#### Question

If \sin \alpha +\sin \beta = \dfrac{3}{5}, \cos \alpha +\cos \beta = \dfrac{4}{5}, find the value of \cos (\alpha -\beta )

Collected in the board: Trigonometry

Steven Zheng posted 5 months ago

\because \sin \alpha +\sin \beta = \dfrac{3}{5}, \cos \alpha +\cos \beta = \dfrac{4}{5}

Square both sides of the two equations and addition

2+ 2\cos \alpha\cos \beta +2\sin \alpha \sin \beta = \dfrac{9}{25}+\dfrac{16}{25}

\therefore \cos (\alpha -\beta ) = \cos \alpha\cos \beta + \sin \alpha \sin \beta

= -\dfrac{1}{2}

Steven Zheng posted 5 months ago

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