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? 
 A: Set $a = \sqrt 2 \sin\theta$ and $b = \sqrt{\cos 2\theta}$. It follows that $a^2 + b^2 = 2\sin^2 \theta + \cos 2\theta = 1$.
If $\alpha = \sin^{-1} a$ where $\alpha \in [-\frac \pi 2, \frac \pi 2]$, we get $\sin(\alpha) = a$, and since $b\ge 0$ we have $b = \sqrt{1 - a^2} = \cos(\alpha) = \sin(\frac \pi 2 - \alpha)$.


*

*If $a \ge 0$, we get $\alpha \in [0, \frac \pi 2]$ and thus $\sin^{-1} b = \frac\pi 2 - \alpha$ which gives us $\sin^{-1} a + \sin^{-1}b = \frac \pi 2$.

*On the other hand, if $a < 0$, we have $\alpha \in [-\frac \pi 2, 0)$ and so $\sin^{-1}b = \pi + \alpha$ so $\sin^{-1}a + \sin^{-1}b$ has no fixed value.

A: 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?
A: If you evaluate that expression at $\theta = 0$, you get $\sin^{-1}(1)$, which is not $1$. So there's no way that the answer is 1. 
