Inverse Trig Functions

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1 Inverse Trig Functions c A Math Support Center Capsule February, 009 Introuction Just as trig functions arise in many applications, so o the inverse trig functions. What may be most surprising is that they are useful not only in the calculation of angles given the lengths of the sies of a right triangle, but they also give us solutions to some common integrals. For example, suppose you nee to evaluate the following integral: b a x x for some appropriate values of a an b. You can use the inverse sine function to solve it! In this capsule we o not attempt to erive the formulas that we will use; you shoul look at your textbook for erivations an complete explanations. This material will simply summarize the key results an go through some examples of how to use them. As usual, all angles use here are in raians. Restrictions on the Domains of the Trig Functions A function must be one-to-one for it to have an inverse. As we are sure you know, the trig functions are not one-to-one an in fact they are perioic (i.e. their values repeat themselves perioically). So in orer to efine inverse functions we nee to restrict the omain of each trig function to a region in which it is one-to-one but also attains all of its values. We o this by selecting a specific perio for each function an using this as a omain on which an inverse can be efine. Clearly there are an infinite number of ifferent restrictions we coul chose but the following are choices that are normally use.

2 Stanar Restricte Domains Function Domain Range sin(x) [ π, π ] [, ] cos(x) [0, π] [, ] tan(x) ( π, π ) (, ) cot(x) (0, π) (, ) sec(x) [0, π ) ( π, π] (, ] [, ) csc(x) [ π, 0) (0, π ] (, ] [, ) Definitions of the Inverse Functions When the trig functions are restricte to the omains above they become one-to-one functions, so we can efine the inverse functions. For the sine function we use the notation sin (x) or arcsin(x). Both are rea arc sine. Look carefully at where we have place the -. Written this way it inicates the inverse of the sine function. If, instea, we write (sin(x)) we mean the fraction sin(x). The other functions are similar. The following table summarizes the omains an ranges of the inverse trig functions. Note that for each inverse trig function we have simply swappe the omain an range for the corresponing trig function. Stanar Restricte Domains Function Domain Range sin (x) [, ] [ π, π ] cos (x) [, ] [0, π] tan (x) (, ) ( π, π ) cot (x) (, ) (0, π) sec (x) (, ] [, ) [0, π ) ( π, π] csc (x) (, ] [, ) [ π, 0) (0, π ] We can now efine the inverse functions more clearly. For the arcsin function we efine y sin (x) if x, y is in [ π, π ], an sin(y) x

3 Note that this is only efine when x is in the interval [, ]. The other inverse functions are similarly efine using the corresponing trig functions. Some Useful Ientities Here are a few ientities that you may fin helpful. cos (x) + cos ( x) π sin (x) + cos (x) π tan ( x) tan (x) Practicing with the Inverse Functions Example : Fin the value of tan(sin ( 5 ). Solution: The best way to solve this sort of problem is to raw a triangle for yourself using the Pythagorian Theorem. 5 θ 6 Here we use θ for the value of sin ( 5 ). Notice that we labele the hypotenuse an the sie opposite θ by using the value of the sin of the angle. We then use the Pythagorian Theorem to get the remaining sie. We now have the information that is neee to fin tan(θ). Since tan(θ) opposite ajacent, the answer is 4 6 Example : Fin the value of sin(cos ( 3 5 )). Solution: Look at the following picture: In this picture we let θ cos ( 3 5 ). Then 0 θ π an cosθ 3 5. Because cos(θ) is negative, θ must be in the secon quarant, i.e. θ π. Using the Pythagorean 3 π θ

4 Theorem an the fact that θ is in the secon quarant we get that sin(θ) Note that although θ oes not lie in the restricte omain we use to efine the arcsin function, the unrestricte sin function is efine in the secon quarant an so we are free to use this fact. Derivatives of Inverse Trig Functions The erivatives of the inverse trig functions are shown in the following table. Function sin (x) cos (x) tan (x) cot (x) sec (x) csc (x) Derivatives Derivative x (sin x) x, x < x (cos x), x < x x (tan x) +x x (cot x) +x x (sec x) x, x > x x (csc x) x x, x > In practice we often are intereste in calculating the erivatives when the variable x is replace by a function u(x). This requires the use of the chain rule. For example, x (sin u) u u x The other functions are hanle in a similar way. u x u, u < Example : Fin the erivative of y cos (x 3 ) for x 3 < Solution: Note that x 3 < if an only if x <, so the erivative is efine whenever x <. 4

