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100 Chapter 3 Functions and Graphs Exercises 2–B 1. The graph of y = ax + b passes through the origin and the asymptotes are x+c the straight line x = −1 and the straight line y = 2. Find the values of the constants a, b and c. √ 2. When the graph of y = kx is translated −3 units towards x-axis and 1 unit towards y-axis, a curb passing through the point (−11, 5) is obtained. Find the value of the constant k. √ 3. Find the value of the constant a such that the function y = − 4 − 2x has the minimum value of −3 in the domain a ≦ x ≦ 2. 4. Find the value of the constant k such that the inverse of the function y = x − 1 gets back to y = x − 1 . 2x + k 2x + k 5. For the function f (x) = ax + b , let the inverse of a function y = f (x) be x−3 y = g(x). When f (2) = 1, g(3) = 4, find the values of the constants a and b. 6. (1) Plot graphs of the functions y = 2x − 5 , y = 5 − x. x−1 (2) As for (1), find the coordinates of the intersection points of the two graphs. (3) Solve the following inequalities with the graphs. 2x − 5 <5−x x−1 7. Solve the following inequalities with the graphs. √ (2) 4 > x (1) x + 2 > x x−3

101 4Chapter Exponential function and logarithmic function § 1 Exponential function 1 1 Radical roots When n is an integer of 2 or greater, for a real number a, a value that becomes a after raised to the n-th power, in other words, a value of x, which can be xn = a, is called n-th root of a. These roots are called a square root, a cube root, a fourth root · · · and all together they are called radical roots. Here, we only deal with the real numbers of n-th roots. Therefore, the n-th root of a is at the x coordinates of common points where a graph of y = xn and a straight line y = a intersect. ( i ) When n is an even number. (n : even number) if a > 0, there are two intersection points. y y = xn The x coordinate of the points within a y=a √ the first quadrant is represented as n a. √ Since n a > 0, the other n-th root of a √ is − n a. If a < 0, there is no n-th root of a. √ O √ x Ex.1 From 24 = (−2)4 = 16, −na na the fourth root of 16 are 2, −2. √√ 4 16 = 2, − 4 16 = −2

102 Chapter 4 Exponential function and logarithmic function ( ii ) When n is an odd number. (n : odd number) y y = xn Regardless of the sign of a, there is only one intersection point and the x coordi- a y=a √x nate is represented as The following is √ na n −a established from the graph. √√ O n −a = − n a −a √√ √ Ex.2 3 8 = 2, 3 −8 = − 3 8 = −2 √ n is called a radical root sign and n is called an exponent of radical root. √ √ √√ 2 can simply be represented as . Regardless of n, n 0 = 0 and n 1 = 1. Q. Represent the following values using radical root signs. (1) The cube root of 15 (2) The fifth root of −5 (3) The fourth root of 2 The following equations are established with respect to the radical root sign by the law of exponents. Properties of radical roots   When a > 0, b > 0, m and n are integers of 2 or greater. √ √√ (I) ( n a)n = a ( II ) ( n a)m = an m (III) (IV) √a  √ √ √ √ na nb n ab √n a = n = nb b  Proof (II) Let x = ( √n a)m. xn = {( √n a)m}n = ( √n a)mn = {( √n a)n}m = am √ Therefore x is n-th root of am. Since x > 0, then x = n am. The others may be proved likewise. √√ √√ √ √√ Ex.3 5 9 ( 5 3)3 = 5 9 5 33 = 5 9 × 33 = 5 35 = ( 5 3)5 = 3 Q. Simplify the following expressions. √ √ (4) 4√80 √ √√ (3) 4 9 √ (1) 5 32 (2) 3 4 3 2 4 36 45 4

§ 1 Exponential function 103 1 2 Extension of exponential distribution When m and n are positive integers, the law of exponents aman = am+n, (am)n = amn, (ab)n = anbn is established, as we learned previously (p. 4) Furthermore, for the positive integers m and n, the division law of exponents am = am−n (a =\\ 0) (1) an is established, when m > n, as we learned previously (p. 19). Let s extend the definition of exponents to the integral numbers, such that the division law of exponents shall be established, when m ≦ n. If (1) is established when m = n, (2) an = an−n = a0 an Since an = 1, an a0 = 1 is defined. Next, if (1) is established when m = 0, a0 = a0−n = a−n an Since a0 = 1 , an an a−n = 1 (3) an is defined.  Extension of exponential distribution (1)  When a =\\ 0, and n is positive integer, a0 = 1, a−n = 1 an   Ex.4 50 = 1 3−2 = 1 = 1 (22)−3 = 1 = 1 = 1 32 9 (22)3 26 64

104 Chapter 4 Exponential function and logarithmic function The division law of exponents can be concluded as below. am = am−n = 1 an an−m In addition, when exponents are any integers, the law of exponents is proved such as (am)−n = 1 = 1 = a−mn = a .m×(−n) (am)n amn Ex.5 a2 × a−3 = a2+(−3) = a−1 = 1 , (a2)−3 = a2×(−3) = a−6 = 1 a a6 a5 = a5−5 = a0 = 1, a2 = a2−5 = a−3 = 1 a5 a5 a3 Exercise Evaluate the following expressions. Note that a =\\ 0 and b =\\ 0. ( )4 1 (ab−1)3 (1) 35 × 9 (2) (a−1b)2 Solution 35 × ( 1 )4 = 35 × (3−2)4 = 35 × 3−8 = 35−8 = 3−3 = (1) 9 1 = 1 33 27 (2) (ab−1)3 = a3 b−3 = a b3−(−2) −3−2 = a5b−5 = a5 // (a−1b)2 a−2 b2 b5 Q. Evaluate the following expressions. Note that a =\\ 0 and b =\\ 0. (1) ( 1 )−3 × 2−7 (2) 352 × 5−4 × ( 1 )2 4 7 (3) (a−1b2)−3 × (a−2b3)3 (2a−2b)4 (4) (5ab−2)2 Next, let’s extend the definition of exponents, such that the law of expo- nents, (ar)s = ars shall be established with respect to the rational numbers r and s, when a > 0. m is an integer and n is an integer of 2 or greater. If the law of exponents is established when r = m , s = n, n (m )n m =×n an n = a am. m Thus a n is defined as n-th root of am.

§ 1 Exponential function 105 Extension of exponential distribution (2)  When a > 0, and m is an integer and n is an integer of 2 or greater. 1√ m = √ = ( √n a)m a n = n a, an m an   Ex.6 √5 √ √1 5− 3 = 1 = (√15)3 = √1 3 25 = 2 3 , 5= 2 5 = 52, 2 55 3 52 When expanding the exponents to the rational numbers as shown above, the following law of exponents is established.  The law of exponents   When a > 0, b > 0, and p and q are rational numbers. ( I ) apaq = ap+q ( II ) ap = ap−q = 1 aq aq−p (III) (ap)q = apq (IV) (ab)p = apbp  √ × √1 = 1 × a− 2 = a 1 − 2 = a− 1 = √1 Ex.7 a 3 2 3 6 a2 a3 2 6a ( 1 )− 2 1 (× − 2 ) a− 1 √1 a 2 3 2 3 3 = a = = 3a Q. Represent the following expressions with the form of ap. Note√that a > 0. √√ (1) √a (2) √1 2 (4) 3 a (3) (a− 3 )−3 4 a a4 3 √ Q. Represent the following expressions with the form of n am. Note that a > 0. (1) a−0.4 (2) a1.25 (3) 1 (4) a × a−3.6 4.3 a−2.5 Q. Simplify the following expressions when a > 0, and p, q, r are the rational numbers. √√ √ (2) 4 a3 × a6 4 (1) a√3 a 6a (3) (ap−q)r(aq−r)p(ar−p)q When a > 0, ap is defined with respect to the irrational number p, and the law of exponents is known to be established when the exponents are any real numbers.

