Let , be such that

Question

Let $\alpha,\beta$ be such that $\pi<\alpha-\beta<3\pi$. 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:

Accepted Answer

Since $\dfrac\pi2<\dfrac{\alpha-\beta}2<\dfrac{3\pi}2$, $\cos\dfrac{\alpha-\beta}2<0$.

$$2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2 = -\frac{21}{65}, \qquad 2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2 = -\frac{27}{65}$$

Squaring and adding:

$$4\cos^2\frac{\alpha-\beta}2 = \frac{21^2+27^2}{65^2} = \frac{1170}{4225} = \frac{36}{130}$$

$$\cos^2\frac{\alpha-\beta}2 = \frac9{130} \implies \cos\frac{\alpha-\beta}2 = -\frac3{\sqrt{130}}$$

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