5 x (cos (x 3 )) (x 3 ) x (x3 ) (3x ) (x 3 ) 3x x 6 Example : Fin the erivative of y tan ( 3x). Solution: x (tan ( 3x)) + ( 3x) x ( 3x) + ( 3x) 3x 3 3 3x ( + 3x) Exercise : For each of the following, fin the erivative of the given function with respect to the inepenent variable. (a) y tan t 4 (b) z t cot ( + t ) (c) x sin t 4 () s t t + cos t (e) y sin x (f) z cot ( y ) y 5

6 Solutions: (a) y tan t 4 y t t tan (t 4 ) + (t 4 ) t (t4 ) 4t3 + t 8 (b) z t cot ( + t ) z t t t cot ( + t ) cot ( + t ) + t cot ( + t ) + ( + t ) (t) t t 4 + t + (c) x sin t 4 x t t sin t 4 ( t 4 ) t ( t 4 ) ( t 4 ) ( t4 ) ( 4t 3 ) + t 4 t 4 ( t3 ) t t 4 ( t3 ) t t 4 6

7 () s t t + cos t s t t t t + t cos t ( t ) t ( t ) ( t) ( + t ) t t + ( t ) t t ( t )( t ) + t ( t )( t ) ( t ) + t ( t ) ( t )( t ) t ( t ) 3 t ( t ) ( t ) ( t ) (e) y sin x y x x sin x ( x) x x x x x( x) 7

8 (f) z cot ( y ) y z y y cot ( y ) y y +( y ) ( y ) +y ( y ) y ( y ) y y ( y ) y ( y ) ( y ) +y ( y ) +y ( y +y ) y +y 4 +y (+y ) y +y 4 ( y ) y ( y) ( y ) ( y ) y ( y) Solving Integrals The formulas liste above for the erivatives lea us to some nice ways to solve some common integrals. The following is a list of useful ones. These formulas hol for any constant a 0 u a u sin ( u a ) + C for u < a u a +u a tan ( u a ) + C for all u u u u a a sec u a + C for u > a > 0 Exercise : Verify each of the equations above by taking the erivative of the right han sie. We now want to use these formulas to solve some common integrals. Example : Evaluate the integral x 9 6x Solution: Let a 3 an u 4x. Then 6x (4x) u an u 4x. We get the following for 6x < 9: 8

9 x 9 6x 4 u a u 4 sin ( u a ) + C 4 sin ( 4x 3 ) + C 4 sin ( 4 3 x) + C Exercise 3: Evaluate the following integrals. (a) (b) (c) () (e) (f) x 5 4x y 36+4y z z 5+z 4 sin x x 0 cos x x 5+4x x 7 x 5 x+4x Solutions: (a) x u 5 4x. For this problem use the formula a 5, u x an u x, giving you (b) y. Use the formula u 36+4y an u y. This gives us y 36+4y a +u a u x 5 4x sin u a + C with u a u sin ( x 5 ) + C a tan ( u a ) + C with a 6, u y u ( a +u )( 6 ) tan ( y 6 ) + C tan ( y 3 ) + C (c) z z. In orer to make the calculations a bit simpler, it is useful to multiply 5+z 4 the numerator an enominator by in orer to get the term 4z 4 instea of z 4 in the enominator. This gives us z z z z. 5+z 4 0+4z 4 Now let u z, u 4z z an a 0 an we have z z 4 z z 5+z 4 u 0+4z 4 ( 0) +u 0 tan ( z 0 ) + C () sin x x 0 cos x. Let u cos x, u sin x x an a 0. Then sin x x 0 cos x ( ) sin x x 0 cos x sin ( cos x 0 ) + C 9

10 (e) form x 5+4x x. We want to transform this expression into something with the u a. To o this we nee to complete the square of the expression in the u enominator as follows: 5 + 4x x x x x x 9 (x 4x + 4) (3) (x ) This gives us x 5+4x x x (3) (x ) sin ( x 3 ) + C (f) 7 x. Again we nee to complete the square. This time we want to transform 5 x+4x the expression into something with the form u. We rewrite the enominator as a +u follows: 5 x + 4x x + 4x (4) + (x 3) Now, letting u x 3 an u x we get 7 x 5 x+4x 7 x 7 (4) +(x 3) 8 tan ( x 3 4 ) + C 0

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