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If b+c=pi prove that 2(1-sinbsinc) = cos(sqr)b+cos(sqr)c?
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LIke this? if b + c = pi, then:\[2(1-\sin b+\sin c)=\cos( \sqrt b)+\cos (\sqrt c)\]?
cos square b + cos square c
or squared?
oh, ok.
i fixed one little mistake in the question
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man i completely read the statement wrong >.<
oh, ok, i thought i was going crazy.
sorry
Basically, since b + c = pi, we have that sin b = sin c, and we exploit that. I guess I could have done the proof the other way around using that:\[2(1-\sin b \sin c)=(1-\sin b \sin c)+(1-\sin b \sin c)\]\[=(1-\sin ^2b)+(1-\sin ^2c)=\cos^2 b +\cos^2c\]
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