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

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

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 $2iy$

$$\displaystyle 2ixy=y^2-1$$

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 $\theta$,

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

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

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

All of the other inverse trig functions can be found using the same process, of course the detials of solving for $\theta$ will vary slightly .

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

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