Let alpha, beta be such that pi

Question

(C) Given: $\cos \alpha + \cos \beta = -\frac{27}{65}$ and $\sin \alpha + \sin \beta = -\frac{21}{65}$.Using sum-to-product formulas:$2 \cos \frac{\al

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$-\frac{3}{\sqrt{130}}$

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(C) Given: $\cos \alpha + \cos \beta = -\frac{27}{65}$ and $\sin \alpha + \sin \beta = -\frac{21}{65}$.Using sum-to-product formulas:$2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} = -\frac{27}{65}$  $(1)$$2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} = -\frac{21}{65}$  $(2)$Squaring and adding $(1)$ and $(2)$:$4 \cos^2 \frac{\alpha - \beta}{2} (\cos^2 \frac{\alpha + \beta}{2} + \sin^2 \frac{\alpha + \beta}{2}) = (-\frac{27}{65})^2 + (-\frac{21}{65})^2$$4 \cos^2 \frac{\alpha - \beta}{2} = \frac{729 + 441}{4225} = \frac{1170}{4225} = \frac{18}{65}$$\cos^2 \frac{\alpha - \beta}{2} = \frac{18}{4 	imes 65} = \frac{9}{130}$Since $\pi In this interval,$\cos \frac{\alpha - \beta}{2}$ is negative in the second quadrant $(\frac{\pi}{2}, \pi)$ and negative in the third quadrant $(\pi, \frac{3\pi}{2})$.Thus,$\cos \frac{\alpha - \beta}{2} = -\sqrt{\frac{9}{130}} = -\frac{3}{\sqrt{130}}$.

Let $\alpha, \beta$ be such that $\pi < \alpha - \beta < 3\pi$. If $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$,then the value of $\cos \frac{\alpha - \beta}{2}$ is

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