MCQ
$\text{If}\ \sin2\theta+\sin2\phi=\frac{1}{2}\ \text{and}\ \cos2\theta\\ \ +\cos2\phi=\frac{3}{2},\text{then}\cos^2(\theta-\phi)=$
- A$\frac{3}{8}$
- B$\frac{5}{8}$
- C$\frac{3}{4}$
- D$\frac{5}{4}$
Solution:$$
Given:
$\sin2\theta+\sin2\phi=\frac{1}{2}....(\text{i})$
and
$\cos2\theta+\cos2\phi=\frac{3}{2}....(\text{ii})$
Squaring and adding (i) and (ii), we get:
$(\sin2\theta+\sin2\phi)^2+(\cos2\theta+\cos2\phi)^2=\frac{1}{4}+\frac{9}{4}$
$\Rightarrow\ \Big[2\sin\Big(\frac{2\theta+2\phi}{2}\Big)\cos\Big(\frac{2\theta-2\phi}{2}\Big)\Big]^2\\ \ \ \ \ +\Big[2\cos\Big(\frac{2\theta+2\phi}{2}\Big)\cos\Big(\frac{2\theta-2\phi}{2}\Big)\Big]^2=\frac{5}{2}$
$\Rightarrow\ 4\sin^2(\theta+\phi)\cos^2(\theta-\phi)\\ \ \ +4\cos^2(\theta+\phi)\cos^2(\theta-\phi)=\frac{5}{2}$
$\Rightarrow\ 4\cos^2(\theta-\phi)[\sin^2(\theta+\phi)+\cos^2(\theta+\phi)]=\frac{5}{2}$
$\Rightarrow\ 4\cos^2(\theta-\phi)=\frac{5}{2}$
$\Rightarrow\ \cos^2(\theta-\phi)=\frac{5}{8}$
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