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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.

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

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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.

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Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

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

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

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

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

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

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

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  • $\begingroup$ @gorgi, How about this? $\endgroup$ – lab bhattacharjee May 14 '14 at 8:24
  • $\begingroup$ im not familiar with the sign $\overline$ but it looks nice. thanks! $\endgroup$ – gorgi May 14 '14 at 23:21
  • $\begingroup$ @gorgi, Sorry for the confusion. Please find the edited version $\endgroup$ – lab bhattacharjee May 15 '14 at 3:52

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