106 Chapter 4 Exponential function and logarithmic function 1 3 Exponential function When a is a positive constant number, not equal to 1, a function of x, which represented in y = ax is called an exponential function of base a. Let s plot a graph of y = ax with several a. x −4 −3 −2 −1 − 1 0 1 1 2 3 4 22 y = 2x 1 1 1 1 √1 1 √ 2 4 8 16 16 8 4 2 2 2 A function of y = 2x is monotonically in- y = (0.4)x y y = 3x y = 2x creasing as you see in the table above, there- fore 2x > 0. The graph passes a point (0, 1) and take the x-axis as an asymptote. The y = (0.6)x 4 same can be said for y = 3x. However, the functions of y = (0.4)x and 2 y = (0.6)x are monotonically decreasing and the graph passes a point (0, 1) and take the −2 −1 O 1 2 x x-axis as an asymptote. The properties of exponential functions in general are explained as below.  Properties of exponential functions  The exponential function y = ax (a > 0, a =\\ 1). ( I ) The domain is all real numbers and the range is all positive real numbers. Therefore, the graph always remains above the x-axis. (II) When a > 1, the graph increases monotonically and when 0 < a < 1, the graph decreases monotonically. (III) The graph passes through the point (0, 1) and (1, a). (IV) The graph takes the x-axis where a straight line y 0 as asymptotes.

§ 1 Exponential function 107 (a > 1) y = ax y = ax (0 < a < 1) y y a x 1 x 1 a O1 O1 Exercise Plot graphs of the following functions (1) y = 2x−1 (2) y = ( 1 )x 2 Solution (1) From the equation on y y = 2x y = 2x−1 4 the p. 89, the graph is the same graph 2 x as y = 2x but translated 1 parallel to- 1 awaprodisntth(e0,x-a1x)isa, nanddtaitkpeatshseesxt-harxoius gahs O 123 2 y y = 2x an asymptote. The graph is plotted as 2 shown in the figure on the right. 11 ( 1 )x (2) It can be transformed to y= y = ( 1 )x = 2−x. 2 2 2x Therefore, from the equation on the −1 O 1 p. 94, the graphs of y = ( 1 )x and 2 y = 2x are symmetric with respect to the y-axis. Thus, the graph is plotted as shown in the figure on the right. // Q. Plot graphs of the following functions. (1) y = (1.5)x (2) y = (2.5)x (3) y = (0.8)x

108 Chapter 4 Exponential function and logarithmic function Q. (2) y = −3x (3) y = −2−x (1) y = 2x 4 Exercise Solve the following equations. √ (2) 9x+1 + 3 × 3x − 2 = 0 (1) 2x−1 = 4 2 √ 15 5 Solution (1) Since 4 2 = 22 · 2 2 = 2 2 , the given equation is 2x−1 = 2 2 , Therefore, x−1= 5 ∴ x= 7. 2 2 (2) Given X = 3x, X > 0 Since 9x+1 = 9 · (32)x = 9 · (3x)2 = 9X2, Then, 9X2 + 3X − 2 = 0 Thus, (3X + 2)(3X − 1) = 0 ∴ X = − 2 , 1 3 3 Since X > 0, X= 1 3 Then, 3x = 1 = 3−1 ∴ x = −1 // 3 Q. Solve the following equations. √ √ (3) 4x − 2x = 56 (1) 32x = 27 (2) 4−x = 3 16 Exercise Solve the following inequality. 9x > 27 Solution The given inequality is 32x > 33 and y = 3x increases monoton- ically. ∴ x> 3 // 2x > 3 2 Q. Solve the following inequalities. (1) 0.5x < 2 (2) 33x+2 > 81

§ 1 Exponential function 109 Exercises 1–A 1. Evaluate the following expressions. Note that a > 0. √ (√ )4 (1) 3 8a−6 (2) 6 a−3 √√ (4) √ ÷ √ (3) a × 3 a 3a 4 a3 2. Arrange the following numbers from the largest to the smallest. (1) 0.7 1 (0.7)−2 (0.7)−3 √ (2) √ 4− 1 1 √ 4− 2 0.7 34 2 3 16 5 3. Find the values of the following equations, when √ √1x = 3. x+ (1) x + x−1 (2) 3 + x− 3 2 x2 4. Plot graphs of the following functions. (1) y = −3−x (2) y = 22−x (3) y = 4x + 1 5. Solve the following equations. (2) 4x − 3 × 2x − 4 = 0 (1) 2x+2 = 64 (3) 9x+1 − 3x − 8 = 0 6. Solve the following inequalities. (2) ( 1 )2−x > 9 (1) 4x < √1 3 2 7. Simplify the following expressions, when a > 0 b > 0. 1 11 1 (1) (a 2 + b 2 )(a 2 − b 2 ) (2) 1 − b− 1 )(a 1 + b− 1 )(a 1 + b− 1 ) 4 4 4 2 2 (a 4 (3) a−b 11 a3 −b3

110 Chapter 4 Exponential function and logarithmic function Exercises 1–B 1. Find the values of the following expressions when a2x = 3. Note that a > 0. (1) (ax − a−x)2 (2) a3x + a−3x ax + a−x 2. Arrange the following numbers from the smallest to the largest. (1) 8− 1 4− 2 √ √√√ 4 5 8 2−7 (2) 27 4 35 3 81 3. Find the value of 2x + 2−x , when 2x − 2−x = 3. 4. Solve the following inequalities and equations. (1) 16x − 5 × 4x + 4 > 0 (2) 2x+2 − 2−x + 3 = 0 (3) 4x + 2x+1 ≦ 24 2x−y+1 = 8 (4) 4x − 4y = 60 5. Solve the following inequality. Note that a > 0, a =\\ 1. a5x−3 > a2 6. Simplify the following expressions when a > 0, b > 0. (1) √ + √ √ − √ + √ (6a 6 b)( 3a 6 ab 3 b) {√ }− 3 √( )−3 3 a2b6 4 × ab−2 (2) 2 7. Prove that the following inequality is valid when a > 0. ax + ay x+y 2 ≧a 2 8. x + 3y − 2 = 0 2x + 8y Find the minimum values of 2x + 8y when x + 3y − 2 = 0.

§ 2 Logarithmic functions 111 § 2 Logarithmic functions 2 1 Logarithm As we find from the graph of an exponential function y = ax, when a is a positive number, not equal to 1, there exists a real number m which can be am = N for any positive number N . (a > 1) (0 < a < 1) y y NN 1 1 x Omx mO This m is called a logarithm of N to base a and represented as follows. m = loga N Then N is called an antilogarithm (also called an antilog) of m to base a.  Definition of Logarithm  When a > 0, a =\\ 1, for N > 0, m = loga N ⇐⇒ am = N   Ex.1 log2 8 = m ⇐⇒ 2m = 8 ∴ m = 3 where log2 8 = 3 log2 1 =m ⇐⇒ 2m = 1 ∴ m = −2 where log2 1 = −2 4 4 4 Q. Find the following values. (1) log3 27 (2) log4 1 (3) log2 1 √ (5) log0.1 10 16 (4) log10 3 10 (6) log0.5 0.125

112 Chapter 4 Exponential function and logarithmic function From the definition of logarithm, when m = loga N , am = N . Therefore, aloga N = N is established. Furthermore, since a0 = 1, a1 = a, when they are rewritten in logarithmic form, the following formulas are obtained.  Properties of Logarithms (1)  When a > 0 a =\\ 1 N > 0, loga 1 = 0, loga a = 1, aloga N = N   In addition, the following formulas are also established.  Properties of Logarithms (2)  When a > 0, a =\\ 1, M > 0, N > 0. If p is a real number, loga M N = loga M + loga N loga M = loga M − loga N N loga M p = p loga M   // Proof Let x = loga M, y = loga N . Since M = ax, N = ay, M N = axay = ax+y is established. From the definition of logarithm, loga M N = x + y = loga M + loga N Likewise, from M = ax = ax−y, N ay loga M = x − y = loga M − loga N N For a given real number p, M p = (ax)p = apx Therefore, loga M p = px = p loga M

