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Dr. Neal, WKU MATH 117 Inverse Trig Functions A function y = f (x) is one–to–one if it is always the case that different x values are assigned to different y values. 3 For example, y = x + 4 is one–to–one, but y = sin x is not one–to–one. The sine function assigns many different x values to the same y . For instance 0, π, 2π, 3π, etc. all have a sine of 0. 3 y = sin x is not one-to-one y = x +4 is one-to-one A one–to–one function f (x) always has an inverse denoted by f !1(x) . The trigonometric functions cos x , sin x , tan x are not one-to-one; thus technically they do not have inverses. However we can make them one-to-one by restricting the domains to only two quadrants so that we have only one segment of the graph. The Inverse Cosine (or Arccos) With cos x , we restrict the domain to [0, π] (the 1st and 2nd quadrants). Then the !1 “inverse cosine” exists: For !1 " x " 1 , cos x is the angle whose cosine is x , where the angle must be in either the 1st or 2nd quadrant and must be written in radians. ! 1 ! –1 f (x) = cos x Dom: 0 ≤ x ≤ π Range: –1 ≤ y ≤ 1 1 –1 !1 f !1(x) = cos x Dom: –1 ≤ x ≤ 1 Range: 0 ≤ y ≤ π We know the cosine values of the standard angles around the unit circle: 5! 6 ! 2! 3! 3 4 ! 2 ! 3 ! 4 cos(0) = 1 " !% 1 cos$ ' = # 3& 2 ! 6 0 " 3! % 2 cos$ ' = – # 4& 2 1 " !% 3 cos$ ' = # 6& 2 " !% cos$ ' = 0 # 2& " !% 2 cos$ ' = # 4& 2 " 2! % 1 cos$ ' = – # 3& 2 " 5! % 3 cos$ ' = – # 6& 2 cos(!) = – Dr. Neal, WKU So we can compute the inverse cosine, or arccos, of the values 1, 1 3 2 1 2 3 , , , 0, ! , ! ,– , –1 : 2 2 2 2 2 2 ! ! 3 2 2 2 ! 1 2 0 1 2 2 2 3 2 1 !1 cos !1 x is the angle between 0 and π whose cosine is x " 3% ! cos !1 $$ '' = # 2 & 6 cos !1(1) = 0 " 2% ! '' = cos !1 $$ # 2 & 4 " 2 %' 3! cos !1 $$ ! ' = # 2 & 4 " 1 % 2! cos !1 $ ! ' = # 2& 3 " 1% ! cos !1 $ ' = # 2& 3 " 3% 5! cos !1 $$ ! '' = # 2 & 6 cos !1(0) = ! 2 cos !1(!1) = ! Evaluating cos(cos!1 x) and cos !1 (cos x) It is always the case that cos(cos !1 x ) = x for !1 " x " 1 . !1 This is because cos x is the angle whose cosine is x ; thus, cos(cos !1 x ) = x . We don't !1 even have to calculate cos x in this case. For example, ( ) cos cos !1 (!1 / 2) = ! ( !1 ) ( Also, cos cos (5) and cos cos the interval !1 " x " 1 . !1 1 2 ( ) and cos cos !1 ( 3 / 2) = 3 2 ) (!4) are undefined because 5 and –4 are not within On the other hand, cos !1 (cos x ) = x only for 0 ! x ! " . In this case, we can start with any angle x . Then !1 " cos x " 1. So cos !1 (cos x ) exists, but cos !1 (cos x ) must be an angle from 0 to π and may not equal the original angle x . Dr. Neal, WKU Example. Compute (a) cos !1 #% $ cos 7" & ( 6 ' (b) cos !1 #% $ cos 5" & ( 3 ' (c) cos # 3" & & % ! (( . cos $ $ 4 '' !1 #% Solution. We first evaluate the cosine of the inner given angle, then evaluate the inverse cosine of the result. The answers must all be angles from 0 to π. (a) cos (b) cos !1 #% $ cos !1 #% $ cos 5" & # 1& " ( = cos !1 % ( = ' $ 2' 3 3 # 7" & 3 & 5" ( = cos!1 % ! (= $ 2 ' 6 ' 6 (c) cos # # 3" & & 2 & 3" % ! ( ( = cos !1 % ! (= cos $ $ 4 '' $ 2 ' 4 !1 #% The Inverse Sine (or Arcsin) With sin x , we restrict the domain to [–π/2, π/2] (the 1st and 4th quadrants). Now the !1 “inverse sine” exists: For !1 " x " 1 , sin x is the angle whose sine is x , where the angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant, then we write it as a negative angle. " 2 1 ! " " 2 2 !1 1 !1 ! f (x) = sin x " ! Dom: ! ≤ x ≤ Range: –1 ≤ y ≤ 1 2 2 " 2 " 3 " 4 " ! 2 ! f !1(x) = sin !1 x Dom: –1 ≤ x ≤ 1 Range: ! " ! ≤y ≤ 2 2 We know the sine values of the standard angles from –π/2 to π/2: " 6 0 " ! 3 " 2 " 4 " ! 6 # "& sin% ! ( = –1 $ 2' # "& 3 sin% ! ( = ! $ 3' 2 # "& 1 sin% ! ( = ! $ 6' 2 " !% 2 sin$ ' = # 4& 2 sin(0 ) = 0 " !% 3 sin$ ' = # 3& 2 # "& 2 sin% ! ( = ! $ 4' 2 " !% 1 sin$ ' = # 6& 2 " !