# Find a formula relating $\operatorname{arcsin}(x)$ and $\operatorname{arccos}⁡(x)$

From the formula $\sin\left(\frac{π}{2}−x\right)=\cos x$, find a formula relating $\operatorname{arcsin}(x)$ and $\operatorname{arccos}⁡(x)$.

I have figured out that the domain of $x$ is $[-1;1]$, but I have no idea how to do this. I've tried letting $y=\cos x$ and the only result I've got is $$\operatorname{arccos}(y)+\operatorname{arcsin}(y)=\frac{π}{2}$$ I need a full answer.

For values of $y\in(0,1)$ you can prove this geometrically:
Denote $\arccos y=\alpha$. Then by definition $\cos \alpha=y$. Draw a right triangle $ABC$ with hypotenuse $AB= 1$, side $AC=y$ and the angle $\angle BAC=\alpha$. Then if you denote $\angle ABC=:\beta$, you get $\sin \beta=\frac{AC}{AB}=\frac{y}{1}=y$. So $\arcsin y=\beta$. But obviously $\alpha+\beta=\frac{\pi}{2}$
In a right angled triangle cos of one acute angle $\alpha = x/\sqrt{x^2+y^2}$ and sine of the other acute angle $\beta = y/\sqrt{x^2+y^2}.$ Find relation between $\alpha,\beta.$
Answer is provided by the direct definition of the trig ratios as : $\alpha+ \beta = \pi/2.$