Inverse Trig Functions - Cornell University

1 Inverse Trig Functionsc A Math Support Center CapsuleFebruary 12, 2009 IntroductionJust as trig Functions arise in many applications, so do the Inverse trig Functions . Whatmay be most surprising is that they are useful not only in the calculation of angles giventhe lengths of the sides of a right triangle, but they also give us solutions to some commonintegrals. For example, suppose you need to evaluate the following integral: ba1 1 x2dxfor some appropriate values ofaandb. You can use the Inverse sine function to solveit! In this capsule we do not attempt to derive the formulas that we will use; you shouldlook at your textbook for derivations and complete explanations.

2 This material will simplysummarize the key results and go through some examples of how to use them. As usual,all angles used here are in on the Domains of the Trig FunctionsA function must be one-to-one for it to have an Inverse . As we are sure you know, thetrig Functions are not one-to-one and in fact they are periodic ( their values repeatthemselves periodically). So in order to define Inverse Functions we need to restrict thedomain of each trig function to a region in which it is one-to-one but also attains all ofits values. We do this by selecting a specific period for each function and using this as adomain on which an Inverse can be defined.

3 Clearly there are an infinite number of differentrestrictions we could chose but the following are choices that are normally Restricted DomainsFunctionDomainRangesin(x)[ 2, 2][ 1,1]cos(x)[0, ][ 1,1]tan(x)( 2, 2)( , )cot(x)(0, )( , )sec(x)[0, 2) ( 2, ]( , 1] [1, )csc(x)[ 2,0) (0, 2]( , 1] [1, )Definitions of the Inverse FunctionsWhen the trig Functions are restricted to the domains above they become one-to-one func -tions, so we can define the Inverse Functions . For the sine function we use the notationsin 1(x) orarcsin(x). Both are read arc sine . Look carefully at where we have placedthe -1. Written this way it indicates theinverse of the sine function.

4 If, instead, we write(sin(x)) 1we mean the fraction1sin(x). The other Functions are following table summarizes the domains and ranges of the Inverse trig that for each Inverse trig function we have simply swapped the domain and range forthe corresponding trig Restricted DomainsFunctionDomainRangesin 1(x)[ 1,1][ 2, 2]cos 1(x)[ 1,1][0, ]tan 1(x)( , )( 2, 2)cot 1(x)( , )(0, )sec 1(x)( , 1] [1, )[0, 2) ( 2, ]csc 1(x)( , 1] [1, )[ 2,0) (0, 2]We can now define the Inverse Functions more clearly. For thearcsinfunction we definey=sin 1(x) if 1 x 1,yis in [ 2, 2], andsin(y) =x2 Note that this is only defined whenxis in the interval [ 1,1].

5 The other Inverse functionsare similarly defined using the corresponding trig Useful IdentitiesHere are a few identities that you may find 1(x) +cos 1( x) = sin 1(x) +cos 1(x) = 2tan 1( x) = tan 1(x)Practicing with the Inverse FunctionsExample 1:Find the value oftan(sin 1(15).Solution:The best way to solve this sort of problem is to draw a triangle for yourselfusing the Pythagorian Theorem. 512 6 Here we use for the value ofsin 1(15). Notice that we labeled the hypotenuse and theside opposite by using the value of thesinof the angle. We then used the PythagorianTheorem to get the remaining side.)

6 We now have the information that is needed to findtan( ). Sincetan( ) =oppositeadjacent, the answer is1 24=12 6 Example 2:Find the value ofsin(cos 1( 35)).Solution:Look at the following picture: SSSSSSS45- 3In this picture we let =cos 1( 35). Then 0 andcos = 35. Becausecos( )is negative, must be in the second quadrant, 2 . Using the Pythagorean3 Theorem and the fact that is in the second quadrant we get thatsin( ) = 52 325= 25 95=45. Note that although does not lie in the restricted domain we used to definethearcsinfunction, the unrestrictedsinfunction is defined in the second quadrant and sowe are free to use this of Inverse Trig FunctionsThe derivatives of the Inverse trig Functions are shown in the following 1(x)ddx(sin 1x) =1 1 x2,|x|<1cos 1(x)ddx(cos 1x) = 1 1 x2,|x|<1tan 1(x)ddx(tan 1x) =11+x2cot 1(x)ddx(cot 1x) = 11+x2sec 1(x)ddx(sec 1x) =1|x| x2 1,|x|>1csc 1(x)ddx(csc 1x) = 1|x| x2 1,|x|>1In practice we often are interested in calculating the derivatives when the variablexisreplaced by a functionu(x).