§ 2 Logarithmic functions 113 Ex.2 log2 32 =√log2 25 = 5 log2 2 = 5√× 1 = 5 (√ ) = log3 5 1 log3 5 − log3 5 1 − log3 5 × √2 = log3 2 2 2 2 5 = log3 5 2 Note log3 2 is not a rational number. Q. Evaluate the following logarithmic expressions. (1) log2 3 − log2 3 (2) log2 1 √ 4 √ 2 12 + log2 3 2 √ (3) log2(3 + 5) + log2(3 − 5) (4) 3 log5 15 − log5 135 Q. Prove the following logarithm equations. Note that L, M, N and a are positive numbers and a =\\ 1, also n is an integer of 2 or greater. (1) loga 1 = − loga N (2) loga √ = 1 loga M N nM n (3) loga L + loga M + loga N = loga LM N Q. Simplify the following logarithmic expressions. Note that A > 0, B > 0 and C > 0. (1) 1 log10 125 + log10 3 − log10 0.3 3 5 (2) loga A + loga B + loga C (a > 0, a =\\ 1) B C A Now, let’s look into how the logarithm is converted by changing the base. Let a, b, and c be positive numbers, and a =\\ 1, c =\\ 1. When letting loga b = x, we have b = ax. Therefore, logc b = logc ax = x logc a From a =\\ 1, logc a =\\ 0. Therefore, x = logc b where loga b = logc b logc a logc a his relationship is called change of base formula.  Transformation formula of base  When a, b, and c are positive numbers, for a =\\ 1, c =\\ 1, loga b = logc b logc a

114 Chapter 4 Exponential function and logarithmic function Exercise Prove the following logarithmic equation. Note that a and b are positive numbers, not equal to 1. loga b = 1 logb a Proof loga b = logb b = 1 // logb a logb a Q. Prove the following logarithmic equation. Note that a, b, and c are positive numbers, not equal to 1. loga b · logb c · logc a = 1 Exercise Simplify (log2 3) × (log9 2). Solution (log2 3) × (log9 2) = log2 3 × log2 2 = log2 3 × 1 = 1 // log2 9 2 log2 3 2 Q. Simplify the following logarithmic expressions. (1) (log4 25) × (log5 8) (2) (log4 3) × (log9 25) × (log5 8) 2 2 Logarithmic functions When a is a positive constant, not equal to 1, the function represented in y = loga x is called a logarithmic function with base a. By switching the roles of x and y in a logarithmic function y = loga x, x = loga y where y = ax Therefore, a logarithmic function y = loga x is the inverse of an exponential function y = ax. y = loga x ⇐⇒ ay = x As we learned on p. 97, the graph of y = loga x is symmetric with respect to the graph of y = ax and the straight line y = x. The graphs are shown in the figures on the next page.

§ 2 Logarithmic functions 115 (a > 1) (0 < a < 1) y=x y y = ax y = x y = ax y a y = loga x 1 1 x 1 ax a O1 Oa y = loga x As shown in the graphs, a logarithmic function has the following properties.  Properties of logarithmic functions  A logarithmic function y = loga x (a > 0, a =\\ 1) ( I ) The domain is x > 0 and the range is all real numbers. (II) When a > 1, it is monotonically increasing. When 0 < a < 1, it is monotonically decreasing. (III) The graph passes through the point (1, 0) and the point (a, 1). (IV) he graph takes the y-axis as an asymptote.   Ex.3 From the table below, the graph of y = log2 x can be plotted as shown in the y y = log2 x figure. x 2 x1 1 1 2 4 8 1 24 42 O1 −1 y −2 −1 0 1 2 3 Since the graph of y = log 1 x is −2 y = log 1 x 2 2 y = log 1 x= log2 x = − log2 x 2 log2 1 2 the graph is symmetric with respect to the graph of y = log2 x and the x-axis.

116 Chapter 4 Exponential function and logarithmic function Q. Plot graphs of the following logarithmic functions. (1) y = log3 x (2) y = log2(−x) (3) y = log5(x − 1) Exercise Find the range of a logarithmic functiony = log2 x in the domain 0.5 < x ≦ 4. Solution Since 0.5 = 2−1 and 4 = 22, y 2−1 < x ≦ 22 2 y = log2 x x Since y = log2 x is monotonically increasing, 0.5 O1 4 log2 2−1 < log2 x ≦ log2 22 −1 ∴ −1 < y ≦ 2 // Q. Find the range of each of the following logarithmic functions in the domains ( ). (1 √) 32 < x < 3 16 (1) y = log2 x (2) y = log0.1 x (0.001 ≦ x < 10) Q. Compare large and small values in each set. (1) log2 0.5 log2 3 log2 5 (2) log 1 4 log 1 0.25 log 1 2 22 2 Exercise Solve the following logarithmic equation. log2 x + log2(x + 2) = 3 Solution Since the antilogarithm is any positive number, we have x > 0 and also x + 2 > 0. ∴ x > 0 The logarithmic equation is transformed to log2 x(x + 2) = 3. From the definition of logarithm, x(x + 2) = 23 = 8. Therefore, x2 + 2x − 8 = 0 ∴ x = 2, −4 Since x > 0, the solution is x = 2 // Note The condition in which the antilogarithm is any positive number is called the antilogarithm condition. Q. Solve the following logarithmic equations. (1) log10 5x − log10(x − 2) = 1 (2) log2(x + 1) + log2(x − 2) = 2

§ 2 Logarithmic functions 117 Exercise Solve the following inequality. log2(2x − 1) ≧ −2 Solution From the antilogarithm condition, 2x − 1 > 0. ∴ x> 1 (1) 2 (2) The given inequality is represented in log2(2x − 1) ≧ log2 2−2. // Since log2 x is monotonically increasing, 2x − 1 ≧ 2−2 ∴ x≧ 5 8 From (1) and (2), the solution is x≧ 5 8 Q. Solve the following inequalities. (1) log3 x ≦ 4 (2) 1 < log10(1 − 10x) < 2 2 3 Common Logarithm The logarithm to base 10 is called a common logarithm. The logarithmic table at the end of the book lists common logarithmic values from 1.00 up to 9.99. Ex.4 log10 2 = 0.3010, log10 3 = 0.4771 log10 120 = log10(1.2 × 102) = log10 1.2 + log10 102 = 0.0792 + 2 = 2.0792 Note Logarithmic values are irrational numbers in general and repre- sented with non-repeating infinite decimals. The logarithmic table lists values rounded off to five decimal places. These values are called approxi- mations. [approximate value] Exercise Find the values for log10 5 and log2 5. Note that log10 2 = 0.3010. Solution log10 5 = log10 ( 10 ) = log10 10 − log10 2 = 1 − 0.3010 = 0.6990 2 log2 5 = log10 5 = 0.6990 = 2.322 // log10 2 0.3010 Q. Let log10 2 = 0.3010 and log10 3 = 0.4771. Find the following values. (1) log2 3 (2) log3 5 (3) log15 12

118 Chapter 4 Exponential function and logarithmic function Exercise Find the largest integer n that satisfies 10n ≦ 1.311000 Solution Notice that a logarithmic function y = log10 x is monotonically increasing. By taking the common logarithm on both sides of the given inequality, n ≦ log10 1.311000 From the logarithmic table, log10 1.31 = 0.1173 log10 1.311000 = 1000 log10 1.31 = 1000 × 0.1173 = 117.3 Therefore, n ≦ 117.3 Since n is the largest integer that satisfies the inequality above, n = 117 // Exercise Find the largest integer n that satisfies 10−n ≧ 1. 4200 Solution By taking the common logarithm on both sides of the given inequality, −n ≧ −200 log10 4 Where, n ≦ 200 log10 4 From the logarithmic table, log10 4 = 0.6021 n ≦ 120.42 Therefore, the solution n is n = 120 // Q. Find the largest integers n that satisfy the following inequalities. (1) 10n ≦ 230 (2) 10n ≦ 330 Q. Find the largest integer n that satisfies 10−n ≧ 4−15.

