For inequalities involving trigonometric functions, we need to consider the signs of the functions when we multiply or divide using that trigonometric function. This is because the sign of a trigonometric function varies as the angle varies from quadrant to quadrant. Example 68 Find the set of values of x which satisfy the inequality sin x , 3 cos x, for 0 x 2π. Solution: sin x , 3 cos x , 0 x 2π. (i) When 0 x 1 2 π, cos x 0, tan x 0. ∴ sin x cos x 3 i.e. tan x 3 ⇒ 0 x 1 3 π tan x < _π 3 _π 2 3 0
81 1 Mathematics Term 1 STPM Chapter 1 Functions (ii) When 1 2 π x π, cos x 0, tan x 0. ∴ sin x cos x 3 i.e. tan x 3 ⇒ no values of x in 1 2 π x π. (iii) When π x 3 2 π, cos x 0, tan x 0. ∴ sin x cos x 3 i.e. tan x 3 ⇒ 4 3 π x 3 2 π (iv) When 3 2 π x 2π, cos x 0, tan x 0. ∴ sin x cos x 3 i.e. tan x 3 ⇒ 3 2 π x 2π Combining the above results, the set of values of x satisfying sin x 3 cos x is {x : 0 x 1 3 π and 4 3 π x 2π}. Example 69 Solve the trigonometric inequality cos 2x 3 sin x + 2 for –π x π. Solution: cos 2x 3 sin x + 2 Using cos 2x ≡ 1 – 2 sin2 x, we have 1 – 2 sin2 x 3 sin x + 2 2 sin2 x + 3 sin x + 1 0 (2 sin x + 1)(sin x + 1) 0 ∴ 2 sin x + 1 0 and sin x + 1 0 or 2 sin x + 1 0 and sin x + 1 0 i.e. sin x – 1 2 and sin x –1 or sin x – 1 2 and sin x –1 This is not possible. ∴ –1 sin x – 1 2 . When x = – 1 2 , x = – π 6 or – 5π 6 . When x = –1, x = – π 2 . Hence, – 5π 6 x – π 6 . The solution to the inequality cos 2x 3 sin x + 2 is 5x : – 5π 6 x – π 6 6. tan x < 0 _π π 2 3 _4 π3 _3 π2 π tan x > 2 π π tan x < _3 2 3
82 1 Mathematics Term 1 STPM Chapter 1 Functions Exercise 1.16 Find the values of x, between –π and π, satisfying the following inequalities. 1. sin x cos x 2. tan x cos x 3. 2 sin x 5 cos x 4. | sin 2x | 1 2 5. | cos 2x | 3 4 6. sin 2x 2 sin x 7. 2 cos 2x 3 cos x – 1 8. | cos x + sin x | 1 Summary 1. A function f from a set A to a set B, written f : A → B, is a rule that assigns to each element x in A exactly one element y in B. The set A is called the domain of f. The range of f, a subset of B, is the set of values of f(x), where y = f(x). 2. If f and g are two functions such that f : x ↦ y and g : y ↦ z for all x, y, z R, then the function that maps x directly to z is called the composite function g ° f, i.e. g ° f : x ↦ z or g ° f(x) = z. 3. If f is a one-to-one function for a given domain and a given range, then the inverse function of f exists and is represented by f–1, i.e. if f : x ↦ y then f–1 : y ↦ x. 4. A polynomial of degree n is P(x) a0 xn + a1 xn – 1 + ……… + ar xn – r + ……… + an – 1 x + a0 with ar R, a0 ≠ 0 and n Z+ . 5. The remainder theorem If the polynomial P(x) is divided by (ax + b), the remainder is P1– b a 2. 6. The factor theorem For a polynomial P(x), (x – a) is a factor of P(x) if and only if P(a) = 0. 7. For any quadratic equation ax2 + bx + c = 0, the roots are given by x = –b ± b2 – 4ac 2a . (a) If b2 – 4ac > 0, the roots are real and distinct. (b) If b2 – 4ac = 0, the roots are real and repeated. (c) If b2 – 4ac < 0, the roots are complex.
