Question

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

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

Using product to sum identity

\sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]

4\sin \theta \cdotp \sin(60\degree -\theta )\cdotp \sin(60+\theta )

=2\sin \theta (\cos2\theta -\cos 120\degree )

=2\sin \theta (\cos2\theta+\dfrac{1}{2} )

=2\sin \theta (\dfrac{3}{2}-2\sin^2 \theta )

=3\sin \theta -4\sin^3 \theta

=\sin 3\theta

Steven Zheng posted 1 month ago

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