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Trignometry

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4 Apr 2008 14:51:51 IST

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If sum of 4 angles be pi .Prove the sum of the products of their cosines taken two together is equal to the sum of product of their sines taken similarly.

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sreeraman nagasubramaniyan

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4 Apr 2008 20:49:55 IST

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$A+B+C+D=\pi \\ \\ \mbox{To prove} \ \Sigma \cos(A)\cos(B) - \Sigma \sin(A)\sin(B) = 0 \\ \\ \mbox{To prove} \ \Sigma (\cos(A)\cos(B)-\sin(A)\sin(B)) = 0 \\ \\ \mbox{To prove} \ \Sigma \cos(A+B) = 0 \\ \\ \mbox{That is to prove} \ [\cos(A+B)+\cos(A+C)+\cos(A+D)+\cos(B+C)+\cos(B+D)+\cos(C+D)]=0 \\ \\ \cos(A+B) = \cos(\pi-(C+D)) = -\cos(C+D) \\ \\ \mbox{Thus we get} \ [\cos(A+B)-\cos(A+B)+\cos(A+C)-\cos(A+C)+\cos(A+D)-\cos(A+D)] \ \mbox{which is} \ 0 \\ \\ \mbox{Thus} \ \Sigma \cos(A)\cos(B) = \Sigma \sin(A)\sin(B)$

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