﻿ Find the value of \dfrac{\sin 7\degree +\sin 8\degree \cdotp \cos 15\degree }{\cos 7\degree

#### Question

Find the value of \dfrac{\sin 7\degree +\sin 8\degree \cdotp \cos 15\degree }{\cos 7\degree -\sin 8\degree \cdotp \sin 15\degree }

Collected in the board: Trigonometry

Steven Zheng posted 1 year ago

\dfrac{\sin 7\degree +\sin 8\degree \cdotp \cos 15\degree }{\cos 7\degree -\sin 8\degree \cdotp \sin 15\degree }

=\dfrac{\sin (15\degree-8\degree ) +\sin 8\degree \cdotp \cos 15\degree }{\cos (15\degree-8\degree ) -\sin 8\degree \cdotp \sin 15\degree }

=\dfrac{\sin 15\degree \cdotp \cos 8\degree }{\cos 8\degree \cdotp \cos 15\degree }

=\tan 15\degree

Using the half angle identity

\tan \dfrac{\alpha }{2}= \dfrac{\sin \dfrac{\alpha }{2}}{\cos \dfrac{\alpha }{2}} = \dfrac{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}{2\cos^2 \dfrac{\alpha }{2}} = \dfrac{\sin \alpha }{1+\cos \alpha }

\tan15\degree =\dfrac{\sin 30\degree }{1+\cos 30\degree }=\dfrac{\dfrac{1}{2} }{1+\dfrac{\sqrt{3} }{2} }

=\dfrac{1}{2+\sqrt{3} }

=2-\sqrt{3}

Steven Zheng posted 1 year ago

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