Question

If $\sin 2A+\sin 2B=\frac{1}{2}$ and $\cos 2A+\cos 2B=\frac{3}{2}$, then the value of $|\cos (A-B)|$ is

• A. $\sqrt{\frac{5}{8}}$
• B. $\sqrt{\frac{3}{8}}$
• C. $\frac{5}{8}$
• D. $\frac{3}{8}$

Answer 1

The correct option is A $\sqrt{\frac{5}{8}}$

$\sin 2A+\sin 2B=\frac{1}{2}\cdots \left(1\right)$
$\cos 2A+\cos 2B=\frac{3}{2}\cdots \left(2\right)$

Squaring and adding equations $\left(1\right)$ and $\left(2\right)$, we get
${\sin }^{2}2A+{\sin }^{2}2B+2\sin 2A\sin 2B+{\cos }^{2}2A+{\cos }^{2}2B+2\cos 2A\cos 2B=\frac{1+9}{4}\Rightarrow 2+2(\cos 2A\cos 2B+\sin 2A\sin 2B)=\frac{5}{2}\Rightarrow 2(1+\cos 2(A-B\left)\right)=\frac{5}{2}$
$\Rightarrow 4{\cos }^2(A-B)=\frac{5}{2}$
$\Rightarrow {\cos }^2(A-B)=\frac{5}{8}\therefore |\cos (A-B)|=\sqrt{\frac{5}{8}}$