Derivation of inverse sine, what is wrong with this reasoning?

Question

I'm trying to find the derivative of $\sin^{-1}(x)$. I know the steps that lead to $\frac{1}{\sqrt{1-x^2}}$, however I don't understand why the following reasoning leads to a wrong answer.

Because

$$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))} $$

If we plug in for $f(x) = \sin(x)$, and because $\frac{d}{dx}\sin(x) = \cos(x)$ we get

$$\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))} $$

Since $\sin(x) = \cos(x-\frac{\pi}{2})$, we can state that $\sin^{-1}(x) = \cos^{-1}(x-\frac{\pi}{2})$. (I suspect this is what is wrong)

Thus,

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

Then, by the definition of inverse function, we have

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

Answer 1

How does $\sin(x) = \cos(x-\frac{\pi}{2})$, imply $\sin^{-1}(x) = \cos^{-1}(x-\frac{\pi}{2})$.

Set $x=0,\sin^{-1}=0,\cos^{-1}\left(-\dfrac\pi2\right)$ is not real as $-\dfrac\pi2<-1$

Similarly for $x=\dfrac\pi2$

Answer 2

Actually if we avoid using $f,f^{-1}$ notations, things will become much more clear:

Let $y=\sin^{-1}x: [-1,1] \mapsto [-\frac{\pi}{2},\frac{\pi}{2}]$, so $x=\sin y$. Thus

$$1=\cos y \cdot y'$$ $$y'=\frac{1}{\cos y}$$

But if $x=\sin y$, we have $\sqrt{1-x^2}=\cos y$, because $\cos y \ge0$ for $y \in [-\frac{\pi}{2},\frac{\pi}{2}]$ (Here, in your question, you need to watch the domain and range when you take the inverse to make sure they match; and as a matter of fact, they do not match - but you really do not need to take the inverse)

Thus

$$y'=\frac{1}{\sqrt{1-x^2}}$$