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Verify: sin4x=8sinxcoscubex-4sinxcosx

Sagot :

We use sin2x = 2sinxcosx and cos2x = 2cos^2 x - 1. We write: sin 2(2x) = 2sin2x(cos2x) = 2(2sinxcosx)(2cos^2 x - 1) = (4sinxcosx)(2cos^2 x - 1) = (8sinxcos^3 x) - (4sinxcosx).
[tex] sin 4x = 8\cdot sin x \cdot cos^3 x - 4\cdot sin x \cdot cos x \\factor \ out \ a \ common \ factor\\\\ 4\cdot sin x \cdot cos x \left ( 2 cos^2 x - 1 \right ) =\\\\=2\cdot 2\cdot sin x\cdot cos x \left ( 2 cos^2 x - 1 \right )=\\\\= 2\cdot sin2x\cdot cos2x=\\\\=sin(2\cdot 2x)=sin4x\\\\double \ angle identit\ : \\ 2\cdot sin x\cdot cos x=sin2x\\2cos^2x-1=cos2x [/tex]