Composition of inverse trigonometric functions with trigonometric functions

function-and-relation-compositioninverse functiontrigonometry

I found in this Wikipedia article a useful table showing the algebraic expressions for the composition of trigonometric functions with inverse trigonometric functions, along with a picture explaining where they come from. Among others,

$$\begin{align}
\sin({\arccos{x}})=\sqrt{1-x^2}, \quad \sin({\arctan{x}})=\frac{x}{\sqrt{1+x^2}}\\
\cos({\arcsin{x}})=\sqrt{1-x^2}, \quad \cos({\arctan{x}})=\frac{1}{\sqrt{1+x^2}}\\
\tan({\arcsin{x}})=\frac{x}{\sqrt{1-x^2}}, \quad \tan({\arccos{x}})=\frac{\sqrt{1-x^2}}{x}\\
\end{align}$$

What would be, instead, the algebraic expressions for the composition of inverse trigonometric functions with trigonometric functions, such as the followings? Could someone provide them or give a source where they can be obtained?

$$\begin{align}
\arcsin({\cos{x}})=?, \quad \arcsin({\tan{x}})=?\\
\arccos({\sin{x}})=?, \quad \arccos({\tan{x}})=?\\
\arctan({\sin{x}})=?, \quad \arctan({\cos{x}})=?\\
\end{align}$$

Best Answer

Hint

Let $$ t=\arcsin(\cos(x))$$

then

$$-\frac{\pi}{2}\le t\le \frac{\pi}{2}$$

and

$$\sin(t)=\cos(x)=\sin(\frac{\pi}{2}+x)$$

thus

$$t=\arcsin(\cos(x))=x+\frac{\pi}{2}+2k\pi$$ or $$t=\frac{\pi}{2}-x+2k\pi$$

we choose the the value in $ [-\frac{\pi}{2},\frac{\pi}{2}]$.

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