Section 8 Inverse Trigonometric Functions Inverse Sine Function Recall that for every function y = f (x) , one may de…ne its INVERSE 1 FUNCTION y = f ¡ (x) as the unique solution of x = f (y) . 1 In other words, the inverse function y = f ¡ (x) "undo" the function y = f (x) : 1 1 f f ¡ (x) = x, f ¡ (f (x)) = x. We also know that eve¡n if a fu¢nction y = f (x) is de…ned every, its inverse 1 function y = f ¡ (x) may not be de…ned everywhere. The su¢cient condi- tion that guarantees existence of inverse functions is called "horizontal line test": a horizontal line can intersect the graph of y = f (x) at most once. For trigonometric functions, for instance the graph of y = sin x intersects horizontal y = 0.6 in…nite many times: Therefore, to de…ne inverse function of y = sin x, we consider the restricted sine function 1 ¼ ¼ i.e., the restriction of y = sin x on , . This function has the domain 2 2 ¼ ¼ ¡ , and the range [ 1, 1] ; it cahptures aill information of the sine function ¡ 2 2 ¡ hover all ireal numbers. Notice that the only di¤erence between the restricted Sine function and the (unrestricted) Sine function is their domains: ¼ ¼ restricted y = sin x de…ned only for x on , ¡ 2 2 y = sin x de…ned for all real numhbers xi ¼ ¼ De…nition: The restricted Sine function de…ned on , is invert- ¡ 2 2 ible, and its inverse function is denoted as h i 1 y = sin¡ x or y = arcsin x. ¼ ¼ The domain of y = arcsin x is [ 1, 1] , and the range y = arcsin x is , . ¡ ¡ 2 2 The graph of y = arcsin x is the re‡ection of the graph for the restrhicted Sinie function about y = x : Example 1 Find the following inverse Sine function values: (a) arcsin 1 (b) arcsin p3 2 ¡ 2 ¡ ¢ ³ ´ 2 1 Solution: (a) By the general de…nition of inverse functions, y = arcsin 2 is the solution of the restricted Sine function for y : ¡ ¢ 1 = sin y 2 The words "restricted Sine" means that ¼ ¼ y . ¡ 2 · · 2 Therefore ¼ 1 ¼ y = , or arcsin = 6 2 6 µ ¶ (Notice that the notation "arcsin" comes from the fact that the length of the arc whose sine is 1/2 is ¼/6.) (b) Similarly, we need to solve p3 = sin y ¡ 2 for ¼ ¼ y . ¡ 2 · · 2 The solution is ¼ p3 ¼ y = , or arcsin = ¡ 3 á 2 ! ¡ 3 Recall that by the de…nition of inverse function, 1 1 f f ¡ (x) = x for x in D f ¡ 1 f¡ (f (x)¢) = x for x in D ¡(f) ¢ Therefore, sin (arcsin x) = x for x in [ 1, 1] ¡ ¼ ¼ arcsin (sin x) = x for x in , ¡ 2 2 h i Example 2 Find following values: 3 ¼ ¼ (a) arcsin sin = sin 1 sin = 4 ¡ 4 ³ ³ 3´¼´ ³ ³ ´´3¼ (b) arcsin sin = sin 1 sin = 4 ¡ 4 µ µ7¼ ¶¶ µ µ7¼ ¶¶ (c) arcsin sin = sin 1 sin = 6 ¡ 6 µ µ ¶¶ µ µ ¶¶ ¼ Solution: (a) According to the formula above, since x = is in ¼ , ¼ 4 ¡ 2 2 ¼ ¼ £ ¤ arcsin sin = 4 4 ³ ³ ´´ 3¼ (b) This time, since x = is not in ¼ , ¼ , the above formula that 4 ¡ 2 2 would have lead to £ ¤ 3¼ 3¼ arcsin sin = doesn’t apply. 