% sin$ ' = 1 # 2& Dr. Neal, WKU So now we can compute the inverse sine, or arcsin, of the values –1, ! 1 1 3 2 2 3 ,! , ! , 0, , , ,1 : 2 2 2 2 2 2 1 3 2 2 2 1 2 0 ! ! !1 3 2 sin !1 x is the angle between ! sin !1( !1) = ! sin !1( 0) = 0 " 3% " sin !1 $$ ! '' = ! # 2 & 3 " 2 ! 2 2 " ! and whose sine is x 2 2 " 2 %' " sin !1 $$ ! ' =! # 2 & 4 " 2% ! '' = sin !1 $$ # 2 & 4 "1 % ! sin !1 $ ' = #2 & 6 1 2 " 1% " sin !1 $ – ' = ! # 2& 6 " 3% ! sin !1 $$ '' = # 2 & 3 sin !1(1) = ! 2 Evaluating sin(sin"1 x) and sin"1 (sin x) It is always the case that !sin(sin !1 x ) = x! for !1 " x " 1 . !1 This is because sin x is the angle whose sine is x ; thus, sin(sin !1 x ) = x . We don't !1 even have to calculate sin x in this case. For example, ( ) 2 2 !1 ) sin sin !1 ( 2 / 2) = ( !1 ) ( ( !1 ) and sin sin (!1 / 2) = ! 1 2 Also, sin sin (!3) and sin sin (2. 4) are undefined because –3 and 2.4 are not within the interval !1 " x " 1 . On the other hand, " " sin !1 (sin x ) = x only for ! # x # . 2 2 In this case, we can start with any angle x . Then !1 " sin x " 1. So sin !1 (sin x ) exists, but sin !1 (sin x ) must be an angle from –π/2 to π/2 and may not equal the original angle x . Dr. Neal, WKU 4 "& !1 # ( Example. Compute (a) sin % sin $ 3 ' 5" & !1 # (b) sin % sin ( $ 4 ' (c) sin # 7" & & % ((. sin $ $! 6 '' !1 #% Solution. We first evaluate the sine of the inner given angle, then evaluate the inverse sine of the result. The answers must all be angles from –π/2 to π/2. # 4 "& 3& " !1 # ( = sin !1 % ! (=! (a) sin %$ sin $ 2 ' 3 ' 3 5" & 2& " !1 # !1 # ( =! (b) sin %$ sin (' = sin % ! $ 2 ' 4 4 # 7" & & " !1 # !1 # 1 & (c) sin % sin % ! ( ( = sin % ( = $ $ 6 '' $2' 6 The Inverse Tangent (or Arctan) With tan x , we restrict the domain to (–π/2, π/2) (the 1st and 4th quadrants). Then the !1 “inverse tangent” exists: For all x , tan x is the angle whose tangent is x , where the angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant, then we write it as a negative angle. f (x) = tan x " ! < x < 2 2 Range: – ∞ < y < ∞ Domain: ! f !1(x) = tan !1 x Domain: – ∞ < x < ∞ " ! Range: ! < y < 2 2 Dr. Neal, WKU " 2 " 3 We know the tangent values of the standard angles from –π/2 to π/2: " 4 " 6 # "& tan % ! ( = – ∞ $ 2' # "& tan % ! ( = ! 3 $ 3' # "& tan % ! ( = –1 $ 4' 0 ! ! " ! 3 " 2 ! " 4 # "& 1 tan % ! ( = ! $ 6' 3 " 6 " !% tan $ ' = 1 # 4& " !% 1 tan $ ' = # 6& 3 tan ( 0) = 0 " !% tan $ ' = 3 # 3& " !% tan $ ' = ∞ # 2& So now we can compute the inverse tangent, or arctan, of the values –∞, ! 3 , !1 , ! 1 1 , 0, , 1, 3 , ∞ : 3 3 ! 3 1 1 3 0 1 3 " "1 " 3 "! tan !1 x is the angle between ! tan !1 (!") = ! tan !1 (0) = 0 " 2 tan !1 (! 3 ) = ! " 1 % ! tan !1 $ ' = # 3& 6 " 3 " ! and whose tangent is x 2 2 tan !1 (!1) = ! tan !1 (1) = ! 4 tan !1 " 1 % " tan !1 $ ! ' = ! # 3& 6 " 4 ( 3 ) = !3 tan !1 (" ) = ! 2 Evaluating tan(tan"1 x) and tan!1 (tan x) It is always the case that ! tan (tan !1 x ) = x for all x . !1 This is because tan x is the angle whose tangent is x ; thus, tan (tan !1 x ) = x . We don't !1 even have to calculate tan x in this case. For example, ( !1 ) ( !1 ) tan tan (20.4) = 20. 4 and tan tan (!0. 56) = !0. 56 Dr. Neal, WKU On the other hand, tan !1 (tan x ) = x only for ! " " <x< . 2 2 In this case, we can start with any angle x . Then ! " < tan x < " . So tan !1 (tan x ) exists, but tan !1 (tan x ) must be an angle between –π/2 and π/2 and may not equal the original angle x . Example. Compute (a) tan !1 #% $ tan 5" & ( 6 ' (b) tan !1 #% $ tan 7" & ( 4 ' (c) tan # 2" & & % (( . tan $ $ ! 3 '' !1 #% Solution. We first evaluate the tangent of the inner given angle, then evaluate the inverse tangent of the result. The answers must all be angles from –π/2 to π/2. (a) tan (b) tan !1 #% $ tan !1 #% $ tan 5" & # 1 & " ( = tan !1 % ! (=! ' $ ' 6 3 6 7" & " ( = tan !1 ( !1) = ! 4 ' 4 (c) tan # 2" & & " ( ( = tan !1( 3 ) = tan % ! $ $ 3 '' 3 !1 #%