§ 2 Logarithmic functions 119 Exercise A certain type of bacteria increases at a constant rate per hour and its number doubles after 5 hours. After how many hours for the first time will the number of bacteria be more than 10 times of the initial count? Solution Let us suppose that the number of bacteria increases at a rate of r of the previous time per hour, Rate 2 1 r5 = 2 Thus, r = 2 5 Then by supposing that the number of bacteria will be more than 10 times of the r2 r initial count after n hours, 1 n rn ≧ 10 where, 2 5 ≧ 10 By taking the common logarithm on both sides, O 12 5 Hour ∴ n ∴ log10 2 5 ≧ log10 10 = 1 n log10 2 ≧ 1 5 n ≧ 5 = 5 = 16.6 · · · · · · log10 2 0.3010 Therefore, it is not until after 17 hours that the number becomes more than 10 times of the initial count. // Q. When light is directed through a certain type of glass plate, 9 % of the incident light is lost. How many glass plates will be stacked such that the incident light is lost down to 30 % of the initial light?

120 Chapter 4 Exponential function and logarithmic function Column Exponential and logarithmic functions An exponential function y = ax (a > 1) such as y = 2x increases rapidly as shown in the graph below. Since 210 = 1024, the increase is very sharp. On the other hand, as you find in the graph, a logarithmic function y = loga x (a > 1) such as y = log2 x increases gently. Since log2 1024 = 10, the increase is very slow. As in y = log2 2x = x, y y = 2x y=x 50 we can limit the growth rate by tak- ing logarithms of a function value 40 that increases rapidly. The values 30 obtained by natural phenomena − such as the energy released by an 20 earthquake corresponding to the size y = log2 x 10 of an earthquake and the number of cancer cells corresponding to time x O 10 20 30 40 50 − tend to increase rapidly like ex- ponential functions. Therefore, by taking logarithms of those values to find the growth rate, the growth rate will increase slowly. For example, magnitude M which represents the size of an earthquake and energy E (joule) released by an earthquake have the following relationship. log10 E = 4.8 + 1.5M . This means, if there is 2 points increase in magnitude, the energy released by an earthquake increases by about 1000 times. In Great Hanshin-Awaji Earth- quake (Jan.17, 1995) and Niigata-Chuuetsu Earthquake (Oct.23, 2004), both measured the magnitude 7. However, in Great East Japan Earthquake which occurred on Mar.11, 2011, it measured the magnitude as large as 9. We find that although there is only 2 points difference in magnitude, we get 1000 times more ground-shaking than the two major great earthquakes.

§ 2 Logarithmic functions 121 Exercises 2–A 1. Find the value of x that satisfies each of the following equations. (1) log4 x = 2 (2) log 1 x = −2 (3) log10 x = 3 3 2 √ (4) logx 8 = 3 (5) logx 10 = 1 2 (6) log√6 216 = x 2. Simplify the following expressions. (2) log10 1 + 1 log10 9 (1) log3 54 − log3 18 15 2 4 (3) (log3 8) × (log2 9) × (log4 2) 3. When log2 3 = m, represent log4 6 and log3 2 with m. 4. Arrange the values in each set from the largest to smallest. (1) 2 log0.5 3 3 log0.5 2 log0.5 41 5 √ (2) log2 7 log2 45 3 1 log2 50 2 5. Plot graphs of the following logarithmic functions. (1) y = log3(x + 2) (2) y = − log5 x (3) y = log2(5 − x) 6. Solve the following logarithmic equation. log2(x − 1) = log2(x2 − 8x − 11) 7. Solve the following inequalities. (2) log4(6 − x) < 1 (1) log3(−2x + 1) > 1 2

122 Chapter 4 Exponential function and logarithmic function Exercises 2–B 1. Simplify the following expressions. (1) 3 log10 3 + log10 24 − 2 log10 9 2 10 (2) (log3 5) × (log2 9) × (log25 4) (3) (log2 9 + log4 3) log27 4 2. Solve the following logarithmic equations. (1) log2(x2 − 6x + 8) − log2(x − 2) = 2 (2) log5(x2 − 5x + 4) = log5(x − 1) + 1 (3) (log2 x)2 − log2 x − 2 = 0 3. When X = loga 4, Y = loga 8, a > 0, a =\\ 1, 3 3 represent loga 3 with X and Y . √ 4. Prove the following equation when 5x = 2y = 10z, x =\\ 0, y =\\ 0, z =\\ 0. 1+1=2 xy z 5. Solve the following inequalities. (1) log2(log3 x) < 1 (2) log2(x + 1) < log2(3 − x) 6. Plot graphs of the following logarithmic functions. (1) y = 2 log2 x (2) y = log2 x2 7. When 1 < b < a, loga b + logb a = 10 , find the value of loga b − logb a. 3

123 5Chapter Trigonometric function § 1 Trigonometric ratios and its applica- tions 1 1 Trigonometric ratios of acute angles As shown in the figure, when ∠XOY = α is an acute angle, draw a perpen- dicular line (PQ) from the point P on the side OY to the side OX. From the similarity conditions of triangles, the ratio values of two sides of the right triangle OPQ are determined purely by the angle α, regardless of the size of △OPQ. Opposite PQ Y Hypotenuse OP = Adjacent = OQ Hypotenuse P Hypotenuse OP Opposite Opposite = PQ α X Adjacent OQ O Adjacent Q Each value is called sin (Sine), cos (Cosine), tan (Tangent) of α and writ- ten as sin α, cos α, tan α respectively, and all together they are called the trigonometric ratios of an angle α. sin α = Opposite , cos α = Adjacent , tan α = Opposite Hypotenuse Hypotenuse Adjacent

124 Chapter 5 Trigonometric function Exercise For △ABC, find the values of sin α, cos α, tan α, when AB = 5, BC = 3, and ∠A = α, ∠B = 90˚. Solution From Pythagorean proposition, √√ AC = 52 + 32 = 34 ∴ sin α = BC = √3 C AC 34 3 √ B cos α = AB = √5 34 AC 34 α tan α = BC = 3 // A AB 5 5 Q. Find the trigonometric ratios of the angle α for each of the following right triangles. (1) (2) (3) 6 5 7 α 49 α α 4 Q. Find the trigonometric ratios of the angle 30˚, 45˚and 60˚from the figures of right triangles below. √ 21 2 60◦ 1 30◦ 45◦ √ 1 3 The right triangles ABC in the figures below, let ∠A = α, then ∠C = 90˚−α. From the definitions of trigonometric ratios, A α cos α = AB AC 90˚−α CB Hypotenuse sin(90˚− α) = AB C Opposite AC Hypotenuse 90˚−α Therefore, sin(90˚− α) = cos α Likewise, A α B Adjacent cos(90˚− α) = sin α tan(90˚− α) = 1 tan α

§ 1 Trigonometric ratios and its applications 125 To summarize, the following formulas are obtained.  Trigonometric ratios of 90˚− α  sin(90˚− α) = cos α, cos(90˚− α) = sin α tan(90˚− α) = 1 tan α   Ex.1 cos 60˚= cos(90˚− 30˚) = sin 30˚ sin 45˚= cos 45˚ Q. Represent the following trigonometric ratios with the trigonometric ratios of the angles less than 45˚. (1) sin 70˚ (2) cos 64˚ (3) tan 87˚ Note The trigonometric values (approximations) of angles 0˚ to 90˚ are shown in the trigonometric function table at the end of the book. Q. Find the trigonometric ratios using the trigonometric function ta- (1b)le. sin 17˚ (2) cos 64˚ (3) tan 3˚ (4) sin 73˚ Exercise The angle of elevation to the top of a tree from a point on the ground 30 m from the base of the tree is 23˚, and the eye level of the observer is 1.5 m. What is the height of this tree (m)? Solution Let the height of the tree be x m, x − 1.5 = tan 23˚ 30 Thus, x x = 30 tan 23˚+ 1.5 1.5 m 23˚ 30 m = 14.24 (m) // Q. When travelled 200 m from A to B on a slope of 9˚, as shown in the figure 200 m B below, how many meters travelled hori- 9˚ A zontally and how many meters raised vertically?

