Question

Prove 8 sin^3 x cos x = 2 sin 2x - sin 4x​

Answer 1

Answer:

steps below

Step-by-step explanation:

$8sin^{3} x*cos x \\= 4sin^{2} x*2sin x cos x\\= 4sin^{2} x* sin 2x\\ = 2sin 2x*(2sin^{2} x)\\ = 2sin 2x*(1-(1-2sin^{2} x))\\ = 2sin 2x*(1-cos 2x)\\ = 2sin 2x - 2sin2x*cos 2x\\ = 2sin 2x - sin 4x$