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.
Best Answer
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.