Verify that $$\sin 3 heta =4\sin heta \cdotp \sin(60\degree - heta )\cdotp \sin(60+ heta ) $$

Question

Verify that $$\sin 3	heta =4\sin 	heta \cdotp \sin(60\degree -	heta )\cdotp \sin(60+	heta )       $$

Uniteasy Admin · Accepted Answer

Using product to sum identity
$$\sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]$$
$$4\sin 	heta \cdotp \sin(60\degree -	heta )\cdotp \sin(60+	heta )  $$
$$=2\sin 	heta (\cos2	heta -\cos 120\degree  ) $$
$$=2\sin 	heta (\cos2	heta+\dfrac{1}{2}   )$$
$$=2\sin 	heta (\dfrac{3}{2}-2\sin^2 	heta   )$$
$$=3\sin 	heta -4\sin^3 	heta  $$
$$=\sin 3	heta $$

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