Problem with sine in a right triangle Given a triangle $ABC$ with angles a,b & c, prove that if $\sin^2(a) + \sin^2(b) + \sin^2(c) = 2$ then the triangle is right angled (has an angle of $90^o$).
If I assume the triangle is right angled and have AB, AC and BC as sides, with BC being the base, then I can say that statement is true, since $\sin^2(a)$  would be 1 where $a=90^o$.
Also, since it's a right angled triangle we can say that $\sin^2(b) = AC^2/BC^2$ and $\sin^2(c) = AB^2/BC^2$. The sum of $\sin^2(b) + \sin^2(c)$ would be equal to $(AC^2+AB^2)/BC^2$ which is 1 (pythagoras theorem). Hence the sum of squares of sines of all 3 angles is indeed 2.
But how do I prove this? As in, I'm only given the starting equation, how do I get from there to a right angled triangle?
 A: We have:
$$\sin^2 A + \sin^2 B+\sin^2 C = 3-\sum_{cyc}\cos^2 A =\frac{1}{2}\left(3-\sum_{cyc}\cos(2A)\right)$$
hence $LHS=2$ is equivalent to $\sum_{cyc}\cos(2A)=-1$ or to:
$$ \cos^2 A+\cos(B+C)\cos(B-C) = 0 \tag{1}$$
but since $A+B+C=\pi$, $\cos(B+C)=\cos(\pi-A)=-\cos(A)$ and the previous line is equivalent to:
$$ \cos(A)=\cos(B-C)\quad\text{or}\quad \cos(A)=0,\tag{2} $$
so $A=\frac{\pi}{2}$ or:
$$ -\cos(B)\cos(C)+\sin(B)\sin(C) = \cos(B)\cos(C)+\sin(B)\sin(C)\tag{3}$$
that gives $\cos(B)\cos(C)=0$.
A: Notice, in $\Delta ABC$ , we have $a+b+c=180^o$
Given that 
$$\sin^2 a+\sin^2 b+\sin^2 c=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (180^o-(a+b))=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (a+b)=2$$ 
$$\implies (\sin a\cos b+\cos a\sin b)^2=2-\sin^2 a-\sin^2 b$$ $$\implies \sin^2 a\cos^2 b+\cos^2 a\sin^2 b+2\sin a\sin b\cos a\cos b=\cos^2 a+\cos^2 b$$ 
 $$\implies  -(1-\sin^2 a)\cos^2 b-(1-\sin^2 b)\cos^2 a+2\sin a\sin b\cos a\cos b=0$$  $$\implies  -\cos^2 a\cos^2 b-\cos^2 a\cos^2 b+2\sin a\sin b\cos a\cos b=0$$
$$\implies -2\cos a\cos b(\cos a\cos b-\sin a\sin b)=0$$  $$\implies \cos a\cos b\cos(a+b)=0$$ $$ \color{blue}{\text{if } \quad \cos a=0 \implies a=90^o}$$ $$ \color{blue}{\text{if } \quad \cos b=0 \implies b=90^o}$$ $$ \color{blue}{\text{if } \quad \cos (a+b)=0 \implies a+b=90^o \iff c=90^o}$$
We find that above three cases show that $\color{blue}{\Delta ABC}$ is $\color{blue}{\text{right angled}}$.   
A: $$F=\sin^2A+\sin^2B+\sin^2C=1-(\cos^2A-\sin^2B)+1-\cos^2C$$
Method $\#1:$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$F=2-\cos(A+B)\cos(A-B)-\cos^2C$$
Method $\#2:$
Alternatively using Double angle & Prosthaphaeresis formula ,
$$\sin^2A+\sin^2B=1-\dfrac{\cos2A+\cos2B}2=1-\cos(A+B)\cos(A-B)$$
Now $\cos(A+B)=\cos(\pi-C)=-\cos c$
$\implies F=2-[-\cos C\cos(A-B)+\cos^2C]$
$=2-\cos C[-\cos(A-B)+\cos C]$
$=2-\cos C[-\cos(A-B)-\cos(A+B)]$ as $\cos(A+B)=-\cos c$
$$\implies\sin^2A+\sin^2B+\sin^2C=2+\cos C[2\cos A\cos B]$$
Can you reach home from here? 
A: $$\sin(a)=\sin(\pi-b-c)=\sin(b+c),$$
Then with obvious shorthands,
$$\begin{align}0&=2-s_a^2-s_b^2-s_c^2\\
&=c_b^2+c_c^2-s_a^2\\
&=c_b^2+c_c^2-(s_a c_b+c_a s_b)^2\\
&=c_b^2+c_c^2-s_b^2c_c^2-2s_b c_b c_c s_c-c_b^2s_c^2\\
&=c_b^2c_c^2+c_b^2c_c^2-2s_b c_b c_c s_c\\
&=2c_b c_c(c_b c_c-s_b s_c)\\
&=2c_bc_cc_{b+c}.\end{align}$$
The product is null with one of
$$b=\frac\pi2,$$
$$c=\frac\pi2,$$
$$b+c=\pi-a=\frac\pi2.$$