83 1 Mathematics Term 1 STPM Chapter 1 Functions 8. Partial fractions Rule 1: For each linear factor (ax + b) in the denominator of a proper rational function, there exists a partial fraction in the form A ax + b . Rule 2: For each quadratic factor ax2 + bx + c in the denominator of a proper rational function, there exists a partial fraction in the form Ax + B ax2 + bx + c . Rule 3: For each repeated linear factor such as (ax + b) 2 in the denominator, there exist two partial fractions in the form A ax + b + B (ax + b) 2 . Rule 4: If f(x) and g(x) are polynomial functions of degrees m and n respectively, where m > n, then the improper rational function f(x) g(x) can be expressed as the sum of a polynomial or a constant plus a proper rational function. This proper rational function can be expressed in partial fractions. 9. Exponents, surds and logarithms For all m, n Q and a R+ , (a) am × an = am + n (b) am ÷ an = am – n (c) (am) n = amn (d) a0 = 1 (e) 1 an = a–n . For all a, b R, (a) ab = a · b (b) a b = a b , b ≠ 0. For all a, x R+ , (a) x = an ⇔ n = loga x (b) loga a = 1 (c) loga 1 = 0 For all x, y, a, p R+ , (a) loga xy = loga x + loga y (b) loga x y = loga x – loga y (c) loga x p = p loga x. For all a, b, c R+ , (a) loga c = logb c logb a (b) loga b = 1 logb a
84 1 Mathematics Term 1 STPM Chapter 1 Functions 10. Trigonometric functions For any angle q, (a) cos2 q + sin2 q = 1 (b) 1 + tan2 q = sec2 q (c) cot2 q + 1 = csc2 q For any angles A and B. (a) sin (A + B) = sin A cos B + cos A sin B (b) sin (A – B) = sin A cos B – cos A sin B (c) cos (A + B) = cos A cos B – sin A sin B (d) cos (A – B) = cos A cos B + sin A sin B (e) tan (A + B) = tan A + tan B 1 – tan A tan B (f) tan (A – B) = tan A – tan B 1 + tan A tan B (g) sin A + sin B = 2 sin 1 2 (A + B) cos 1 2 (A – B) (h) sin A – sin B = 2 cos 1 2 (A + B) sin 1 2 (A – B) (i) cos A + cos B = 2 cos 1 2 (A + B) cos 1 2 (A – B) (j) cos A – cos B = –2 sin 1 2 (A + B) sin 1 2 (A – B) STPM PRACTICE 1 1. The function f is defined by f : x ↦ 4x3 + 3, x R. Find the corresponding definition for f–1. State the relation between the graph of f and the graph of f –1. 2. The functions f and g are defined by f : x ↦ ln x, x R+ , g : x ↦ 1 x , x R+ . Obtain the definition for f ° g and f–1 and state a relation between the graphs of (a) f and f ° g, (b) f and f–1 3. (a) The function f is defined by 2x2 + 4x, –3 x , –1, f(x) = 2x3 , –1 x , 1, 4 – 2x, 1 x 3. Sketch the graph of f and state its range. (b) The function g with domain {x : x –1} is defined by g : x ↦ x2 + 2x + 5 Find the inverse function of g, i.e. g–1.
85 1 Mathematics Term 1 STPM Chapter 1 Functions 4. The sets A and B are given by A = {x : 0 x 1} B = {x : 1 x 2} The functions f and g are defined by f : x ↦ x2 – 2x + 2 g : x ↦ x + 2 x + 1 , where f : A → B, g : B → C with C as the range of the function g. (a) Find the inverse function of f. (b) Find the range of the composite function g ° f : A → C. (c) Determine if the function g ° f has an inverse. 5. The function f is defined by f(x) = sin x, – π 2 x π 2 . Explain briefly why f has an inverse function f–1. State the domain and range of f –1. Find f–1 1 1 2 2. Given that g(x) = cos x for x R, find (g º f–1)1 3 2 2 and (f–1 ° g)1 π 3 2. 6. The function f and the composite function g ° f are defined as follows: (a) f : x ↦ x + 1 and g ° f : x ↦ x2 + 2x + 2, (b) f : x ↦ sin x and g ° f : x ↦ 1 sin2 x + 1 . Find the function g, in a similar form, in each case. 7. The function g and the composite function g ° f are defined as follows: (a) g : x ↦ ln x and g ° f : x ↦ 2 ln (x + 1), (b) g : x ↦ ex and g ° f : x ↦ ex2 – 1. Find the function f, in a similar form, in each case. 8. The functions f, g and h are defined as follows: f : x ↦ sin x, x R, g : x ↦ x + 1 – x , 0 < x < 1, h : x ↦ ln x, x . 0. State, in each case, whether the function is one-to-one and give the range of each function. For any function that is not one-to-one, give two distinct values of x which have the same image. 9. Find the inverse of each of the following functions. (a) f(x) = x + 1 (b) g(x) = x3 – 3 (c) h(x) = 2x (d) k(x) = log10 x In each case, sketch the graph of the function and its inverse. 10. Functions f and h are defined by f : x ↦ ex + 1 where x ∈ R and h : x ↦ x2 – 4 where x > 0. (a) Sketch, on separate diagram, the graphs f and h. (b) (i) Explain why f–1 and h–1 exist. (ii) Determine f–1 and h–1 and state their domains. (c) Find the composite function f ° h and state its domain and range. (d) Given that g(x) = (ex + 1)2 – 4. Express g(x) as a composition of functions which involves f and h.
