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  4. If sin 2 θ+ sin 2 varphi=(1/2) and cos 2 θ+ cos 2 varphi=(3/2), then cos 2(θ- varphi)=

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

TS EAMCET 2020

Solution:

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