How to derive compositions of trigonometric and inverse trigonometric functions? To prove:
$$\begin{align}
\sin({\arccos{x}})&=\sqrt{1-x^2}\\
\cos{\arcsin{x}}&=\sqrt{1-x^2}\\
\sin{\arctan{x}}&=\frac{x}{\sqrt{1+x^2}}\\
\cos{\arctan{x}}&=\frac{1}{\sqrt{1+x^2}}\\
\tan{\arcsin{x}}&=\frac{x}{\sqrt{1-x^2}}\\
\tan{\arccos{x}}&=\frac{\sqrt{1-x^2}}{x}\\
\cot{\arcsin{x}}&=\frac{\sqrt{1-x^2}}{x}\\
\cot{\arccos{x}}&=\frac{x}{\sqrt{1-x^2}}
\end{align}$$
 A: Here is astart
$$ \sin({\arccos{x}})= \sqrt{1-\cos(\arccos(x))^2}=\sqrt{1-x^2},  $$
since $\cos(\arccos(x))=x.$ You need to use some trig. identities.
A: I'll be ignoring domains and possible roots of negative numbers. (If you let $\mbox{$x\in \textbf{]}0,\pi/2[$}$ everything works fine).
Given $f\circ g$, the trick is too relate $f$ with $g^{-1}$. 
I did some. You should be able to handle the remaining ones.
$\bullet \sin (\arccos (x))=\sqrt {1-(\cos (\arccos (x) ))^2}=\sqrt {1-x^2}$
$\bullet \sin (\arctan (x))=\dfrac{\tan (\arctan (x))}{\sqrt {1+(\tan (\arctan (x)))^2}}=\dfrac{x}{\sqrt {1+x^2}}$
For this one I used 
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\
&\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\
&\implies 1-(\sin(x))^2=\dfrac{1}{1+(\tan(x))^2}\\
&\implies \sin (x)=\dfrac{\tan(x)}{\sqrt{1+(\tan(x))^2}}\end{align}$$
$\bullet \tan (\arcsin (x))=\dfrac{\sin (\arcsin (x))}{\sqrt{1-(\sin(\arcsin (x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$
For this one I used 
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\
&\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\
&\implies \tan(x)=\dfrac{\sin(x)}{\sqrt{1-(\sin(x))^2}}\end{align}$$
$\bullet \cot (\arcsin(x))=\dfrac{\sqrt{1-(\sin (\arcsin(x)))^2}}{\sin (\arcsin(x))}=\dfrac{\sqrt{1-x^2}}{x}$
For this one I used 
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\
&\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\
&\implies \cot(x)=\dfrac{\sqrt{1-(\sin(x))^2}}{\sin (x)}\end{align}$$
$\bullet \cot (\arccos(x))=\dfrac{\cos (\arccos(x))}{\sqrt{1-(\cos (\arccos(x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$.
For this one I used 
$$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\
&\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\
&\implies (\cot(x))^2=\dfrac{(\cos(x))^2}{1-(\cos(x))^2}\\
&\implies \cot(x)=\dfrac{\cos(x)}{\sqrt{1-(\cos(x))^2}}\end{align}$$
A: I agree with the comment about right angled triangles. For example, by definition, $\arccos(x)$ is the angle $\theta$ in a right triangle with hypotenuse $1$ occurring adjacent to a leg of length $x$, and by the Pythagorean theorem and the definition of sine, the sine of such an angle is $\sqrt{1-x^2}/1$ (draw the picture). The others can be shown similarly.
I think this is perfectly rigorous (and can be extended by symmetry or differentiation arguments to the whole domain of the arccosine function) -- it's just a matter of what you take as the starting definition of the trig functions.
