Question

If $sin2\theta +sin2\varphi =\frac{1}{2}$ and $cos2\theta +cos2\varphi =\frac{3}{2}$, then $co{s}^2(\theta -\varphi )=$

• A. $\frac{3}{8}$
• B. $\frac{5}{8}$
• C. $\frac{3}{4}$
• D. $\frac{5}{4}$

Answer 1

The correct option is B

$\frac{5}{8}$

Given:
$sin2\theta +sin2\varphi =\frac{1}{2}\dots \left(i\right)$
and
$cos2\theta +cos2\varphi =\frac{3}{2}\dots \left(ii\right)$
Squaring and adding (i) and (ii), we get
$(sin2\theta +sin2\varphi {)}^2+(sin2\theta +sin2\varphi {)}^2=\frac{1}{4}+\frac{9}{4}\Rightarrow \left[2sin\left(\frac{2\theta +2\varphi }{2}\right)cos{\left(\frac{2\theta -2\varphi }{2}\right)}^2\right]+\left[2cos\left(\frac{2\theta +2\varphi }{2}\right)cos{\left(\frac{2\theta -2\varphi }{2}\right)}^2\right]=\frac{5}{2}$

$\Rightarrow 4si{n}^2(\theta +\varphi )co{s}^2(\theta -\varphi )+4co{s}^2(\theta +\varphi )+co{s}^2(\theta -\varphi )=\frac{5}{2}\Rightarrow 4co{s}^2(\theta -\varphi )+\left[si{n}^2\right(\theta +\varphi )+co{s}^2(\theta +\varphi )+\frac{5}{2}\Rightarrow 4co{s}^2(\theta -\varphi )+[si{n}^2(\theta +\varphi )+co{s}^2(\theta +\varphi )+\frac{5}{2}\Rightarrow 4co{s}^2(\theta -\varphi )=\frac{5}{2}\Rightarrow co{s}^2(\theta -\varphi )=\frac{5}{2}$