Let alpha, beta be such that pi

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(D) Given $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$.Squaring and adding both equations:$(\sin \alpha

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

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(D) Given $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$.Squaring and adding both equations:$(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2$$2 + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \frac{441 + 729}{4225}$$2 + 2\cos(\alpha - \beta) = \frac{1170}{4225} = \frac{18}{65}$$2(1 + \cos(\alpha - \beta)) = \frac{18}{65}$$4 \cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{65}$$\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{260} = \frac{9}{130}$$\cos\left(\frac{\alpha - \beta}{2}\right) = \pm \frac{3}{\sqrt{130}}$Since $\pi In this interval,the cosine function is negative.Therefore,$\cos\left(\frac{\alpha - \beta}{2}\right) = -\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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