Question

Let α and β be such that π < α – β < 3π. If sin α + sin β = ${\displaystyle -\frac{21}{65}}$ and cos α + cos β = ${\displaystyle -\frac{27}{65}}$, then the value of ${\displaystyle \cos \frac{(\alpha -\beta )}{2}}$ is ______.

Options

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

• ${\displaystyle \frac{3}{\sqrt{130}}}$

• ${\displaystyle \frac{6}{65}}$

• ${\displaystyle -\frac{6}{65}}$

Answer 1

Let α and β be such that π < α – β < 3π. If sin α + sin β = ${\displaystyle -\frac{21}{65}}$ and cos α + cos β = ${\displaystyle -\frac{27}{65}}$, then the value of ${\displaystyle \cos \frac{(\alpha -\beta )}{2}}$ is ${\displaystyle \underline{-\frac{3}{\sqrt{130}}}}$.

Explanation:

Given, sin α + sin β = ${\displaystyle -\frac{21}{65}}$  ...(i)

and cos α + cos β = ${\displaystyle -\frac{27}{65}}$  ...(ii)

On squaring and adding equations (i) and (ii), we get

(sin α + sin β)² + (cos α + cos β)² = ${\displaystyle {(-\frac{21}{65})}^2+{(-\frac{27}{65})}^2}$

${\displaystyle \Rightarrow }$ sin²α + sin² β + 2 sin α sin β + cos² α + cos² β + 2 cos α cos β = ${\displaystyle \frac{441}{4225}+\frac{729}{4225}}$

${\displaystyle \Rightarrow }$ 2 + 2 cos (α – β) = ${\displaystyle \frac{1170}{4225}}$

${\displaystyle \Rightarrow }$ ${\displaystyle 2\left[2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)\right]=\frac{1170}{4225}}$

${\displaystyle \Rightarrow {\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{1170}{4\times 4225}=\frac{9}{130}}$

${\displaystyle \Rightarrow \cos \left(\frac{\alpha -\beta }{2}\right)=-\frac{3}{\sqrt{130}}}$  ...[∵ π < α – β < 3π]