Inverse Trig Functions


Since we have expressions for the trigonometric functions in terms of complex exponentials, we can invert these expressions to find explicit expressions for the inverse trig functions! For example, we know

\displaystyle\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}.

If we can find an expression for \theta, then we have found the inverse sine function.  Symbolically, if x=\sin\theta then

\displaystyle\theta(x)=\sin^{-1}(x).

As an example, let's try finding \sin^{-1}(x):

let y=e^{i\theta}.  This implies that y^{-1}=e^{-i\theta}, and thus

\displaystyle x=\sin\theta=\frac{y-y^{-1}}{2i}.

Multiplying the equation by 2iy yields:

\displaystyle 2ixy=y^2-1

or

\displaystyle 0=y^2-2ixy-1.

We can solve for y using the quadratic formula,

\displaystyle y=\frac{2ix\pm\sqrt{4-4x^2}}{2}=ix\pm\sqrt{1-x^2}.

Thus,

\displaystyle e^{i\theta}=ix\pm\sqrt{1-x^2}.

Taking the logarithm of both sides and isolating \theta:

\displaystyle \theta=-i\ln\left(ix\pm\sqrt{1-x^2}\right).

This yields two values of \theta for each x.  We can make the function single-valued by taking only the positive root,

\displaystyle\theta=-i\ln\left(ix+\sqrt{1-x^2}\right)=\sin^{-1}(x).

In order to obtain a numerical result, one can compute the Taylor expansion of the expression.

All of the other inverse trig functions can be found using the same process, of course the details of solving for \theta will be a bit different in each case.

These expressions can be used to compute the inverse trig derivatives without using the trick that I demonstrated here.

For example,

\displaystyle\frac{d}{dx}\sin^{-1}(x)=\frac{d}{dx}\left[-i\ln\left(ix+\sqrt{1-x^2}\right)\right].

Upon differentiating, simplifying a bit, and defining \alpha=\sqrt{1-x^2}, we get

\displaystyle\frac{d}{dx}\sin^{-1}(x)=-\frac{1}{ix+\alpha}-\frac{ix}{\alpha xi+\alpha^2}=\frac{1}{\alpha}.

Thus,

\displaystyle\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}.