Calculus of One and Several Variables

5.4 Circular Functions and Inverses

Here in this section, the derivatives of the circular functions are derived and then the derivatives of their inverse functions are considered.

Proof: Consider the picture where here h is small

From Corollary 2.3.10

(1 − cos(h))+ sin (h) ≥ h ≥ sin(h)

It follows that

1-−-cos(h) + 1 ≥--h---≥ 1 (5.4) sin (h ) sin(h)

(5.4)

Now

which obviously converges to 0 as h → 0.

lim 1−-cos(h)= 0 (5.5) h→0 sin(h)

(5.5)

Passing to a limit in 5.4 yields

lim --h---= lim sin(h)= 1 (5.6) h→0 sin(h) h→0 h

(5.6)

Now

From what was shown above 5.5, 5.6, it follows that the second term on the last line converges to 0 as h → 0 and the first term converges to cos. Therefore,

lim sin(x+-h)−-sin(x)= cos(x). h→0 h

Now consider the other claim. cos

= sin and so, from the above and the chain rule,

cos′(x) = cos(x + π-) = cos(x)cos(π-)− sin (x )sin( π) = − sin(x). 2 2 2

You could also do a similar argument to what was used to find the derivative of the sine function. ■

The sine function is one to one on

taking all values between −1 and 1 and so one can define

[ π π] arcsin : [− 1,1] → −-,- 2 2

as an inverse function for the sine restricted to

. In words, arcsin is the angle whose sine is y which lies in . Letting sin = y, you can find by using the chain rule. Thus

dx cos(x)---= 1 dy

and so

dx 1 dy-= cos(x)

Now for x ∈

, cos ≥ 0 and so the above equation reduces to

′ dx 1 1 arcsin (y) ≡ dy-= ∘------2---= ∘-----2- 1 − sin (x) 1− y

The cosine function is one to one on

taking all values between −1 and 1 and so one can define

arccos : [− 1,1] → [0,π]

as an inverse function for the cosine restricted to

. In words, arccos is the angle whose cosine is y which lies in . Letting y = cos, you can find using the chain rule. As explained, this is the derivative of the inverse function just described.

dx 1 = − sin(x)--- dy

and so

dx- ---1---- dy = − sin (x )

For x ∈

, sin ≥ 0 and so −sin = − = −. Thus

arccos′(y) ≡ dx-= −∘-----1----- = −∘--1---- dy 1 − cos2(x) 1 − y2

The tangent function is one to one on

and maps onto , all of ℝ. Thus one can define arctan as x ∈ where tan = y as the above. Now applying the quotient rule to find tan^′,

cos2 (x) − (− sin (x ))sin(x) 1 tan′(x) = -----------2-----------= --2----= sec2(x) = 1 + tan2 (x) cos (x) cos (x)

the last being a well known identity which says essentially that cos²

+ sin² = 1. Then as before, y = tan,

( 2 ) dx ( 2) dx 1 = 1 + tan (x ) dy-= 1 + y dy-

and so

′ --1--- arctan (y) = 1 +y2

You can do all the other trigonometric functions and their inverses the same way. Of course none of them have inverses unless their domains are restricted as above. For example, x ∈

in order that sin will be one to one. The choice of interval on which the function is one to one is somewhat arbitrary. One could have done the same thing for arcsin if the interval had been instead of for example. However, it is traditional to pick the interval to which the function is restricted to be that interval closest to 0 such that the function is one to one and maps onto its maximum range. The following table sumarizes the derivatives of the trigonometric functions and their inverses. In the table D will be the domain of the function and R will be the range of the function.

|---------|--------------Table-of Derivatives---------|-------------| |f-(x)-----|-Domain-------------------|Range-----------|f′(x)--------| |sin-(x-)---|-ℝ------------------------|[[− 1,1]]--------|cos(x)-------| |arcsin(x)-|-[−-1,1]--------------------|-−-π2, π2---------|-√11−x2-------| -cos(x)-----ℝ-------------------------[− 1,1]----------−-sin-(x)----- |tan(x) | all except odd multiples of π2|ℝ |sec2(x) | |arctan-(x-)|-ℝ------------------------|(−-π, π)--------|--12---------| |sec(x)---|-all except odd-multiples of-π|[1,∞2)2∪-(− ∞,-−-1]|s1+exc(x)tan(x) |arcsec(x-)|-[1,∞-)∪-(−-∞,-− 1]-------2-|[0, π)∪-(π,π]---|--√-1--------| -----------------------------------------2----2---------|y|-y2−-1-----| (5.7)

(5.7)

For the line corresponding to arcsec, I have picked the domain to be [1,∞) ∪ (−∞,−1] to correspond to restricting sec

to [0,) ∪ (,π]. There is no consensus on how to do this. I have done it this way because the interval on which cos is one to one which we use in defining the inverse cosine is and I have tried to make it as similar to this as possible.

One can do similar things for csc, and arccsc and cot and arccot but the above is likely enough of this tedium to cover the situations of interest.