4 4 µ µ ¶¶ We thus solve this problem, we need to …nd an angle x in ¼ , ¼ such that ¡ 2 2 3¼ £ ¤ sin x = sin . 4 µ ¶ This can be easily done as follows: 3¼ ¼ ¼ sin = sin ¼ = sin . 4 ¡ 4 4 µ ¶ ³ ´ Thus 3¼ ¼ ¼ arcsin sin = arcsin sin = . 4 4 4 µ µ ¶¶ 7¼ ³ ³ ´´ (c) This time again is outside of the domain of the restricted sine 6 function ¼ , ¼ . according to the method in (b), ¡ 2 2 £ ¤ 7¼ ¼ ¼ ¼ sin = sin ¼ + = sin = sin . 6 6 ¡ 6 ¡ 6 µ ¶ ³ ´ ³ ´ ¼ Now since is in ¼ , ¼ , ¡ 6 ¡ 2 2 £ 7¤¼ ¼ ¼ arcsin sin = arcsin sin = . 6 ¡ 6 ¡ 6 µ µ ¶¶ ³ ³ ´´ 4 Example 3 Find the following values 2 2 (a) sin arcsin = sin sin 1 = 3 ¡ 3 µ µ ¶¶ 1µ µ ¶¶ (b) sin (arcsin 2) = sin sin¡ (2) = 2 Solution: (a) Obviousl¡y, x = i¢s in the domain of arcsin x, [ 1, 1] . 3 ¡ 2 2 sin arcsin = 3 3 µ µ ¶¶ (b) x = 2 is not in the [ 1, 1] , arcsin 2 is unde…ned. So sin (arcsin 2) is not de…ned. ¡ 1 1 Example 4 (a) What is cos sin¡ x ? (b) What is sin 2 sin¡ x ? ¡1 ¢ ¡ ¢ Solution: (a) Set µ = sin¡ x which is de…ned for any 1 x 1. This means ¡ · · ¼ ¼ sin µ = x and µ ¡ 2 · · 2 Thus cos µ 0, and ¸ cos µ = 1 sin2 µ = p1 x2. ¡ ¡ (b) p sin 2µ = 2 sin µ cos µ = 2xp1 x2 ¡ Inverse Cosine function For y = cos x 5 we de…ne the restricted Cosine function as y = cos x for x in [0, ¼] so that its graph is the piece in [0, ¼] Apparently, the restricted Cosine function passes the horizontal line test and thus is invertible. We call the inverse function of the restricted Cosine function inverse Cosine and is denoted by 1 y = cos¡ x or y = arccos x. Analogous to the inverse sine function, there are some basic facts for y = 1 cos¡ x : 1 y = cos¡ x has domain [ 1, 1] and range [0, ¼] ¡ 6 1 cos cos¡ x = x for x in [ 1, 1] 1 ¡ cos¡ (cos x¢) = x for x in [0, ¼] Example 5 Find the values: 1 1 1 (a) cos¡ (0) , (b) cos¡ ¡2 1 µ ¶ Solution: (a) µ = cos¡ (0) is the angle in [0, ¼] satisfying ¼ cos µ = 0 = µ = ) 2 1 1 (b) µ = cos¡ is the solution of ¡2 µ ¶ 1 cos µ = , µ in [0, ¼] ¡2 ¼ 1 Since cos = , 3 2 2¼ ¼ ¼ 1 cos = cos ¼ = cos = . 