How to evaluate $\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$ $$\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$$
to evaluate the above equation, I used the formula: $$\sin^{-1} (x) + \sin^{-1} (y) = \sin^{-1}[x(1-y^2) + y(1-x^2)]$$
therefore I have got, 
$$\sin^{-1} (\sqrt{2} \sin \theta \sin2 \theta + \cos \theta )$$
The result is supposed to be 1. 
How can I do this? 

Let $\sin^{-1}(\sqrt2\sin\theta)=x\implies\sqrt2\sin\theta=\sin x$ and $-\dfrac\pi2\le x\le\dfrac\pi2$
Now $\sqrt{\cos2\theta}=\sqrt{1-\sin^2x}=|\cos x|$
As $\cos x\ge0,|\cos x|=+\cos x\implies\sin^{-1}\sqrt{\cos2\theta}=\sin^{-1}(\cos x)=\dfrac\pi2-\cos^{-1}(\cos x)$
Now $\cos^{-1}(\cos x)=\begin{cases} x &\mbox{if } 0\le x\le\dfrac\pi2\iff\sin\theta\ge0\\
-x & \mbox{ if } -\dfrac\pi2\le x<0\end{cases} $
Can you take it from here?
