University Calculus I. Worksheet # 8 Mar b. sin tan e. sin 2 sin 1 5. b. tan. c. sec sin 1 ( x )) cos 1 ( x )) f. csc. c.

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1 MATH 6 WINTER 06 University Calculus I Worksheet # 8 Mar The topic covered by this worksheet is: Derivative of Inverse Functions and the Inverse Trigonometric functions. SamplesolutionstoallproblemswillbeavailableonDL, Thursday 0 th.pleasereportanytypos,omissionsanderrorsto aiffam@ucalgary.ca 0. Let fx be a function such that f 4 and f. Assuming that fx hasan inverse, find f Find f, if a. fx x x b. fx x x Evaluate each of the following, when possible. a. sin π/ b. tan c. cos / d. sin / e. sec f. sec / 04. Evaluate a. cos sin 8 b. sin tan d. cos sec e. sin sin c. sec cos 4 f. cos cot 0. Given 0 <x<, simplify each of the following a. sin cos x b. tan sin x c. sin tan x 06. Given <x<, simplify a. cos sin x b. tan d. sin cos x e. cot sin x c. sec sin x cos x cos x f. csc 0. Find the exact value of sin sin in each of the following cases a. 9 π b. 6 π c. π 08. Find the exact value of cos cos in each of the following cases a. 0 π b. π c. 8 π

2 09. Compute the exact value of tan tan in each of the cases a. 6 π b. 9 π c. π 0. Compute y and simplify your answer. x + x + a. y sin b. y tan c. y sec x. Compute y and simplify your answer. a. y x 4 + tan x x b. y x sin x+ x c. y sec x + d. y x tan x. Compute y and simplify your answer. a. y x + tan x + ln x + b. y sin x +x 4 cos x x x 4 c. y x sin x + x d. y x / sin x 4 x / + 4 x Sample Solutions Recall the derivative of the inverse function f of the function f, at the point x c, is given by f c f f c provided f f c 0 In order to apply the above formula, you will need to evaluate a f c fa c compute the derivative f x evaluate f a

3 0. We have f 4. We need to evaluate f f f 4. 4 To do so, we start with f 4. Applying f to both sides, we get f f f 4 f 4 Thus f 4 f f 4 f 0a. We have f To evaluate this expession, we need a f f f, f x, and f a. To compute a f, apply f to both sides fa f f a a a a +a +60 since a +a +6 0 a Thus f a. Now, fx x x f x x, and f f f 9 4 It follows f f f f 4 0b. Start by finding a f fa f f equivalent to { a a +8 9 a>0 a a +8 This last equation is { a 4 +8a { 90 a a +9 0 a>0 a>0 { a a +a +90 a a>0 Hence f. Next we compute f x. We have x f x x +8+x x x x +8 + x +8+ x +8 x +8 x +8 x +8 x +8 Then f f f f 0 0

4 The table below lists the domain, range and derivative of the six inverse trigonometric functions. Domain Range Useful Formula Derivative sin x [, ] [ π, π ] sin x sin x x cos x [, ] [0,π] cos x π sin x x tan x, + π, π tan x tan x x + cot x, + 0,π cot x π tan x x + sec x, ] [, + [0, π π,π] sec x cos x x x csc x, ] [, + [ π, 0 0, π ] csc x sin x x x 0a. Clearly π/ >. Therefore π/ is not in the domain of sin. We conclude that sin π/ is not defined. 0b. First, notice that belongs to the domain of tan. Setting tan and applying tan, we tan get with π π/ <<π/ 0c. Here belongs to the domain of cos. Setting cos / and applying cos, we get cos / with π 0 π 0d. Here / belong to the domain of cos. Setting sin / and applying sin, we get sin / with π π/ π/ 6 0e. do belong to the domain of sec. To evaluate, we make use the identity sec x cos /x, to get sec cos /. Setting cos /, and applying cos to both sides leads to cos / with π 0 π 4 0f. Clearly / is not in the domain of sec x. Hence sec / can not be evaluated. 4

5 The idea when evaluating the trig of an inverse trig, say tan sin a, is to b set sin a b apply sin to both sides and simplify to get sin a b next, draw a right triangle with angle, and label the sides so that sin a b a b b a use pythagoras theorem to solve for the missing side read out tan sin a b tan opposite adjacent a b a All this works, provided 0 < < π a b > 0. If that is not the case, i.e., a b < 0, then the formula sin x sin x should be used first. 04a. Set sin sin. Next make use of the right triangle cos sin cos 4 4 to get 8 04b. Set tan 8 get sin tan 8 tan 8. Next make use of the right triangle sin 8 to 04c. Setting cos 4 sec cos 4 sec cos 4.Itfollows cos cos 4/ 0a. Set cos x cos x. Next make use of the right triangle to get sin cos x sin x x / 0b. Set sin x sin x. Next make use of the right triangle to get tan sin x tan x x / x x / x /