86 1 Mathematics Term 1 STPM Chapter 1 Functions 11. The diagram below shows the graph of y = f(x). Sketch the graph of (a) y = 1 f(x) , (b) y2 = f(x), (c) y – 1 = f(x), (d) y = –f(x), (e) y = f(x + 2) 12. Given that y = x2 + 1 x2 + x + 1 , (a) state the value of the limit of y (i) as x → ∞, (ii) as x → –∞ (b) show that y is finite for all x R. (c) determine the set of values of y for all x R. (d) determine the value of x when the function is stationary. (e) sketch its graph. 13. Sketch, on separate diagrams, the graph of each of the following, showing clearly any turning points and asymptotes parallel to the coordinate axes. (a) y = (x – 1)(x – 3) (b) y = 1 (x – 1)(x – 3) , where x ≠ 1, x ≠ 3 14. Sketch, on separate diagrams, the graph of each of the following. (a) y = 1 x2 – a2 (b) y = 1 x2 – a2 (c) y2 = 1 x2 – a2 where a is a positive integer. State the equation of any asymptotes and the coordinates of any stationary points. 15. The polynomial p(x) = ax4 + bx3 – 6x2 + x – 26, where a and b are constants, leaves a remainder of 8 when divided by x – 2 and a remainder of –12 when divided by x + 2. (a) Determine the values of a and b. (b) Express the polynomial p(x) in the form (x2 – 4)f(x) + g(x), where f(x) is quadratic and g(x) is linear. (c) Express f(x) in a completed square form a(x + b) 2 + c. (i) Deduce that f(x) is always positive for all real values of x. (ii) Deduce the minimum value of f(x) and the corresponding value of x. (d) Find the set of values of x for which p(x) . 5x – 2. 16. Given that x2 – 1 is a factor of the polynomial f(x) = 3x4 + ax3 + bx2 – 7x – 4, find the values of a and b. Hence, factorise f(x). Sketch the graph of y = f(x) and find the set of values of x such that f(x) 0. Sketch also the graph of y = |f(x)|. 17. The function g is defined by 2ex – 1, x , 0, g(x) = 2 – (1 – x) 2 , x . 0, (a) Sketch the graph of g. (b) State the range of g. (c) State whether g is a one-to-one function or not. Give a reason for your answer. y x 0 1 2 4 –1 y = f(x)
87 1 Mathematics Term 1 STPM Chapter 1 Functions 18. Sketch the graphs of y = 1 |x| and y = |2x – 1| on the same diagram. Obtain the set of values of x such that |2x – 1| 1 |x| . 19. Solve the equation 4|x| = |x – 1|. On the same diagram, sketch the graphs of y = 4|x| and y = |x – 1|. Hence, solve the inequality 4|x| |x – 1|. 20. Find the values of A, B and C such that 2x3 + x2 – 5x + 7 ≡ (x + 2)(Ax2 + Bx + C) + 5 21. Find F(x) if 6x4 + 11x3 + 8x + 5 ≡ (2x + 1)F(x). 22. Express 9x (2x + 1)2 (1 – x) in partial fractions. 23. Express x – 2 (x2 + 1)(x – 1)2 in partial fractions. 24. Express 7x + 4 (x – 3)(x + 2)2 in partial fractions. 25. If f(x) = x6 – 5x4 – 10x2 + p, find the value of p such that (x – 1) is a factor of f(x). With this value of p, find another factor of f(x) in the form (x + a), where a is a constant. 26. If f(x) = ax2 + bx + c has remainder 1, 25 and 1 when divided by (x – 1), (x + 1) and (x – 2) respectively, show that f(x) is a perfect square. 