Question

prove that sin3theta=3sin theta-4sin^3theta​

Answer 1

Answer:

your proof is as follows

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Step-by-step explanation:

$to \: prove : \\ sin \: 3θ = 3.sin \: θ - 4.sin {}^{3} θ \\ \\ sin \: 3θ = sin(2θ + θ) \\ \\ = sin \:2 θ.cos \: θ + sin \: θ.cos \: 2θ \\ \\ =( 2.sin \: θ.cos \: θ.cos \: θ) + sin \: θ(1 -2 sin {}^{2} θ) \\ \\ = 2.sin \: θ.cos {}^{2} θ + sin \: θ - 2.sin {}^{3} θ \\ \\ = 2.sin \: θ(1 - sin {}^{2} θ) - 2.sin {}^{3} θ + sin \: θ \\ \\ = 2.sin \: θ - 2.sin {}^{3} θ - 2.sin {}^{3} θ + sin \: θ \\ \\ = 3.sin \: θ - 4.sin {}^{3} θ$