Question

Let $a$ and $b$ be real numbers such that $\sin a+\sin b=\frac{1}{\sqrt{2}}$ and $\cos a+\cos b=\frac{\sqrt{6}}{2}$ then the value of $\sin (a+b)$ is :

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}$

$\sin a+\sin b=\frac{1}{\sqrt{2}}$
$\Rightarrow 2\sin \frac{a+b}{2}\cos \frac{a-b}{2}=\frac{1}{\sqrt{2}}\cdots \left(1\right)$
$\cos a+\cos b=\frac{\sqrt{6}}{2}$
$\Rightarrow 2\cos \frac{a+b}{2}\cos \frac{a-b}{2}=\frac{\sqrt{6}}{2}\cdots \left(2\right)$

Dividing $\left(1\right)$ and $\left(2\right)$, we get
$\tan \frac{a+b}{2}=\frac{1}{\sqrt{3}}$
$\Rightarrow a+b=\frac{\pi }{3}$

$\therefore \sin (a+b)=\frac{\sqrt{3}}{2}$