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

Answer 1

Answer: (B)

$\frac{\sqrt{3}}{2}$

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