27. Given that (x – 3) and (2x + 1) are factors of f(x) ≡ ax4 + bx3 + 13x2 + 30x + 9, find the values of a and b. With these values of a and b, show that f(x) 0 for all x R. 28. The polynomial P(x) has remainder 1 when divided by (x – 1), and remainder 3 when divided by (x + 1). Find the remainder when P(x) is divided by (x2 – 1). 29. Find the value of k such that (2x – 1) is a factor of f(x) ≡ 4x4 + 8x3 + kx2 – 11x + 6 Assuming this value of k, factorise f(x) completely. 30. When the polynomial P(x) is divided by (x2 – 1), its remainder is (ax + b), where a and b are constants. Given that (x + 1) is a factor of P(x), and that the remainder is 4 when P(x) is divided by (x – 1), find the values of a and b. 31. If the function ax2 + bx + c has a minimum value of –5 when x = –1, and 0 when x = –2, find the values of a, b and c. Find the range of values of x such that ax2 + bx + c 75. 32. Show that –x2 + px + q 0 for all values of x R if –q 1 4 p2 . Show also, that (24 – x)(x – 8) – k 0 for all values of x R if k 64. Hence, show that (6 + y)(4 – y)(y + 4)(y – 2) – 65 0 for all values of y R.
88 1 Mathematics Term 1 STPM Chapter 1 Functions 33. If x = 2 is a root of the equation a2 x2 + 2(2a – 5)x + 8 = 0, find the possible values of a. Find the corresponding roots with these values of a. 34. (a) Show that the equation x2 + (3 – 2)x + ( – 1) = 0 has real roots for all values of R. (b) Show that x2 – x + 1 has the same sign for all values of x. 35. Show that the roots of the equation px2 + (p + q)x + q = 0 are real for all values of p and q. 36. Find the value of p if x2 + (p + 3)x + 2p + 3 is an expression in the form of a perfect square. 37. The polynomial 2x4 – ax3 + 19x2 – 20x + 12 has a factor in the form (x – k) 2 , where k Z+ . Find the values of k and a and show that the polynomial is non-negative for all x R. 38. (a) Find the value of the constant k such that (x + 1) is a factor of 2x3 + 7x2 + kx – 3. With this value of k, solve the equation 2x3 + 7x2 + kx – 3 = 0 (b) Find the values of the constants A, B, C and D such that x4 + Ax3 + 5x2 + 3 ≡ (x2 + 4)(x2 – x + B) + Cx + D. 39. If f(x) = x4 + 2x3 + 5x2 – 16x – 20, show that f(x) can be expressed in the form (x2 + x + a) 2 – 4(x + b) 2 , where a and b are constants to be determined. Hence, find both the real roots of the equation f(x) = 0. Find also the set of values of x such that f(x) 0. 40. Determine the condition to be satisfied by k so that the expression 2x2 + 6x + 1 + k(x2 + 2) is positive for all x R. 41. Find the set of values of x which satisfies each of the following inequalities. (a) x(x + 2) x – 3 x + 1 (b) x2 + 4x – 3 x2 + 1 x 42. Without using tables or a calculator, show that (a) 3 13 4, and deduce that 0 13 – 3 1, (b) ( 13 + 3)4 + ( 13 – 3)4 = 1904, (c) 1903 ( 13 + 3)4 1904 43. Express ( p + q r) 2 in the form a + b c. Without evaluating the square root or using tables or a calculator, show that 10 + 2 2 6. 44. If p and q are positive numbers, prove that (1 – p)(1 – q) 1 – p – q. If p, q, r R+ , with at least one of them less than unity, prove that (1 – p)(1 – q)(1 – r) 1 – p – q – r.
