Can expressions such as $\sin^{-1}(\sin^2(x))$ be simplified analytically? I have the expression $\sin^{-1}(\sin^2(x))$ from attempting to solve another equation $\sin(x^2) = \sin^2(x)$. I am aware of the addition rules with the inverse trigonometric functions - two that could prove useful here would be $\sin^{ – 1}(a) + \sin^{ – 1}(b) = \sin^{ – 1}(a\sqrt {1 – {b^2}} + b\sqrt {1 – {a^2}} )$ and $\sin^{ – 1}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) = 2\tan^{ – 1}(a)$ - however I am not aware of any rules related to powers, exponentials or products with the inverse trigonometric functions, or the normal trigonometric functions. I suppose one painful, not very analytical, way to do this would be to expression the powers/multiplications as additions - then put them in one of the two equations but that seems like it would complicate the expression rather than simplify it - is there anything that can be done about this? What about using the complex variants of the trig functions?
 A: We can prove that $\color{brown}{\text{for any}\,x\!\in\Bbb R}$
$\color{blue}{\arcsin\left(\sin^2\!x\right)=\dfrac{\pi}2-2\arcsin\left(\!\dfrac{\sqrt2}2\big|\cos x\big|\right)}\,.\quad\color{blue}{(*)}$
For any $\,x\in\Bbb R\;$ it results that
$\arcsin\left(\sin^2\!x\right)=\dfrac{\pi}2-\arccos\left(\sin^2\!x\right)\,.$
Let $\;\alpha=\arccos\left(\sin^2\!x\right).$
It implies that $\,\alpha\!\in\![0,\pi]\,$ and $\,\dfrac\alpha2\!\in\!\left[0,\dfrac\pi2\right],\;$ moreover ,
$\sin\left(\dfrac\alpha2\right)=\sqrt{\dfrac{1-\cos\alpha}2}=\sqrt{\dfrac{1-\sin^2\!x}2}=\dfrac{\sqrt2}2\big|\cos x\big|\;,$
consequently ,
$\dfrac{\alpha}2=\arcsin\left(\dfrac{\sqrt2}2\big|\cos x\big|\right)\;,$
$\alpha=2\arcsin\left(\dfrac{\sqrt2}2\big|\cos x\big|\right)\,.$
Therefore ,
$\arcsin\left(\sin^2\!x\right)=\dfrac\pi2-\alpha= \dfrac{\pi}2-2\arcsin\left(\!\dfrac{\sqrt2}2\big|\cos x\big|\right)\,.$
In this way we have proved the equality $\,(*)\,.$
In particular, we have proved that $\color{brown}{\text{for any}\,x\!\in\left[-\dfrac\pi2,\dfrac\pi2\right]}$
$\color{blue}{\arcsin\left(\sin^2\!x\right)=\dfrac{\pi}2-2\arcsin\left(\!\dfrac{\sqrt2}2\cos x\right)}\,.$
