6.6 Inverse Trigonometric Functions

Transcription

1 Inverse Trigonometric Functions We recall the following definitions from trigonometry. If we restrict the sine function, say fx) sinx, π x π then we obtain a one-to-one function. π/, /) π/ π/ Since f is one-to-one, f exists. Notice also that f is differentiable with f x) cosx. So, for example, the slope of the tangent line through the point shown is cosπ/ /. Now by the Inverse Function theorem, f is differentiable and we expect that ) f ) /) /

2 6.6 Definition. ) Inverse Sine also called arcsine) y sin x iff siny x x and π y π π/ π/ Notice that the inverse sine appears to be differentiable everywhere except at the end points. Why? Example. Evaluate the following limit. lim x +arcsinx π/

3 6.6 We give the definitions of two more inverse trig functions. Definition. ) Inverse Tangent also called arctangent) y tan x iff tany x π < y < π π/ π/ Notice that the domain for the inverse tangent is, ) and that the function appears to be differentiable everywhere. Example. Notice that End Behavior of Arctangent lim x tan x π/

4 6.6 4 Definition. 4) Inverse Secant y sec x iff secy x x and y [0,π/) [π,π/) Also, see Figure in section 6.6 of the text. π Example. Evaluate the limit below. lim x sec x π

5 6.6 5 Example 4. Evaluate each of the following without using a calculator. a. sin ) b. arctan c. sinsec x), x Observe that 5) sin sec ) ) sinπ 0 Now suppose that x < and let θ sec x. Then secθ x and π/ < θ < π why?). So let θ be the reference angle of θ. Then θ is acute and secθ x. It follows that sin sec x ) sinθ sinθ, since π/ < θ < π) 6) x x x x Notice that 6) agrees with 5) when x. Thus, for x, we have sin sec x ) x x d. costan z) +z Remark. Example 4 is very important. Please be sure that you understand it.

6 6.6 6 The Derivative of Arcsine If we let f x) sin x, x,) then fy) siny, y π/,π/) is differentiable and never zero. So by the Inverse Function theorem, y sin x is differentiable. In other words, y is a differentiable function of x in the expression below. siny x Now differentiate by sides of the equation with respect to x to get In other words, d cosy dy dy cosy cos sin x ) sin x ) x x So for example, π/ f ) /) /) /,π/). as we saw in ).

7 6.6 7 The Derivative of Arctangent We repeat the procedure for arctangent. Let y tan x. Then y is a differentiable function of x in the expression below. Thus tany x sec y dy dy sec y sectan x)) +x ) +x The Derivative of Arcsecant In a similar manner, we can also derive the following formula d sec x ) x x

8 6.6 8 We now summarize the general derivative formulas for the primary inverse trig functions. 7) 8) d d sin u ) u tan u ) du +u du, u < 9) d sec u ) u u du, u > Example 5. Find the derivatives of each of the following. a. gx) arctanln x) g x) +lnx) x b. Fx) sin x 0 lnt+)dt By the Fundamental Theorem, F x) ln sin x+ ) d ln sin x+ ) x sin x )

9 6.6 9 c. y sin / x) dy d ) / x / x) ) /x x / x x An alternative approach: We rewrite the original equation. xsiny Now differentiate both sides with respect to x to obtain Now cosy x )/x why?), hence as we saw above. 0 siny x + xcosy dy dy siny x x xcosy x x x x x

10 6.6 0 d. w costan z) Although we can use the chain rule, we might try rewriting the expression first. cos tan z ) +z It follows that dw dz z) +z / ) z +z ) / It is worthwhile to compare the above result with a direct computation. Thus as we saw above! dw dz sin tan z ) +z z +z +z

11 6.6 The Integration Formulas The derivative formulas above immediately yield the following integration formulas. If a 0 then 0) du a u sin u a +C, u < a ) ) du a +u a tan u a +C du u u a u a sec +C, u > a > 0 a

12 6.6 Example 6. Evaluate the integrals. a. 9x Let u x. Then du and 9x u du sin u+c sin x)+c Remark. It is incorrect to use 0) directly. du, with u x) 9x u This is wrong! Do you see why? Instead, we must rewrite as 9x, x) sin x)+c

13 6.6 b. 9+x Before we try u-substitution, let s rewrite the expression as 9 + x/ ) Now let u x/. Then du / so that 9 +u du 9 tan u+c 9 tan x/ ) +C Let s check this one. d 9 tan x/ ) ) 9 9 d 9+x tan x/ )) + x/ )

14 6.6 4 c. A look ahead. Let s revisit the last example with a eye towards a different type of substitution. This time let x tanθ. Then sec θdθ and 9+x 9+ tanθ ) sec θdθ sec θ 9+9tan θ dθ sec θ 9 +tan θ dθ sec θ 9 sec θ dθ dθ 9 9 θ +C Now what? Observe that θ tan x/ ). So, as we saw above, we get 9+x 9 θ +C 9 tan x/ ) +C We will have more to say about this technique, called trigonometric substitution, in the next chapter.

15 6.6 5 d. 5+6x x Completing the square under the radical) yields 4 x ) x )/) du, with u x )/) u sin u+c sin x ) +C Once again we should check. d )) x sin x ) 4 x ) 5+6x x

16 6.6 6 Example 7. We end this section with a curiosity. Can you identify the function below? ) ) fx) iln x ix Suppose for a moment that we could treat i just like any other real constant. Then we might try differentiating f. f x) i x ix ) d x ix) ) i x x ix ) i x ) i x i x x ix ) x ) x ix ) x ix x x Now, by a corollary of the Mean Value Theorem, we must conclude that fx) and arcsinx differ by a constant. Thus arcsin0+c f0) iln 0 i0 ) 0 It follows that C 0 and that ) 4) arcsinx iln x ix Now it is not even clear that 4) is real-valued. Nor is it clear that one can define the logarithm on the complexes, let alone that the subsequently defined function is differentiable. We will say more about this later.