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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.

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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 $\alpha+\beta=\frac{\pi}{2}$

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Your question is like:

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. $

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