Question

For some particular values of $A$ and $B,$ $-\sin A+\sin B=\frac{\sqrt{3}}{2\sqrt{2}}$ and $\cos A+\cos B=\frac{1}{2\sqrt{2}}.$ Then the value of $\cos (A+B)$ is

• A. $-\frac{3}{2}$
• B. $-\frac{3}{4}$
• C. $-\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B $-\frac{3}{4}$

$-\sin A+\sin B=\frac{\sqrt{3}}{2\sqrt{2}}$
Squaring both the sides
${\sin }^{2}A+{\sin }^{2}B-2\sin A.\sin B=\frac{3}{8}...\left(1\right)$

$\cos A+\cos B=\frac{1}{2\sqrt{2}}$
Squaring both the sides
${\cos }^{2}A+{\cos }^{2}B+2\cos A.\cos B=\frac{1}{8}...\left(2\right)$

Adding eqn$\left(1\right)$ and $\left(2\right)$, we get
$1+1+2\cos A.\cos B-2\sin A.\sin B=\frac{3}{8}+\frac{1}{8}$
$\Rightarrow \cos (A+B)=-\frac{3}{4}$