Solve system of equations: $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$ $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$
I tried to use tangent half-angle substitution, tried to sum these two equations and get this
$\sin(x+ \pi/4) = \sqrt3 / \sqrt2 \sin(y+\tan^{-1}(\sqrt5))$
I stuck.
Any Help is appreciated!
 A: Hint:$$2\sin^2(x)+2\cos^2(x)=2=\sin^2(y)+5\cos^2(y)\implies 4\cos^2(y)=1\implies ?$$
A: Square both equations, then add them up. $2 \sin^2 x = \sin^2 y$ and $2 \cos^2 x = 5 \cos^2 y$, so $2 = \sin^2 y + 5 \cos^2 y$, and hence $1 = 4 \cos^2 y$. From here, $\cos y = \pm \frac1{2}$, and you should substitute back in to check if both work. You'll also get multiple values of $y$ from this, which you need to check with the $\sin$ relations.
A: Substitute $u=\sin x;\;v=\sin y$. Then square both side of the two equations
$
\left\{
\begin{array}{l}
 2 u^2=v^2 \\
 2 \left(1-u^2\right)=5 \left(1-v^2\right) \\
\end{array}
\right.
$
Solutions need to be checked. Solution in $[...]$ are not solutions of the given equation
$$\sin x = -\frac{\sqrt{\frac{3}{2}}}{2},\sin y = -\frac{\sqrt{3}}{2},\left[\sin x = -\frac{\sqrt{\frac{3}{2}}}{2},\sin y = \frac{\sqrt{3}}{2}\right],\\\left[\sin x = \frac{\sqrt{\frac{3}{2}}}{2},\sin y = -\frac{\sqrt{3}}{2}\right],\sin x = \frac{\sqrt{\frac{3}{2}}}{2},\sin y = \frac{\sqrt{3}}{2}$$
Hope this helps