89 1 Mathematics Term 1 STPM Chapter 1 Functions 45. Solve each of the following inequalities. (a) 2 |x + 2| |4 – x| (b) |3x + 1| – 4|x + 1| 0 (c) |x2 – 3x – 2| 2 (d) 6 |x| + 1 |x| 46. Sketch on the same coordinate axes, the graphs of y = |2 – 3x| and y = 2 – x2 . Hence solve the inequality |2 – 3x| > 2 – x2 . 47. The function f and h are defined by f(x) = 1 + e4x , x ∈ R and h(x) = a ln (x – 1), x . 1, where a is a constant. The function h is the inverse of function f. (a) Determine the value of a. (b) Sketch the graph of f and h on the same axes. 48. Solve the equation 4x – 3 · 52 – 3x = 20 giving your answer correct to 3 significant figures. 49. Solve the equation 32x + 1 = 10(3x ) – 3 50. If 3x + 1 · 4x + 2 = 5x + 3, find the value of x, correct to 3 significant figures. 51. Find the exact values of x such that 9(32x ) – 28(3x ) + 3 = 0 52. Show that the equation 22x + 64(2–x ) = 32 has a real root apart from the root x = 2. Find this root, correct to 3 significant figures. 53. If a2x – 1 = b1 – 3y and a3x – 1 = b2y – 2, show that 13xy – 7x – 5y + 3 = 0. 54. Given that 2y = ax + a–x, where a 1 and x 0, show that ax = y + (y2 – 1). Similarly, if 2z = a3x + a–3x , show that z = 4y3 – 3y. 55. Solve the following equations. (a) x + 5 = 7 – x (b) x – 2 + x + 3 = 5 (c) 2x + 11 + x + 2 = 5x + 17 (d) 5 x + 1 – 3 x + 1 + 2 = 0 56. Find the value of x, in surd form, if log3 (x2 ) + log3 x = log9 27. 57. Find the set of x such that (log8 2x)(3 – log2 x) = 1. 58. Prove that logy x = 1 logx y , x, y R+ . Find the possible values of x if 2 (log9 x + logx 9) = 5.
90 1 Mathematics Term 1 STPM Chapter 1 Functions 59. (a) If a, b and c are any positive numbers not equal to 1, show that logb a · logc b · loga c = 1 (b) If a2 + b2 = 7ab, show that 2 log10 1 a + b 3 2 = log10 a + log10 b 60. If log10 (n + 1) – log10 n log10 1.01, find the smallest positive integral value for n. 61. If 1 + loga (7x – 3a) = 2 loga x + loga 2, find, in terms of a, the possible values of x. 62. Find the smallest integral value of n such that (0.9)n 0.0001. 63. (a) If loga N = x and logb N = y, show that logab N = xy x + y . (b) If a = log10 1 10 9 2 , b = log10 1 25 242 and c = log10 1 81 802 , show that log10 2 = 7a – 2b + 3c. 64. Sketch the graph of (a) y = x e–x (b) y = e–x x (c) y = e–x sin x, –π x π 65. Sketch the graphs of y = x2 e –x and y = x3 e–x on separate diagrams, showing any stationary points. (Assume that for both curves, y → 0 as x → ∞). 66. On the same axes, draw the graphs of y = 5 sin x and y = 1 – 2 sin 2x for 0° x 180°. From the graphs, obtain the values of x , in the domain, that satisfy the equation 2 sin 2x + 5 sin x = 1. Without further graphical work, deduce two other values of x, outside the domain, which satisfy the equation. 67. The expression 12 cos x – 5 sin x is denoted by f(x). Find the value of r and , where r is positive and is acute, such that f(x) = r cos (x + ) for all values of x. State the value of in degrees, to the nearest 0.1°. (a) Obtain the largest and smallest values of f(x), giving your answers to the nearest 0.1°, for 0° x 360°. (b) Sketch the graph of y = f(x) for 0° x 360°. (c) Find the values of x, where 0° x 360°, satisfying 12 cos x – 5 sin x + 4 = 0, giving your answers correct to the nearest 0.1°. 68. Sketch the graph of y = cos 2x in the range 0 < x < 2π. Hence solve the inequality |cos 2x| . 1 2 where 0 < x < 2π. 69. Sketch the graph of y = tan x for values of x between 0 and 2π. Use your sketch graph and an additonal graph of a straight line to solve each of the following problems. (a) If a 0 and b 0, find the number of roots of the equation tan x = ax + b for 0 x 2π. (b) If m can take all real values, find the set of values of m such that the equation tan x = mx + π has only three roots between 0 and 2π.