3 ¡ 3 ¡ 3 ¡2 ³ ´ So 1 1 2¼ µ = cos¡ = ¡2 3 µ ¶ Example 6 compute 2 (a) cos cos 1 , (b) arccos (cos 4) ¡ 3 µ ¶ 2 Solution: (a) Since x = is in the domain of arccos x, 3 2 2 cos cos 1 = ¡ 3 3 µ ¶ (b) Note that x = 4 > ¼ 3.14, which is beyond the domain of the restricted cosine. So the formul'a arccos (cos x) = x doesn’t apply. 7 we need to …nd an angle µ in [0, ¼] such that cos µ = cos 4. To this end cos 4 = cos (2¼ 4) , µ = 2¼ 4 is in [0, ¼] ¡ ¡ and arccos (cos 4) = arccos (cos µ) = µ = 2¼ 4 ¡ 1 1 1 Example 7 (a) What is sin (cos¡ x)?(b) What is sin¡ x + cos¡ x? 1 Solution: (a) Set µ = cos¡ x so µ is in [0, ¼] where sin µ > 0. Thus 1 2 sin cos¡ x = sin µ = p1 cos µ ¡ = p1 x2. ¡ ¢ ¡ 1 (b) Set ® = sin¡ x. Then x = sin ®. Draw a right triangle whose one angle is ®, hypotenuse = 1, and the opposite side is a = x. Then this side x ¼ is adjacent to the third angle in the triangle, i.e., ¯ = ®, and cos ¯ = x/1, 2 ¡ 1 or ¯ = cos¡ x. We conclude ¼ sin 1 x + cos 1 x = ®+ ¯ = ¡ ¡ 2 Inverse function for Tangent Function Since the graph of y = tan x is and the period is ¼. So one entire period 8 passes the horizontal test and thus invertible. The inverse function of this ¼ ¼ restricted piece in , is denoted as ¡ 2 2 ³ ´ 1 y = tan¡ x or y = arctan x. Some basic facts: 1 ¼ ¼ y = tan¡ x has the domain ( , ) and range , ¡1 1 ¡ 2 2 ³ ´ 1 ¼ ¼ tan¡ (tan x) = x only for x in , ¡ 2 2 1 tan tan¡ x = x for all x. ³ ´ ¡ 1 ¢ 1 Example 8 Find (a) tan¡ ( 1) , (b) tan tan¡ p5 ¡ ¡ ¢ 9 ¼ 1 ¼ Solution: (a) Since tan = 1, tan¡ ( 1) = ¡ 4 ¡ ¡ ¡ 4 1 (b) tan tan¡ p5 = p³5 ´ Example 9¡ (a) Give¢n tan µ = x/2, 0 < µ < ¼/2. Express tan 2µ as a func- tion of x. 1 (b) Simplify sec (tan¡ x) Solution: (a) Using double-angle formula, x 2 tan µ 2 4x tan 2µ = = 2 = 1 tan2 µ ³ x´ 2 4 x2 ¡ 1 ¡ ¡ 2 ³ ´ 1 (b) Set µ = tan¡ x. x = tan µ. 1 2 2 sec tan¡ x = sec x = 1 + tan µ = p1 + x ¡ ¢ p Homework: In Exercise 1-10, evaluate each of the quantities that is de…ned. If a quantity is unde…ned, explain it. 1 p3 1. sin¡ = à 2 ! 1 2. cos¡ ( 1) = ¡ 1 3. tan¡ p3 = 1 5 4. sin sin¡ = ¡4 µ µ ¶¶ 4¼ 5. cos 1 cos = ¡ 5 µ µ ¶¶ 1 ¼ 6. tan¡ tan = ¡ 5 ³ ³ ´´ 4¼ 7. sin 1 sin = ¡ 3 µ ¶ 10 3 8. cos cos 1 = ¡ 7 µ µ ¶¶ 3¼ 9. sin sin 1 = ¡ 2 µ µ ¶¶ 1 10. cos (cos¡ 2) = In Exercise 11-15, …nd and simplify the exact value of each quantity. 4 11. tan sin 1 ¡ 5 µ µ ¶¶ 4 12. cos arcsin 9 µ µ ¶¶ 1 13. sin (tan¡ 2) 1 14. sin cos 1 ¡ 4 µ µ ¶¶ 4 15. cos tan 1 ¡ 3 µ ¶ Simplify the following expressions. 1 16. tan (cos¡ x) 1 17. sin (tan¡ x) 1 1 18. cos sin¡ x + cos¡ x ¡ ¢ 11.