126 Chapter 5 Trigonometric function 1 2 Trigonometric ratios of obtuse angles Let’s expand an approach of the trigonometric ratios of acute angles into the case of right angles and obtuse angles. Taking an origin O as a center, draw a semicircle with radius r on a coordi- nate plane and plot a point P(X, Y ) on it. Let y the angle of which the segment OP makes on the r P(X,Y) positive direction of x-axis be α. The trigonomet- ric ratio is expressed by the coordinates (X, Y ) α rx O of the point P and the radius r. −r When α is an acute angle, from the definition on the p123, sin α = Opposite = Y P(X,Y) y Hypotenuse r r cos α = Adjacent = X Hypotenuse r α tan α = Opposite = Y −r O rx Adjacent X These values are determined purely by the angle α, regardless of the length of radius r. The same can be said when α is an obtuse angle, so the trigonometric ratios of the angle α are defined as below.  Definition of trigonometric ratios  sin α = Y , cos α = X , tan α = Y r r X   Exercise Find the trigonometric ratio of an angle 135˚. Solution When considering a semicircle y with radius 1, as shown in the figure, P √1 2 since OP = 1, the coordinates of the point ( ) 135˚ P are − √1 , √1 . Therefore, − √1 x 22 O 1 2 sin 135˚= √1 , cos 135˚= − √1 , tan 135˚= −1 // 22

§ 1 Trigonometric ratios and its applications 127 Note When the angle α is an obtuse angle, since X < 0, Y > 0, sin α > 0 cos α < 0 tan α < 0. Note A circle centered at the origin with a radius of 1 is called a unit circle. By thinking with a unit circle, sin α = Y cos α = X Therefore, the coordinates of the point P are (cos α, sin α). Q. Find the trigonometric ratios of the angles of 120˚, 150˚. The trigonometric ratios are defined when the angle α is 0˚, 90˚and 180˚. When α = 90˚, the coordinates of the point P are (0, r). y P(0, r) sin 90˚= r = 1 (−r, 0) x r O (r, 0) cos 90˚= 0 = 0 r These are defined. tan 90˚is not defined. Likewise, the values of the trigonometric ratios of the angle 0˚and 180˚are defined as follows. sin 0˚= 0, cos 0˚= 1, tan 0˚= 0 sin 180˚= 0, cos 180˚= −1, tan 180˚= 0 Let s look into the relationship between the trigonometric ratios of the angle α and 180˚− α. When α is an acute angle, the points P and Q, corresponding to α and 180˚−α, are symmetric with respect to the y-axis, and taking (X, Y ) as the coordinates of P and (−X, Y ) as the coordinates of Q. sin(180˚− α) = Y = sin α y r r P(X, Y ) cos(180˚− α) = −X = − cos α Q(−X, Y ) r −r 180◦− α tan(180˚− α) = Y = − tan α rx −X α O The same relationship is established when α is not an acute angle, and the following formulas are obtained.

128 Chapter 5 Trigonometric function  The trigonometric ratios with the angle of 180˚− α  sin(180˚− α) = sin α, cos(180˚− α) = − cos α tan(180˚− α) = − tan α   Ex.2 √ 3 sin 120˚= sin(180˚− 60˚) = sin 60˚= 2 tan 158˚= tan(180˚− 22˚) = − tan 22˚= −0.4040 Q. Find the trigonometric ratios of the following using the trigonomet- ric function table at the end of the book. (1) sin 165˚ (2) cos 142˚ (3) tan 116˚ Summary of the main trigonometric ratios from 0˚to 180˚. α 0˚ 30˚ 45˚ 60˚ 90˚ 120˚ 135˚ 150˚ 180˚ √√ √√ 1 23 3 21 sin α 0 1 222 0 cos α 1 22 2 √√ 1 1 √√ 2 2 3 32 0 − − 2 − 2 −1 222 tan α 0 √1 1 √ √ −1 − √1 0 3 3 −3 3 With respect to the point P(X, Y ) on the unit y circumference, let the angle of which the segment OP makes on the positive direction of the x-axis 1 YP be α. α x X1 cos α = X, sin α = Y, tan α = Y −1 O X Therefore, whenα =\\ 90˚, X =\\ 0, tan α = sin α cos α Next, from Pythagorean proposition, X2 + Y 2 = 1 (cos α)2 + (sin α)2 = 1 Then, divide this formula with (cos α)2, 1 + (tan α)2 = 1 . (cos α)2

§ 1 Trigonometric ratios and its applications 129 Note From now onward, (sin α)2, (sin α)3 and others will be represented as sin2 α, sin3 α and so on. This would be the same for cos α and tan α.  Mutual relationships between the trigonometric ratios  tan α = sin α , cos2 α + sin2 α = 1, 1 + tan2 α = 1 cos α cos2 α   Exercise Find the values of cos α and tan α, when α is an obtuse angle and sin α = 1 . 3 Solution Since cos2 α + sin2 α = 1, cos2 α = 1 − sin2 α = 1 − ( 1 )2 = 8 √ 39 ∴ cos α = ± 2 3 2 √ 2 2 Since α is an obtuse angle, cos α < 0 thus, cos α = − (√ ) √ 3 And, tan α = sin α = 1 ÷ − 232 =− 2 // cos α 3 4 Alternative solution Since α is an ob- y tuse angle, X < 0. Draw a circle with a 3 radius of 3 as shown in the figure below. Draw a perpendicular line from the point P(X, 1) 1 α 3 −3 H O x 3 P and let the intersection point of the line and the x-axis be H. By applying Pythagorean proposition to the right triangle OPH, OH2+PH2 =OP2 X2 + 12 = 32, since X < 0, √ √ X = −2 2 √ 2 X = − 232 Y 1√ 4 Thus, cos α = r tan α = X = −2 2 =− // Q. Find the values of cos α and sin α, when α is an obtuse angle and tan α = −2. Q. Find the values of sin α and tan α, when α is an obtuse angle and cos α = − 1 . 7

130 Chapter 5 Trigonometric function 1 3 Applications to the triangles ⋄ Sine rule (theorem) A For △ABC, let the angles of ∠A, ∠B, ∠C be c A A, B, C, and the length of the opposite side of b the vertex A, B, C be a, b, c respectively. B C B C a A circle that passes through each of the triangle’s three vertices is called a circumcircle. The relationship between the lengths of the three sides of a triangle and the sine of its vertices, the following rule is established.  Sine rule (theorem)  When R stands for the radius of the circumcircle of △ABC a = b = c = 2R sin A sin B sin C   Proof Here, proving the case when A is an acute angle. Let the segment BA′ be a diameter as shown in the figure. A Since A = A′ and ∠A′CB = 90˚, A′ O sin A = sin A′ = BC = a 2R BA′ 2R ∴ a = 2R // C sin A B The same are established for B and C. Exercise Find the radius R and the value of b of the circumcircle, when a = 9, B = 45˚, C = 75˚. Solution Since A = 180˚− (45˚+ 75˚) = 60˚, A From the sine rule, 9 = b = 2R √ B 45◦ 75◦ C sin 60˚ sin 45˚ =3 3 ∴ R= 9 = 9√ 9 ∴ 2 sin 60˚ 2 × 3 2R sin 45˚= 2· 2 √ √ 2 √ b = 3 3· 2 =3 6 //

