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Use an angle sum identity to derive each double-angle identity.

COS 2(theta) = 2 COS^2(theta) -1

S.O.S

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\ [cos(2\theta)]'=(2cos^2\theta-1)'=2(cos^2\theta)'-(1)'\\\\=2\cdot2cos\theta\cdot(-sin\theta)-0=-2\cdot2sin\theta cos\theta=\boxed{-2sin2\theta}\\\\Used:\\\ [f(x)-g(x)]'=f'(x)-g'(x)\\\{f[g(x)]\}'=f'[g(x)]\cdot g'(x)\\(cosx)'=-sinx\\sin2x=2sinxcosx
User Arnonuem
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