Question

Verify \tan \dfrac{3}{2}x-\tan \dfrac{1}{2}x =\dfrac{2\sin x}{\cos x+\cos 2x}

Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 2 weeks ago


Answer

Using the sum to product identity for tangent function,


\tan \dfrac{3}{2}x-\tan \dfrac{1}{2}x

=\dfrac{\sin(\dfrac{3}{2}x-\dfrac{1}{2}x )}{\cos\dfrac{3}{2}x\cos \dfrac{1}{2}x }

Using product to sum identity

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

\tan \dfrac{3}{2}x-\tan \dfrac{1}{2}x

=\dfrac{2\sin x}{\cos x \cos 2x}

Steven Zheng Steven Zheng posted 2 weeks ago

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