# 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.

• Have you used that $\alpha +\beta +\gamma=\pi?$ – mfl May 13 '14 at 19:28
• do you mean: $\alpha +\beta +\gamma=\pi$? and yes i did – gorgi May 13 '14 at 19:29
• @AidanF.Pierce - but i need to prove directly.. – gorgi May 13 '14 at 19:41

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.

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.

$$\sin^2A+\sin^2B+\sin^2C=1-(\cos^2A-\sin^2B)+1-\cos^2C$$

$$=2-\cos(A-B)\cos(A+B)-\cos C\cos C$$

$$=2-\cos(A-B)\cos(\pi-C)-\cos\{\pi-(A+B)\}\cos C$$

$$=2+\cos(A-B)\cos C+\cos(A+B)\cos C\text{ as }\cos(\pi-x)=-\cos x$$

$$=2+\cos C[\cos(A-B)+\cos(A+B)]$$

$$=2+\cos C[2\cos A\cos B]$$

• im not familiar with the sign $\overline$ but it looks nice. thanks! – gorgi May 14 '14 at 23:21