prove that if $\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma=2$ then the triangle has a right angle prove that if $\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma=2$ then the triangle has a right angle. 
$\alpha,\beta,\gamma$ are the angles of the triangle.
I tried to use all kinds of trigonometric identities but it didn't work for me. it's to complex form me
Thanks. 
 A: Let $\alpha$, $\beta$ and $\gamma$ be the angles of a triangle; then it holds
$$
\boxed{\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2+2\cos\alpha\cos\beta\cos\gamma}
$$
If this equals $2$, we conclude
$$
\cos\alpha\cos\beta\cos\gamma=0
$$
so one of the angles is a right angle.
Proof of the claim
Let's use that $\gamma=\pi-\alpha-\beta$, so $\cos\gamma=-\cos(\alpha+\beta)$. Then
\begin{align}
\cos^2\alpha+\cos^2\beta+\cos^2\gamma&=
\cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)\\
&=\cos^2\alpha+\cos^2\beta+\cos^2\alpha\cos^2\beta+\sin^2\alpha\sin^2\beta\\
&\qquad-2\sin\alpha\sin\beta\cos\alpha\cos\beta\\
&=\cos^2\alpha+\cos^2\beta+1-\cos^2\alpha-\cos^2\beta+2\cos^2\alpha\cos^2\beta\\
&\qquad-2\sin\alpha\sin\beta\cos\alpha\cos\beta\\
&=1+2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\\
&=1-2\cos\alpha\cos\beta\cos\gamma
\end{align}
giving the final relation
$$
\boxed{\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1-2\cos\alpha\cos\beta\cos\gamma}
$$
Now
\begin{align}
\sin^2\alpha+\sin^2\beta+\sin^2\gamma
&=3-\cos^2\alpha-\cos^2\beta-\cos^2\gamma\\
&=2+2\cos\alpha\cos\beta\cos\gamma
\end{align}
as claimed at the beginning.
A: Rearranging, we have
$$\sin^2(a) = \cos^2(b) + \cos^2(c)$$
Since $a=\pi-(b+c)$, we get
$$\sin^2(b+c) = \cos^2(b) + \cos^2(c)$$
Hence,
$$(\sin(b)\cos(c) + \cos(b) \sin(c))^2 = \cos^2(b) + \cos^2(c)$$
This gives us
$$\sin(b)\sin(c)\cos(b)\cos(c) = \cos^2(b) \cos^2(c)$$
This means
$$\cos(b) \cos(c) = 0 \text{ or }\cos(b+c) = 0$$
Now conclude what you want.
A: Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\sin^2A+\sin^2B+\sin^2C=1-(\cos^2A-\sin^2B)+1-\cos^2C$$
$$=2-\cos(A-B)\cos(A+B)-\cos C\cos C$$
$$=2-\cos(A-B)\cos(\pi-C)-\cos\{\pi-(A+B)\}\cos C$$
$$=2+\cos(A-B)\cos C+\cos(A+B)\cos C\text{ as }\cos(\pi-x)=-\cos x$$
$$=2+\cos C[\cos(A-B)+\cos(A+B)]$$
$$=2+\cos C[2\cos A\cos B]$$
