Question

If $\sin 2\theta +\sin 2\phi =\frac{1}{2}$ and $\cos 2\theta +\cos 2\phi =\frac{3}{2}$, then ${\cos }^2(\theta -\phi )=$

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

Answer 1

Answer: (B)

Given
$\sin 2\theta +\sin 2\phi =\frac{1}{2}\dots$ (i)
$\cos 2\theta +\cos 2\phi =\frac{3}{2}\dots$ (ii)
Squaring on both Eqs. (i) and (ii), we get
$(\sin 2\theta +\sin 2\phi {)}^2=\frac{1}{4}$
$\Rightarrow {\sin }^{2}2\theta +{\sin }^{2}2\phi +2\sin 2\theta \sin 2\phi =\frac{1}{4}$
$\Rightarrow {\sin }^{2}2\theta +{\sin }^{2}2\phi +2\sin 2\theta \sin 2\phi =\frac{1}{4}$ ... (iii)
$(\cos 2\theta +\cos 2\phi {)}^2=\frac{9}{4}$
$\Rightarrow {\cos }^{2}2\theta +{\cos }^{2}2\phi +2\cos 2\theta \cos 2\phi =\frac{9}{4}$ ....(iv)
On adding Eqs. (iii) and (iv), we get
$\Rightarrow {\sin }^{2}2\theta +{\cos }^{2}2\theta +{\sin }^{2}2\phi +{\cos }^{2}2\phi$
$+2(\sin 2\theta \sin 2\phi )+\cos 2\theta \cos 2\phi )=\frac{10}{4}$
$\Rightarrow 2+2\{\cos (2\theta -2\phi )\}=\frac{10}{4}$
$\Rightarrow 2\{1+\cos 2(\theta -\phi )\}=\frac{10}{4}$
$\Rightarrow 1+\cos 2(\theta -\phi )=\frac{10}{8}$
$\Rightarrow 2{\cos }^2(\theta -\phi )=\frac{10}{8}$
$\Rightarrow {\cos }^2(\theta -\phi )=\frac{5}{8}$