Q. 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

A) $- \frac{3}{\sqrt{130}}$
B) $\frac{3}{\sqrt{130}}$
C) $\frac{6}{65}$
D) $- \frac{6}{65}$
Explanation

(a)
On squaring an adding given equations, we get
$\left( \sin{\alpha + \sin\beta} \right)^{2} + \left( \cos{\alpha + \cos\beta} \right)^{2} = \left( - \frac{21}{65} \right)^{2} + \left( - \frac{27}{65} \right)^{2}$
$\Rightarrow \sin^{2}{\alpha + \sin^{2}{\beta + 2\sin{\alpha\sin{\beta + \cos^{2}{\alpha + \cos^{2}{\beta + 2\cos{\alpha\cos{\beta = \frac{1170}{4225}}}}}}}}}$
$\Rightarrow \ \ 2 + 2\cos(\alpha - \beta) = \frac{1170}{4225}$
$\Rightarrow \ \ 2\left\lbrack 2\cos^{2}\left( \frac{\alpha - \beta}{2} \right) \right\rbrack = \frac{1170}{4225}$
$\Rightarrow \ \cos^{2}{\left( \frac{\alpha - \beta}{2} \right) = \frac{1170}{4 \times 4225} = \frac{9}{130}}$
$\Rightarrow \ \cos{\left( \frac{\alpha - \beta}{2} \right) = - \frac{3}{\sqrt{130}}}\ \ \ \ \ \ \lbrack\because\ \ \pi < \alpha - \beta < 3\pi\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.