§ 1 Trigonometric ratios and its applications 131 Q. △ABC is an isosceles triangle with A = 120˚. find the values of a and b, when the radius of the circumcircle is 2. C Q. For △ABC with AB = 30, A = 68˚, B = 73˚ as shown in the figure. Draw a perpendicular line CH from C to the segment AB. Find the lengths of AC and CH. 68˚ 73˚ ⋄ Cosine rule (theorem) A HB 30 Let’s look into the relationship between the lengths of the three sides of a triangle and the cosine of its interior angles. C For △ABC, suppose A and B are both acute ba angles. When drawing a perpendicular line CH from the vertex C to the segment AB, H CH = b sin A, AH = b cos A AcB Since △BCH is a right triangle with ∠BHC = 90˚, a2 = CH2 + HB2 = (b sin A)2 + (c − b cos A)2 = b2 sin2 A + c2 − 2bc cos A + b2 cos2 A = b2 + c2 − 2bc cos A The same can be said when either A or B is an obtuse angle. Therefore, a2 = b2 + c2 − 2bc cos A The same equalities are derived for b2 and c2, and the following cosine rule (theorem) is obtained.  cosine rule (theorem)  a2 = b2 + c2 − 2bc cos A, b2 = c2 + a2 − 2ca cos B c2 = a2 + b2 − 2ab cos C   Ex.3 When b = 3, c = 7, A = 60˚, from the cosine rule, a2 = 9 + 49 − 2 · 3 · 7 cos 60˚= 37 √ ∴ a = 37

132 Chapter 5 Trigonometric function Q. Find the value of b, when a = 5, c = 7, B = 60˚. Q. Prove the first formula of the cosine rule for the following situations. (1) When B is an obtuse angle. (2) When A is an obtuse angle. C C ba ba A c BH HA c B ∠CBH = 180˚− B ∠CAH = 180˚− A The following equalities are obtained from the cosine rule. cos A = b2 + c2 − a2 , cos B = c2 + a2 − b2 2bc 2ca cos C = a2 + b2 − c2 2ab Q. Find the values of cos A, cos B, cos C, When a = 2, b = 3, c = 4. Q. For △ABC, prove the followings. (1) The necessary and sufficient condition for A to be a right angle is a2 = b2 + c2. (2) The necessary and sufficient condition for A to be an acute angle is a2 < b2 + c2. (3) The necessary and sufficient condition for A to be an obtuse angle is a2 > b2 + c2. ⋄ The area of a triangle Let s derive the formulas for finding the area of a triangle. Draw a perpendicular line CH from the vertex C to the segment AB, and let the area of △ABC be S, C S= 1 c · CH. ba 2 H When A is an acute angle, since sin A = CH , cB b CH = b sin A A Thus, 1 1 2 2 S= c · CH = bc sin A

§ 1 Trigonometric ratios and its applications 133 In the same way, the following formulas for finding the area of a triangle are obtained. The area of a triangle   S = 1 bc sin A = 1 ca sin B = 1 ab sin C 22 2   Ex.4 The area of an equilateral triangle with a side-length of a is √ S = 1 a2 sin 60˚= 3 a2. 24 Q. Find the areas of the following triangles. (1) a = 3, b = 4, C = 30˚ √ (2) c = 5, a = 2 3, B = 120˚ Q. For an acute triangle ABC, find the value of C, when a = 10, b = 6 √ and the area is 15 3. Exercise Find the following values, when a = 5, b = 6, c = 7. (1) cos A (2) sin A (3) The area S of △ABC Solution (1) From the cosine rule, A 76 cos A = b2 + c2 − a2 = 62 + 72 − 52 = 5 B5C 2bc 2·6·7 7 // (2) From sin2 A + cos2 A = 1 and sin A > 0, √ √ ( 5 )2 √ 1 1 26 sin A = − cos 2 A = − = 77 (3) S= 1 bc sin A = 1 ·6·7· √ √ 2 2 26 =6 6 7 Same as the exercise above, the following Heron’s formula can be derived.

134 Chapter 5 Trigonometric function  Heron’s formula  By letting s = a + b + c , the area S of a triangle is 2 √ S = s(s − a)(s − b)(s − c)   By using Heron’s formulas, the area of a triangle can be found from the lengths of three sides of a triangle. Ex.5 When a = 5, b = 6, c = 7, s = 5+6+7 = 9. Thus, √ √2 S = 9 (9 − 5) (9 − 6) (9 − 7) = 6 6 Q. The distances between the three points A, B and C were measured and the results were as follows. BC = 24.3 m, CA = 46.2 m, AB = 30.8 m. Find the area of △ABC.

§ 1 Trigonometric ratios and its applications 135 Exercises 1–A 1. Find the lengths of the side x and y for the following figures. (1) 8 (2) x 36˚ x 59˚ y y 8.6 2. Find the value of 5 sin α − 2 , when α is an obtuse angle and cos α = − 3 . 6 tan α + 7 5 3. Prove the following equalities. (1) sin4 θ − cos4 θ = 1 − 2 cos2 θ (2) tan2 θ + (1 − tan4 θ) cos2 θ = 1 4. For △ABC, answer the following questions. √ (1) Find the value of C, when a = 4, b = 3, c = 13. (2) Find the values of a and c, when C = 80˚, A = 32˚, b = 12. 5. For △ABC, when the equality sin C = 2 sin A cos B is established, answer the following questions. (1) Convert the equality sin C = 2 sin A cos B into the relational expressions of a, b, c, R. Note that R is a radius of a circumcircle of △ABC. (2) What kind of triangle is △ABC ? 6. Find the length of the side AB and the area of the trapezoid in the figure below. A3D 57˚ 5 68˚ B C

136 Chapter 5 Trigonometric function Exercises 1–B 1. Let s look into the isosceles triangle ABC, of which apex angle A is 36˚and the length of the side AB is 1. A The point D denotes the intersection point of the bisector of ∠B and the side AC. 36˚ (1) Let BC = x and find the value of x. 1 (2) Find the value of sin 18˚. D 2. For △ABC, prove the following equalities. BxC (1) (b − c) sin A + (c − a) sin B + (a − b) sin C = 0 (2) a(b cos C − c cos B) = b2 − c2 3. What kind of triangle is it, when △ABC satisfies a cos A + b cos B = c cos C ? 4. When looking at the top of a tree from point A on a plane surface, the angle of elevation is 27˚. Then, 10 meters closer to the tree, the angle of elevation increases to 30˚. What is the height of this tree (m) ? 5. By letting the area of △ABC be S, prove the following formula is established. S = a2 sin B sin C 2 sin(B + C) 6. Prove the Heron s formula with the following procedures. √ (b + c + a)(b + c − a)(a + b − c)(a − b + c) (1) Prove that sin A = 2bc . (2) By letting the area of △ABC be S and s = a + b + c , 2 √ S = s(s − a)(s − b)(s − c)

§ 2 Trigonometric function 137 § 2 Trigonometric function 2 1 General angle When a half-line OP centered at the point O rotates on a plane surface, the half-line is called radius and a line OX, where the radius starts, is called an initial line. There are two rotational directions centered at the point O, counter-clockwise is called positive direction P and clockwise is called negative direction. For the rotational amount of radius OP, the positive rotation is represented by a positive sigh O X and the negative rotation is represented by a neg- ative sigh. These are the angles that the radius OP makes. The rotational amount increase 360 degrees in every complete rotation of the radius OP. Therefore, even if the position of the radius OP is determined, the angle may not be found. For example, if the radius OP makes the angle α, as shown below, the radius takes the same position in all the cases. α + 360˚, α + 360˚× 2, α − 360˚× 2 PPP α X α X α X O O O α + 360˚ α + 360˚× 2 α − 360˚× 2 In general, by letting α stands for the angle that the radius OP makes, the angle α can be expressed by the formula below. α + 360˚× n (where n is an integer)