91 1 Mathematics Term 1 STPM Chapter 1 Functions 70. If tan A = 2 3 and tan B = 1 2 , find the value of cos 2A + cos 2B sin 2A – sin 2B . 71. Express 12 cos x + 9 sin x in the form r cos (x – a) where r > 0 and 0 , a , π 2 . (a) State the minimum and maximum values of 12 cos x + 9 sin x and find the corresponding values of x in the interval 0 < x < 2π. (b) Solve the equation 12 cos x + 9 sin x = 0 for 0 < x < 2π. (c) Sketch the graph of y = 12 cos x + 9 sin x for 0 < x < 2π, and determine the range of values of x in this interval satisfying the inequality 0 < 12 cos x + 9 sin x < 10. 72. Prove that sec x + tan x = tan 1 π 4 + x 2 2, and obtain a similar expression for sec x – tan x. Find the values of tan 3 8 π and tan 1 12 π, expressing them in surd form. 73. (a) Prove that sin4 q + cos4 q = 1 4 (3 + cos 4q). Hence, find the largest and the smallest possible values of sin4 q + cos4 q. (b) Using the identity csc2 q – cot2 q ≡ 1, find the exact value of tan q if csc q + cot q = 4. 74. Find the values of x in the range 0° x 360° for each of the following equations. (a) 3 sin 2x = 2 tan x (b) 4 cos x – 6 sin x = 5 75. Find all the values of x, between 0° and 360° inclusive, satisfying each of the following equations. (a) 3 tan 2x + 2 tan x = 0 (b) cos 3x + cos x = cos 2x 76. The angles θ and ø are acute such that sin θ = 24 25 and cos ø = 12 13 . Find, without the use of tables or a calculator, the exact values of sin (θ + ø) and cos (θ + ø). Deduce that 1 2 π θ + ø π. 77. Find all the possible values of x, between 0° and 360° inclusive, that satisfy the equation 2 sin 2x + cos 2x = 2. 78. Express 2 sin θ + 12 cos θ in the form r sin (θ + a) where r > 0 and 0 a π 2 . (a) Find the minimum and maximum values of 2 sin θ + 12 cos θ, and the corresponding values of θ in the interval 0 < θ < 2π. (b) Sketch the graph of 2 sin θ + 12 cos θ for 0 < θ < 2π. (c) Solve the equation |2 sin θ + 12 cos θ| = 2 for 0 < θ < 2π and determine the range of values of θ in the interval [0, 2π] which satisfies the inequality |2 sin θ + 12 cos θ| 2. 79. Solve the equation 3 cos x + sin x = 1 for 0° < x < 360°.
2 CHAPTER SEQUENCES 2 AND SERIES Subtopic Learning Outcome 2.1 Sequences (a) Use an explicit formula and a recursive formula for a sequence. (b) Find the limit of a convergent sequence. 2.2 Series (a) Use the formulae for the nth term and for the sum of the first n terms of an arithmetic series and of a geometric series. (b) Identify the condition for the convergence of a geometric series, and use the formula for the sum of a convergent geometric series. (c) Use the method of differences to find the nth partial sum of a series, and deduce the sum of the series in the case when it is convergent. 2.3 Binomial expansions (a) Expand (a + b) n , where n ∈ Z+ . (b) Expand (1 + x) n , where n ∈ Q, and identify the condition |x| , 1 for the validity of this expansion. (c) Use binomial expansions in approximations. approximation – penghampiran arithmetic – aritmetik binomial expansion – kembangan binomial converge – menumpu explicit – tak tersirat geometric – geometri limit – had sequence – jujukan series – siri recursive – rekursif Bilingual Keywords
