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If $A+B+C=\pi$ then prove:$$\sin^2A+\sin^2B+\sin^2C=2-2\cos A\cos B\cos C$$

I am completely lost on this, please help.

Sawarnik
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  • Have you tried using the law of cosines (Al-Kashi Theorem) ? A, B and C represent the angles of a triangle – T_O Mar 28 '14 at 13:58
  • @T_O Where do you have that A,B,C are the angles of triangles? – Sawarnik Mar 28 '14 at 14:01
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    Their sum is Pi (180°), and the sum of the angles of any triangle is Pi – T_O Mar 28 '14 at 14:02
  • @T_O Not every angle A,B,C which sum to pi, aren't of a triangle. – Sawarnik Mar 28 '14 at 14:04
  • It's a way of transposing your problem. If you define (draw) a triangle $XYZ$ for which the angles are A, B and C, you didn't change your problem but now you can apply nice properties due to the fact that you work on a triangle – T_O Mar 28 '14 at 14:05
  • @T_O I am not getting what you say. But lets assume its a triangle, then how would you do it? – Sawarnik Mar 28 '14 at 14:07
  • I will try to give you an answer within the next 2 hours, I am currently busy and cannot write a proper proof, but I believe combining http://en.wikipedia.org/wiki/Law_of_sines and http://en.wikipedia.org/wiki/Law_of_cosines should give you an elegant answer. It may not work but it's worth a try – T_O Mar 28 '14 at 14:09
  • Did you mean $A+B+C=\pi$? – robjohn Mar 28 '14 at 15:09
  • @robjohn Oh yes! – Sawarnik Mar 28 '14 at 15:25
  • @Sawarnik sorry for being late. Actually my proof work but is actually longer than the ones used in the answers below, and uses the same formulas so it is too redundant – T_O Mar 31 '14 at 13:53
  • @T_O Ok, no prob :) – Sawarnik Mar 31 '14 at 13:54

2 Answers2

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$A+B+C = \pi \implies A + B = \pi - C \implies \sin(A+B)=\sin(C) \land \cos(A+B)=-\cos(C)$

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

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

Now, $\displaystyle\cos^2A-\sin^2B=\cos(A+B)\cos(A-B)=\cos(\pi-C)\cos(A-B)$ $=-\cos C\cos(A-B)$

So, $\displaystyle\cos^2C+\cos^2A-\sin^2B=\cos^2C-\cos C\cos(A-B)$

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

$\displaystyle=\cos C[-\cos(A+B)-\cos(A-B)]$ as $A+B=\pi-C$

Now, $\displaystyle\cos(A+B)+\cos(A-B)=?$

lab bhattacharjee
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