138 Chapter 5 Trigonometric function As mentioned in the previous page, the extended angle is called a general angle and it can simply be called an angle. In general, it shall be 0˚≦ α < 360˚or −180˚< α ≦ 180˚. Ex.1 The general angle that the radius OP Q makes in the figure on the right is 840˚ 120˚ X −30˚+ 360˚× n (where n is an integer) O −30˚ Also, since 840˚= 120˚+ 360˚× 2 , the radius which makes the angle of 840˚ P corresponds to the radius OQ which makes the angle of 120˚. Q. Draw the radiuses that make the following angles. (1) 500˚ (2) 1210˚ (3) −310˚ (4) −400˚ (5) −1520˚ On a coordinate plane, O takes the origin and the initial line OX takes a positive direction on the x-axis. When the radius OP is on either of the first, second, third or fourth quadrant, the angle that the radius OP makes is called the first, second, third or fourth quadrant angle respectively. Q. Which quadrant angles are these ? (5) −570˚ (1) 470˚ (2) −315˚ (3) −410˚ (4) 1280˚ 2 2 Trigonometric function of general angle Let s extend the approach of trigonometric P(X,Y) y ratios into the case of general angle. r On a coordinate plane, draw a circle cen- −r θ rx tered at the origin O with radius r. When the O initial line is at the positive direction on the x-axis, let the position where the radius with −r general angle θ intersects with the circle be P(X, Y ). Then, the followings are defined, the same as the case of trigonometric ratios on the p126. sin θ = Y , cos θ = X , tan θ = Y (1) r rX

§ 2 Trigonometric function 139 Especially, when thinking with a unit circle, where r = 1, they will be sin θ = Y cos θ = X, −1 ≦ sin θ ≦ 1, −1 ≦ cos θ ≦ 1 will be found. The functions of θ defined at (1) are called trigonometric functions. The signs of the trigonometric functions are determined depending on the quadrant angles of θ, as shown below. sin θ cos θ tan θ y yy ++ x −+ x −+ x −− −+ +− Exercise Find the value of trigonometric function of 300˚. Solution Draw a unit circle. Let the position where the radius with the angle of 300˚intersects with the circle be P. y Then, draw a perpendicular line from P to 1 the x-axis and let the line be PH. Since, △OPH is a right triangle with ∠POH = 60˚, 300˚ O H x P 1 OP = 1. −1 // √ 1 3 OH = 2 , PH = 2 . ( √) −1 1 3 The coordinates of point P are 2 , − 2 Thus, √ √ 3 1 tan 300˚= − 3. sin 300˚= − 2 cos 300˚= 2 Q. Find the values of the following trigonometric functions. (1) sin 210˚ (2) cos 495˚ (3) tan 315˚ (4) sin(−840˚) (5) cos(−450˚) (6) tan(−570˚)

140 Chapter 5 Trigonometric function 2 3 Circular measure (Radian system) So far, we have been using degree to describe the magnitude of an angle. This method is called a Sexagesimal method where units (degree, minute, second) are determined with a base of 60. On the other hand, there is a method to represent the magnitude of an angle based on the length of an arc. An arc AB, which is the same length as the radius r, is given on a circum- ference of a circle centered at the point O with radius r. Let ∠AOB = α˚. r : 2πr = α : 360 B Where, r α = 360 × r α˚ A 2πr Or = 180 = 57.2957 (˚) π Therefore, the magnitude of the angle α is fixed, regardless of the radius r. The magnitude of an angle is called 1 radian (circular measure) and the method of representing the angle using this unit is called circular measure (radian system). Let the center angle be θ radian which subtends l r the arc length l on a circumference of a circle with (1) radius r. θ1 Or θ : 1 = l : r. Where, θ= l. r Thus, θ is the ratio value of the arc length l and the radius r. Therefore, the unit of an angle “radian” is commonly omitted in the circular measure. Ex.2 Since the length of the circumference is 2πr, 360˚= 2πr = 2π(radian) r π π 180˚= π, 90˚= 2 , 45˚= 4 Q. Represent the following angles with circular measure(radian). (1) 15˚ (2) 60˚ (3) 120˚ (4) 270˚ (5) −135˚

§ 2 Trigonometric function 141 Let s find the converting expressions between the sexagesimal method and the circular measure. Let α˚= θ (radian), where 180˚= π. α : 180 = θ : π. From this, the following formulas are obtained.  The relationship between the sexagesimal method   Let α˚= θ (radian), and the circular measure. θ= π α, α= 180 θ 180 π  Q. Represent the following angles with the sexagesimal method. (1) π (2) 5 π (3) 7 π (4) − π (5) 11 π 4 6 4 5 6 Find the arc length l and the area S of a fan-like shape, which has a center angle θ (radian) and radius r. First, from (1) on the p140, Bl A l = rθ S θ An area of a fan-like shape in a circle is pro- portional to the magnitude of the center angle. Or S : πr2 = θ : 2π Where, S = 1 r2θ 2 To summarize, the following formulas are obtained.  The arc length and the area of a fan-like shape  The arc length l and the area S of a fan-like shape, which has a center angle θ (radian) and radius r. ( ) l = rθ, S= 1 r2θ = 1 rl  22  Q. Find the arc length and the area of a fan-like shape, which has a center angle 2 π and a radius of 9 cm. 3 Q. Find the center angle that subtends the given arc length of 3π cm in a circle with a radius of 10 cm.

142 Chapter 5 Trigonometric function Q. In a circle O with radius r, find the O B area of the segment enclosed by the arc AB r that subtends the center angle θ (radian) and Aθ the chord AB. Note that 0 < θ < π. 2 4 Properties of trigonometric function Since the value of the trigonometric function is determined by the angle, there will be no difference for the angle to be represented by the sexagesimal method or the circular measure. √ π 3 Ex.3 sin 3 = sin 60˚= 2 √ 5 3 cos 6 π = cos 150˚= cos(180˚− 30˚) = − cos 30˚= − 2 Q. Find the values of the following trigonometric functions. (1) sin π (2) cos 3 π (3) tan 2 π 4 4 3 The mutual relationships between the trigonometric ratios on the p129 are established in the case of general trigonometric functions. tan θ = sin θ , cos 2 θ + sin2 θ = 1, 1 + tan2 θ = 1 cos θ cos 2 θ Exercise Prove the following equality. sin θ + 1 + cos θ = 2 1 + cos θ sin θ sin θ Proof Left part = sin2 θ + (1 + cos θ)2 = sin2 θ + 1 + 2 cos θ + cos2 θ (1 + cos θ) sin θ (1 + cos θ) sin θ = 2(1 + cos θ) = 2 = Right part // (1 + cos θ) sin θ sin θ Q. Prove the following equalities. (1) 1+ 1 = 1 (2) sin θ − cos θ = 1 tan2 θ sin2 θ 1 − cos θ sin θ sin θ (3) sin3 θ − cos3 θ = (sin θ − cos θ)(1 + sin θ cos θ)

§ 2 Trigonometric function 143 Exercise Find the values of cos θ and tan θ, when θ is the third quad- rant angle and sin θ = − 2 . 3 Since sin2 θ + cos2 θ = 1, cos 2 sin2 ( 2 )2 5 −− Solution θ = 1 − θ = 1 = √ 39 5 ∴ cos θ = ± 3 Here, since θ is t√he third quadrant angle, where cos θ < 0. 5 then, cos θ = − 3 and tan θ = sin θ 2 ( √ ) √2 cos θ 3 ÷− 5 5 = − = // 3 Alternative solution Since θ is the third quadrant angle, where X < 0, Y < 0. Draw a circle with a radius of 3 as shown on the right. Y = −2, r = 3 y 3 For the right triangle OHP, OH2+PH2 =OP2 θ X2 + (−2)2 = 32 since X < 0, H O x −3 3 3 √ X=− 5 // √ 5 √2 P Thus, cos θ = − 3 , tan θ = 5 (X, −2) −3 Q. Find the values of sin θ and cos θ, when θ is the fourth quadrant angle and tan θ = −2. Let s look into the properties of trigonometric function that aren t mutually related. First, when n is an integer, the radius with the angle θ corresponds to the radius with the angle θ + 2nπ. So the values of these trigonometric functions are the same. sin(θ + 2nπ) = sin θ, cos(θ + 2nπ) = cos θ tan(θ + 2nπ) = tan θ

