On-line Math 21

3.1  Trigonometric functions

3.1.4  Derivatives of inverse trig functions

Inverse Trigonometric Functions.

For someone uncomfortable with trigonometric functions and all their identities, the thought of having to worry about inverse-trigonometric functions may be a little unnerving. But don't let it be. These functions are simply additions to our ``zoo'' of available functions. There actually are some inverse-trigonometric identities, but they are not that useful and we will not worry about them.

So why study inverse trigonometric functions at all? Well, we want to be able to integrate and differentiate just about every ``normal'' function we can. I'm not considering inverse trig functions as ``normal'' in that sense. What we will see today, though, is that the derivatives of these inverse trig functions are algebraic, sometimes even rational functions, so in order to antidifferentiate those functions, we need to talk about inverse trig functions.

The Inverse sine

sin^-1 is the inverse of sin. By saying that, we have to mean that, if you restrict the domain of sin(x) to the right interval so that it is one-to-one, then there is an inverse, which we call sin^-1(x) . It is also called arcsin(x) or Arcsin(x) . Maple understands arcsin(x) . Technically, sin^-1(x) is defined by:

sin-1(x): = yÛ y Î [-p/2,p/2],  and sin(y) = x.

To find the derivative of sin^-1(x) , we do as we did before: define y to be y = sin^-1(x) . Then we want dy/dx . By definition, sin(y) = x , and so:

sin(y) = x.  Now, differentiate implicitly: cos(y) dy dx = 1.  Solve for   dy dx : dy dx = 1 cos(y) .

Here it is a little complicated to solve for dy/dx in terms of x directly.

The Inverse cosine

You can think about how you define the inverse cosine, just as for sin^-1 . The only really new thing is the restriction on y , which can't be the same as for sin^-1 since it isn't one-to-one there.

cos-1(x): = yÛ y Î [0,p]  and  cos(y) = x.

If you were to go through the calculations again, you would have that:

(cos-1(x))¢ = -1 Ö 1-x2 .

Since the derivatives are almost the same (they differ by just the sign), we don't worry too much about cos^-1x . Actually, there is an inverse-trigonometric identity behind this.

The Inverse tangent

This one is perhaps easier. We start out as before:

tan-1(x): = yÛ y Î (-p/2,p/2) and tan(y) = x.

And:

tan-1(x) : = y x = tan(y) 1 = sec2y dy dx cos2y = dy dx .

Now, look at the right triangle again, this time with the side opposite y having length (by scaling) x , and the adjacent side 1 .

The Inverse secant

Three more to go. Actually, though, after this one the rest of the inverse trig functions will not give any new integration (antidifferentiation) formulas. The reason is the same as for the inverse cosine. So, this is the last one.

sec-1(x): = yÛ y Î [0,p/2) or y Î [p,3p/2),  and  sec(y) = x.

You should note that the domain here for sec^-1(x) is a bit nonstandard in the choice of domain.

And:

sec-1(x) : = y x = sec(y) 1 = sec(y)tan(y) dy dx 1 sec(y)tan(y) = dy dx .

Here it is a little messier. For the right triangle with standard angle y ,

I must confess. I cheated. I described these pictures, with the right triangles and all, forgetting that the domain of these functions include some x < 0 . But there the geometry is invalid. You actually only have to see how the signs of the trig functions change when x < 0 to see how to extend them. But that is usually unneeded, since the signs won't change. For sec^-1x , however, it matters.

tan(y) = Ö   x2-1

for y Î [p,3p/2) . But then sec(y) = x as well, since both are negative. So, really:

(sec-1(x))¢ = 1 x Ö x2-1 .

Using a different domain for the definition of the inverse secant, such as y Î [0,p/2)È(p/2,p] , which is actually more common than the definition used here, then

d dx sec-1(x) = 1 |x| Ö x2-1 .

Copyright (c) by David L. Johnson.