7 This requires the use of thechain rule. For example,ddx(sin 1u) =1 1 u2dudx=dudx 1 u2,|u|<1 The other Functions are handled in a similar 1:Find the derivative ofy=cos 1(x3) for|x3|<1 Solution:Note that|x3|<1 if and only if|x|<1, so the derivative is defined whenever|x|< (cos 1(x3)) = 1 1 (x3)2 ddx(x3)= 1 1 (x3)2 (3x2)= 3x2 1 x6 Example 2:Find the derivative ofy=tan 1( 3x).Solution:ddx(tan 1( 3x)) =11 + ( 3x)2 ddx( 3x)=11 + ( 3x)2 12 3x 3=32 3x(1 + 3x)Exercise 1:For each of the following, find the derivative of the given function with respectto the independent variable.(a)y=tan 1t4(b)z=t cot 1(1 +t2)(c)x=sin 1 1 t4(d)s=t 1 t2+cos 1t(e)y=sin 1 x(f)z=cot 1(y1 y2)5 Solutions.

8 (a)y=tan 1t4dydt=ddttan 1(t4)=11 + (t4)2 ddt(t4)=4t31 +t8(b)z=t cot 1(1 +t2)dzdt=ddtt cot 1(1 +t2)=cot 1(1 +t2) +t 11 + (1 +t2)2 (2t)=cot 1(1 +t2) 2t2t4+ 2t2+ 2(c)x=sin 1 1 t4dxdt=ddtsin 1 1 t4=1 1 ( 1 t4)2 ddt( 1 t4)=1 1 (1 t4) 12 (1 t4) 12 ( 4t3)=1 1 1 +t4 1 1 t4 ( 2t3)=1t2 1 1 t4 ( 2t3)= 2t 1 t46(d)s=t 1 t2+cos 1tdsdt=ddtt 1 t2+ddtcos 1t=( 1 t2) 1 t 12(1 t2) 12 ( 2t)( 1 t2)2+ 1 1 t2= 1 t2+t2 1 t2(1 t2) 1 1 t2=( 1 t2)( 1 t2) +t2( 1 t2)(1 t2) (1 t2)(1 t2) 1( 1 t2)=(1 t2) +t2 (1 t2)( 1 t2)(1 t2)=t2(1 t2)32(e)y=sin 1 xdydx=ddxsin 1 x=1 1 ( x)2 ddx x=1 1 x 12x 12=12 x(1 x)7(f)z=cot 1(y1 y2)dzdy=ddycot 1(y1 y2)= 11+(y1 y2)2 ddy(y1 y2)= 1(1 y2)2+y2(1 y2)2 ddy(y1 y2)= (1 y2)2(1 y2)2+y2 (1 y2) 1 y ( 2y)(1 y2)2= 1(1 y2)2+y2 (1 y2) 1 y ( 2y)1= 1(1 y2+2y2)1 2y2+y4+y2= (1+y2)1 y2+y41 Solving IntegralsThe formulas listed above for the derivatives lead us to some nice ways to solve somecommon integrals.

9 The following is a list of useful ones. These formulas hold for anyconstanta6= 0 du a2 u2=sin 1(ua) +Cforu2< a2 dua2+u2=1atan 1(ua) +Cfor allu duu u2 a2=1asec 1|ua|+Cfor|u|> a >0 Exercise 2:Verify each of the equations above by taking the derivative of the right now want to use these formulas to solve some common 1:Evaluate the integral dx 9 16x2 Solution:Leta= 3 andu= 4x. Then 16x2= (4x)2=u2anddu= 4dx. We get thefollowing for 16x2<9:8 dx 9 16x2=14 du a2 u2=14sin 1(ua) +C=14sin 1(4x3) +C=14sin 1(43x) +CExercise 3:Evaluate the following integrals.(a) dx 25 4x2(b) dy36+4y2(c) z dz5+2z4(d) sin x dx 10 cos2x(e) dx 5+4x x2(f) 7dx25 12x+4x2 Solutions:(a) dx 25 4x2.

10 For this problem use the formula du a2 u2=sin 1ua+Cwitha= 5,u= 2xanddu= 2dx, giving you dx 25 4x2=12 du a2 u2=12sin 1(2x5) +C(b) dy36+4y2. Use the formula dua2+u2=1atan 1(ua) +Cwitha= 6,u= 2yanddu= gives us dy36+4y2=12 dua2+u2= (12)(16)tan 1(2y6) +C=112tan 1(y3) +C(c) z dz5+2z4. In order to make the calculations a bit simpler, it is useful to multiplythe numerator and denominator by 2 in order to get the term 4z4instead of 2z4in thedenominator. This gives us z dz5+2z4= 2z dz10+ letu= 2z2,du= 4z dzanda= 10 and we have z dz5+2z4=12 4z dz10+4z4=12 du( 10)2+u2=12 10tan 1(2z2 10) +C(d) sin x dx 10 cos2x. Letu=cos x,du= sin x dxanda= 10.