Mathematics Term 1 STPM Chapter 2 Sequences and Series 93 2 2.1 Sequences Sequences Consider the set of numbers 3, 7, 13, 21, …, 111, … This set of numbers can be written as 1 + 1 × 2, 1 + 2 × 3, 1 + 3 × 4, 1 + 4 × 5, …, 1 + 10 × 11, … The nth term, un, can be written as un = 1 + n(n + 1), n Z+ 1 1 1 For the set of numbers 1, — , — , — , …, the nth term, vn, can be written as 3 5 7 1 vn = , n Z+ 2n – 1 A set of numbers u1 , u2 , u3 , …, un which is arranged with each term un as a function f(n), n Z+ , is known as a sequence. A sequence is usually denoted by {un}. If the nth term of a sequence is given, say un = f(n), n Z+ , then we can find the successive terms of 1 1 1 1 this sequence. For example, if un = — , n Z+ , then the sequence is 1, —, —, — , … n2 4 9 16 We say that this sequence is defined by the 1 explicit formula, un = — , n Z+ . n2 There is a type of sequence where the value of each term is related to its preceding terms, with the initial terms being given. For example, if the first two terms of a sequence are u1 = 2 and u2 = 6, and un + 2 = un + 1 + un, then we can find the successive terms of the sequence, i.e. u3 = u2 + u1 = 6 + 2 = 8, u4 = u3 + u2 = 8 + 6 = 14, u5 = u4 + u3 = 14 + 8 = 22 and so on. Hence, the sequence obtained is 2, 6, 8, 14, 22, 36, … We say that this sequence can be defined by the recursive formula, un + 2 = un + 1 + un, with u1 = 2 and u2 = 6. Example 1 Write down the first five terms of the sequence with its nth term un = n n + 1 , n Z+ Solution: un= n n + 1 , n Z+ Hence, u1 = 1 1 + 1 = 1 2 u2 = 2 2 + 1 = 2 3 u3 = 3 3 + 1 = 3 4 u4 = 4 4 + 1 = 4 5 u5 = 5 5 + 1 = 5 6 Hence, the first five terms of the sequence are 1 2 , 2 3 , 3 4 , 4 5 and 5 6 . Sequences INFO
94 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Example 2 If a sequence with un + 1 = 2un , n Z+ , and u1 = 1 is given, find the first five terms of this sequence. State the value of the nth term in terms of n. Solution: un + 1 = 2un , n Z+ . u1 = 1 Hence, u2 = 2u1 = 2—1 2 u3 = 2u2 = 2—1 2 + —1 4 u4 = 2u3 = 2—1 2 + —1 4 + —1 8 u5 = 2u4 = 2—1 2 + —1 4 + —1 8 + —1 16 Hence, the first five terms of the sequence are 1, 2—1 2 , 2 —1 2 + —1 4 , 2 —1 2 + —1 4 + —1 8 and 2 —1 2 + —1 4 + —1 8 + —1 16 or 1, 2—1 2 , 2—3 4 , 2—7 8 and 2 15 —16. The nth term, un = 2 —1 2 + —1 4 + —1 8 + … + 1 –––––– 2n – 1 A sequence with terms which are repeated after a certain fixed number of terms is known as a periodic sequence. In trigonometry, the range of values for the graphs of sine and cosine functions are between –1 and 1. This range repeats itself after an interval of 2π radians. For example, the sequence with un = cos nπ 2 , n Z+ , has the terms 0, –1, 0, 1, … which repeats itself after four terms. Therefore, this sequence is a periodic sequence. Consider the sequence for un = (–1)n , n Z+ , which has the terms –1, 1, –1, 1, … This sequence oscillates finitely between –1 and 1. Therefore, this sequence is also a periodic sequence. The sequence for un = (–10)n , n Z+ , has terms –10, 100, –1000, 10 000, …, and oscillates infinitely between –∞ and ∞. Therefore, this sequence is not a periodic sequence. Convergent and divergent sequences Consider the sequence 1.1, 1.01, 1.001, 1.0001, …, 1 + (0.1)n , … The nth term of this sequence is un = 1 + 1 1 102 n When n→∞, 1 1 10 2 n → 0 and 1 + 1 1 102 n → 1. We write lim x → ∞ un = lim n → ∞ 31 + 1 1 102 n 4 = 1. We say that this sequence converges to the value 1. In general, if un is the nth term of a sequence and lim n → ∞ un exists, then this sequence is called a convergent sequence and the value of the limit of un is known as the limiting value or limit of sequence. A sequence which is not convergent is called a divergent sequence.