Question

# Let $$\alpha,\beta$$ be such that $$0<\alpha-\beta-\pi<2\pi.$$ if $$\sin \alpha+\sin \beta=-\dfrac{21}{65}$$ and $$\cos \alpha+\cos \beta=-\dfrac{27}{65}.$$ Then the value of cos $$\dfrac{\alpha-\beta}{2}$$ is a

A
665
B
3130
C
3130
D
665

Solution

## The correct option is B $$\dfrac{-3}{\sqrt{130}}$$$$\left (\sin \alpha +\sin \beta \right )^{2}+\left (\cos \alpha +\cos \beta \right )^{2}$$$$\Rightarrow 2+2 \cos \left (\alpha -\beta \right )=\dfrac{18}{65}$$$$\Rightarrow \cos^{2}\dfrac{\alpha -\beta }{2}=\dfrac{9}{130}, As \pi < \alpha -\beta < 3\pi \Rightarrow \dfrac{\pi }{2}< \dfrac{\alpha -\beta }{2}< \dfrac{3\pi }{2}$$$$\Rightarrow \cos \dfrac{\alpha -\beta }{2}=\dfrac{-3}{\sqrt{130}}$$Mathematics

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