144 Chapter 5 Trigonometric function Since the radius with the angle θ and the radius with the angle −θ are symmetric with respect to the x-axis, the following formulas are established as you can see on the figure. y sin(−θ) = − sin θ r P(X, Y ) cos(−θ) = cos θ −r θ rx tan(−θ) = − tan θ O −θ Since the radius with the angle θ and the −r Q(X, −Y ) radius with the angle θ + π are symmetric with respect to the origin. y r P(X, Y ) sin(θ + π) = − sin θ π+θ θ rx −r O cos(θ + π) = − cos θ Q(−X, −Y ) −r tan(θ + π) = tan θ The property of the trigonometric ratio of y r Q(Y, X) 90˚− α is established as the property of the trigonometric function of π − θ. θ P(X, Y ) ( = 2 ) −r θ rx sin π −θ O 2 cos θ π ( ) −r cos 2 − θ = sin θ π ( ) 1 y tan −θ = Q(−X, Y ) r 2 tan θ P(X, Y ) The same can be said for π − θ. −r θθ rx sin(π − θ) = sin θ O cos(π − θ) = − cos θ tan(π − θ) = − tan θ −r ( π ) { ( π )} ( π ) 2 π 2 θ 2 − θ = cos θ Ex.4 sin θ + = sin − − = sin ( ) Thus, sin θ + π = cos θ 2 Q. (Prove the)following equalities. ( ) (1) cos θ + π = − sin θ (2) tan θ + π = − 1 2 2 tan θ

§ 2 Trigonometric function 145 2 5 Graphs of trigonometric functions Here, the angle is represented by x (radian) and the value of trigonometric function is represented by y with respect to x. Let s look into the graphs of y = sin x, y = cos x and y = tan x. First, think about the graph of y = sin x. The values of sin x derived by the trigonometric function table are as follows. x 0 π π 3π 2π π 3π 7π 4π 9π π 10 5 10 5 2 5 10 5 10 sin x 0.00 0.31 0.59 0.81 0.95 1.00 0.95 0.81 0.59 0.31 0.00 Since sin(−x) = − sin x, y = sin x is an odd function. Therefore, the graph is symmetric with respect to the origin. The graph is as shown below and it is called a sine curve. y 1 − π 3π 2 2 x −π Oπ π 2π 2 −1 Secondly, think about the graph of y = cos x. Since cos(−x) = cos x, y = cos x is an even function and the graph is symmetric with respect to the y-axis. The graph is as shown below and it is called a cosine curve. y 1 −π π x 3π 2π − π Oπ 2 2 2 −1 Note Since cos x = ( π) from Ex.4 on the p144, the graph of sin x + 2 y = cos x is the same as the graph of y = sin x but translated − π parallel 2 towards the x-axis.

146 Chapter 5 Trigonometric function Since sin(x + 2π) = sin x, cos(x + 2π) = cos x, the graph repeats the same shape every 2π. In general, for a function f (x), when f (x + p) = f (x) has a nonzero positive constant p such that the function is established with all the possible values of x, the function f (x) is called a periodic function, and the smallest possible constant p is called a period. Both sin x and cos x are periodic functions with a period of 2π. Let’s look into the graphs of y = 2 sin x and y = sin 2x. The following graphs are plotted by using the table below. x 0 π π 3π 2π π 3π 7π 4π 9π π 10 5 10 5 2 5 10 5 10 sin x 0.00 0.31 0.59 0.81 0.95 1.00 0.95 0.81 0.59 0.31 0.00 2 sin x 0.00 0.62 1.18 1.62 1.90 2.00 1.90 1.62 1.18 0.62 0.00 sin 2x 0.00 0.59 0.95 0.95 0.59 0.00 −0.59 −0.95 −0.95 −0.59 0.00 y y = 2 sin x y = sin 2x 2 1 Oπ π 2π x 2 y = sin x By comparing the graphs of y = 2 sin x and y = sin 2x with the graph of y = sin x, the followings are found. ( I ) The period of y = 2 sin x doesn t change, but the maximum and minimum values are doubled. (II) The period of y = sin 2x becomes 1 , but the maximum and minimum 2 values remain the same.

§ 2 Trigonometric function 147 From the formulas on the p95, when a is a positive constant, the graph of y = sin ax is the same as the graph of y = sin x but stretched or compressed 1 times towards the x-axis. Therefore, the period is 2π · 1 = 2π . a a a Ex.5 The period of y = sin 3x is 2π · 1 = 2π . 3 3 The period of y = sin 1 x is 2π · 1 = 2π × 2 = 4π. 2 1 2 Exercise Find the periods of the following functions and plot the graphs. x ( π ) 3 sin 2x − 3 (1) y = 2 cos (2) y = Solution (1) Since the graph is the same as the graph of y = cos x but stretched 3 times towards the x-axis and 2 times towards the y-axis, the period is 6π and the graph is as shown below. − 3π y 3π 9π 2 2 2 3π 2 x O −2 ( π ) ( π ) sin 2x − 3 sin 2 x − 6 , (2) Since it can be transformed into = the graph is the same as the graph of y = sin 2x but translated parallel toward the x-axis. Therefore, the period is π and the graph is as shown below. // y 1 Oπ 2π x 6 3 7π 6 −1 √ 3 − 2

148 Chapter 5 Trigonometric function Q. Find the periods of the following functions and plot the graphs. (1) y = − sin x (2) y = cos 2x ( ) ( ) = sin x − π cos 2x + π (3) y 4 (4) y = 2 Lastly, think about the graph of y = tan x. The values with respect to x are shown below. x 0 π π 3π 2π π 3π 7π 4π 9π π 10 5 10 5 2 5 10 5 10 tan x 0.00 0.32 0.73 1.38 3.08 −3.08 −1.38 −0.73 −0.32 0.00 tan π is not defined. When looking into the values of tan x close to π , 2 2 you will find that the straight line x = π takes an asymptote of the graph. 2 Since tan(x + π) = tan x, tan x is a periodic function with a period of π. Furthermore, since tan(x + π) = tan x, y = tan x is an odd function and the graph is symmetric with respect to the origin. The graph is as shown below and it is called a tangent curve. y 1 − 5π − π − π π 3π 3π x 4 2 4 2 2 4π − 3π −π − 3π Oπ 5π 2 4 4 4 −1

§ 2 Trigonometric function 149 Q. Find the periods of the following functions and plot the graphs. (1) y = tan 2x (2) y = tan x 2 Exercise Solve the following equation and inequality, when 0 ≦ x < 2π. (1) sin x = − 1 (2) sin x > − 1 2 2 Solution (1) As shown in the figure, Y 1 Let the position where the unit circle in- tersects with the straight line Y = − 1 −1 H′ O H1 X 2 Q be P and Q. The angle resulting from the radius OP and OQ, when within the P Y = − 1 2 given range, is the value of x we are look- −1 ing for. Let the perpendicular line from P to the x-axis be PH and Q to the x-axis be QH′, then ∠POH = ∠QOH′ = π 6 thus x= 7 π, 11 π 6 6 (2) Find the range of x where the Y coordinate is greater than − 1 on a unit 2 circumference ( the blue line on the figure ) 0≦x< 7 π, 11 π < x < 2π // 6 6 Alternative solution (2) Plot a sine curve y = sin x and the straight line y = − 1 within the range of 0 ≦ x < 2π. The x coordinates of the 2 intersection points of two graphs are, from (1), x = 7 π, 11 π. 6 6 y 1 y = sin x π 7π 3π 11π O 2 6 2 6 2π x π −1 y = − 1 2


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