If $\alpha+\beta-\gamma=\pi$, show that $$\sin^2\alpha+\sin^2\beta-\sin^2\gamma=2\sin\alpha\sin\beta\cos\gamma$$ Please don't ask about ideas because I really don't have one. I was initially thinking about a triangle with the given angles... But no chance EDIT. Sorry guys, now is correct.
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1This seems incorrect - take $\alpha=\beta=\gamma=\frac{\pi}3$. The RHS is irrational but the LHS is not – Sep 29 '18 at 08:47
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Are you sure of it? Shouldn't it be $$\sin(\alpha)^2+\sin(\beta)^2+\sin(\gamma)^2$$? – Dr. Sonnhard Graubner Sep 29 '18 at 08:48
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You needed $\cos\gamma$ on the RHS. – Henno Brandsma Sep 29 '18 at 08:55
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Looks from the duplicate as though you need $\cos \gamma$ in the product – Mark Bennet Sep 29 '18 at 08:56