﻿ Find the value of \tan (\alpha +\beta) such that \sin \alpha +\sin \beta = \dfrac{1}{4}

#### Question

Find the value of \tan (\alpha +\beta) such that \sin \alpha +\sin \beta = \dfrac{1}{4} and \cos \alpha +\cos \beta =\dfrac{1}{3}

Collected in the board: Trigonometry

Steven Zheng posted 2 years ago

\sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

\sin \alpha +\sin \beta =2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}= \dfrac{1}{4}
(1)
\cos \alpha +\cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}=\dfrac{1}{3}
(2)
(1) over (2) yields

\tan \dfrac{\alpha +\beta }{2} = \dfrac{3}{4}
(3)

Double angle identity

\tan (\alpha +\beta) = \dfrac{2\tan\dfrac{\alpha +\beta }{2} }{1-\tan^2\dfrac{\alpha +\beta }{2} }

=\dfrac{2\cdotp \dfrac{3}{4} }{1-(\dfrac{3}{4} )^2}

=\dfrac{24}{7}

Steven Zheng posted 2 years ago

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