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}}$
$\Rightarrow \,(\sin \,a \cos \,a+\sin \,b \cos\, b)+(\sin \,a \cos \,b + \cos\, a \sin b)=\frac{\sqrt{3}}{2}$
$\Rightarrow \, \frac{1}{2}(\sin 2 a+\sin 2 b)+\sin (a +b)=\frac{\sqrt{3}}{2}$
$\Rightarrow \sin (a +b) \cos (a-b)+\sin (a +b)=\frac{\sqrt{3}}{2}$
$\Rightarrow 0+\sin (a +b)=\frac{\sqrt{3}}{2}$
$[$ From Eq. $(v), \cos (a-b)=0]$
$\Rightarrow \sin (a +b)=\frac{\sqrt{3}}{2}$