How do you prove sin3θ=3sinθ−4sin3θ?
hsubramani42p98axj:
is it sin3 or sin cube theta?
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Proof: sin 3A
= sin (2A + A)
= sin 2A cos A + cos 2A sin A
= 2 sin A cos A ∙ cos A + (1 - 2 sin^2 A) sin A
= 2 sin A (1 - sin^2 A) + sin A - 2 sin^3 A
= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A
= 3 sin A - 4 sin^3 A
Therefore, sin 3A = 3 sin A - 4 sin^3
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