6 0c. Making use of the fact that both sin and tan are odd functions, we can write sin tan x sin tan x sin tan x Setting tan x tan x. x / +x / Making use of the right triangle shown on the right, we have sin tan x sin tan x sin x +x 06. a. cos sin x cos sin x cos sin x sin sin x sin sin x x b. tan sin x sin sin x cos sin x x x Wehavemadeuseofquestiona. above c. sec sin x cos sin x x Wehavemadeuseofquestiona. above The above was an analytic way of solving the problem. Its advantage over the geometric way used in the previous problem, is the fact that you do not need to assume x>0. Of course you could solve the problem by setting x sin x, x > 0 sin x, and make use of the triangle shown to the right. But then you will have to deal the case x<0, separately. x 0. The idea is to reduce the angle to an angle in [ π/,π/] so that the cancelation identity sin sinx x may be used. Start by dropping the multiple of π part of your angle. If the resulting angle, say α is in the first quadrant, then you are done. If α lies in the second or third quadrant, then use the identity sin α sin π α [ to bring it to π, π ]. Finally, if α [ is in the fourth quadrant, subtract π to bring it to π ], 0 9π 0a. sin 6π 0b. sin sin 6 π + π π sin. Hence 9π sin sin sin 4 π + 4π sin 4 π π sin sin π. Here identity sin α sin π α to get 6π 4 π sin sin Hence 6π sin sin 4 π, lies in the third quadrant. We make use the sin π 4 π sin π sin sin π π 6

7 π 0c. sin from the angle, to get 08. Hence sin 0 π + π sin π π sin sin. Here π, lies in the fourth quadrant. We subtract π π π sin π sin π π sin sin sin sin π π The idea is to reduce the angle to an angle in [0,π], so that the cancelation identity cos cosx x may be used. Start by dropping the multiple of π part of your angle. If the resulting angle, say α is in the first or second quadrants, then you are done. If α lies in the third or fourth quadrant, then use the identity cos α cos π α [ ] to bring it to 0,π. 0π 08a. cos cos 6 π + π cos π 08b. cos π 0π cos cos cos 4 π + π cos π identity cos α cos π α to get Hence π cos cos 8π 08c. cos cos 8 π + 4π 4 π cos 09. identity cos α cos π α to get Hence, with 0 < π cos cos <π. Hence π π. Here π, lies in the fourth quadrant. We make use of the π cos π π π cos π cos cos π cos cos π. Here 4 π, lies in the third quadrant. We make use of the 8π 4 π cos cos cos π 4 π π cos 8π cos cos cos cos π π The idea is to reduce the angle to an angle in π/,π/, so that the cancelation identity tan tanx x may be used. Start by dropping the multiple of π part of your angle. If the resulting angle, say α is in the first quadrant, then you are done. If α lies in the second quadrant, then use the identity tan α tan α π ] to bring it to π/, 0.

8 6π 09a. tan 9 π 09b. tan tan + π tan π + π π tan. Hence 6π tan tan tan + 4 π tan π + 4 π 9 π tan tan π tan tan π tan 4 π tan tan 4 π tan π π π 09c. tan π tan π π + π tan. It follows tan tan π π tan tan π 0a. y x + 0b. y x + + x + x + 4 x x 4 4 x +4x + 0c. y x x x x x 4 x x x 4 tan π. Hence a. y 4x tan x + x x x x 4x tan x b. y sin x+x x + x x sin x c. We have y x + x + x + 4 x +4x x + 4x + x x + 4 x + x You can also rewrite y as y cos x + x + x + x, and differentiate. 8

9 d. Rewrite y as y e lnx tan x e lnx tan x.itfollows y e lnx tan x lnx tan x x tan x x tan x tan x x You can also use logarithmic differentiation: y x tan x Differentiating, leads to + lnx +x ln y ln y y x + lnx+tan x x y y x tan x+lnx +x x tan x ln y tan x lnx y x tan x lnx x + + tan x x lnx x + + tan x x a. y x x + tan x + + x + x + x + x x + tan x + x + +x + x x + x x + tan x + x x + x + x b. y x x 4 +8x cos x +x 4 x x x 4 x 4 x x 4 x 4 x x 4 +8x cos x 4 x x x4 + x x 4 x 4 x 4 x x 4 +8x cos x + 4 x x x 4 +x x 4 x x 4 +8x cos x + x x 4 8x cos x c. y x sin x + x sin x x + x x sin x x x x + x 9

10 d. y x / sin x + x / x / 4 x + 4 x x sin x + x x + 4 x 4 x x x sin x + 6 x x 4 x x sin x + 4 x + 4 x 4 x x sin x 4 x + 4 x x sin x 0