If sin 2 θ + sin 2 ϕ

Solution:

(b) $\frac{5}{8}$

Given:

$\sin 2 \theta+\sin 2 \phi=\frac{1}{2}$        ...(i)

and

$\cos 2 \theta+\cos 2 \phi=\frac{3}{2}$       ....(ii)

Squaring and adding (i) and (ii), we get:

$(\sin 2 \theta+\sin 2 \phi)^{2}+(\cos 2 \theta+\cos 2 \phi)^{2}=\frac{1}{4}+\frac{9}{4}$

$\Rightarrow\left[2 \sin \left(\frac{2 \theta+2 \phi}{2}\right) \cos \left(\frac{2 \theta-2 \phi}{2}\right)\right]^{2}+\left[2 \cos \left(\frac{2 \theta+2 \phi}{2}\right) \cos \left(\frac{2 \theta-2 \phi}{2}\right)\right]^{2}=\frac{5}{2}$

$\Rightarrow 4 \sin ^{2}(\theta+\phi) \cos ^{2}(\theta-\phi)+4 \cos ^{2}(\theta+\phi) \cos ^{2}(\theta-\phi)=\frac{5}{2}$

 

$\Rightarrow 4 \cos ^{2}(\theta-\phi)\left[\sin ^{2}(\theta+\phi)+\cos ^{2}(\theta+\phi)\right]=\frac{5}{2}$

$\Rightarrow 4 \cos ^{2}(\theta-\phi)=\frac{5}{2}$

 

$\Rightarrow \cos ^{2}(\theta-\phi)=\frac{5}{8}$