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

Answer: (C)

Given,
$\sin a+\sin b=\frac{1}{\sqrt{2}}...(i)$
$\cos a+\cos b=\frac{\sqrt{6}}{2}...(ii)$
On squaring both sides in Eq. (i), we get
${\sin }^{2}a+{\sin }^{2}b+2\sin a\sin b=\frac{1}{2}...(iii)$
And on squaring both sides in Eq. (ii), we get
${\cos }^{2}a+{\cos }^{2}b+2\cos a\cos b=\frac{6}{4}=\frac{3}{2}...(iv)$
Now, by adding Eqs. (iii) and (iv), we get
$\left({\sin }^{2}a+{\sin }^{2}b+2\sin a\sin b\right)$
$+\left({\cos }^{2}a+{\cos }^{2}b+2\cos a\cos b\right)=\frac{1}{2}+\frac{3}{2}$
$\Rightarrow \left({\sin }^{2}a+{\cos }^{2}a\right)+\left({\sin }^{2}b+{\cos }^{2}b\right)$
$+2(\sin a\sin b+\cos a\cos b)=\frac{4}{2}$
$\Rightarrow 1+1+2\cos (a-b)=2$
$\therefore \cos (a-b)=0$...(v)
On multiplying Eqs. (i) and (ii), we get
$(\sin a+\sin b)(\cos a+\cos b)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{6}}{2}$
$\Rightarrow \sin a\cos a+\sin a\cos b+\sin b\cos a+\sin b\cos b=\frac{\sqrt{6}}{2\sqrt{2}}$
$\Rightarrow (\sin a\cos a+\sin b\cos b)+(\sin a\cos b+\cos a\sin b)=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{1}{2}(\sin 2a+\sin 2b)+\sin (a+b)=\frac{\sqrt{3}}{2}$
$\Rightarrow \sin (a+b)\cos (a-b)+\sin (a+b)=\frac{\sqrt{3}}{2}$
$\Rightarrow 0+\sin (a+b)=\frac{\sqrt{3}}{2}$
$[$ From Eq. $(v),\cos (a-b)=0]$
$\Rightarrow \sin (a+b)=\frac{\sqrt{3}}{2}$