Q96 / 225 · JEE Main 2004 · Mathematics

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

  • Correct answer
    A
    $-\frac{3}{\sqrt{130}}$
  • B
    $\frac{3}{\sqrt{130}}$
  • C
    $\frac{6}{65}$
  • D
    $-\frac{6}{65}$

Solution

$\sin \alpha+\sin \beta=-\frac{21}{65}$ and $\cos \alpha+\cos \beta=-\frac{27}{65}$. Squaring and adding, we get $$ \begin{aligned} & 2+2 \cos (\alpha-\beta)=\frac{1170}{(65)^2} \\ & \Rightarrow \cos ^2\left(\frac{\alpha-\beta}{2}\right)=\frac{9}{130} \Rightarrow \cos \left(\frac{\alpha-\beta}{2}\right)=\frac{-3}{\sqrt{130}} \quad\left(\because \frac{\pi}{2} < \frac{\alpha-\beta}{2} < \frac{3 \pi}{2}\right) . \end{aligned} $$

Hence, the correct answer is option: 'A'.

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