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