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If $\sin ^{2} \theta+\sin ^{2} \phi=1 / 2, \cos ^{2} \theta+\cos ^{2} \phi=3 / 2,$ then $\cos ^{2}(\theta-\phi)$ is equal to
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$5 / 8$
Using cosine formula
$2 \sin (\theta+\phi) \cos (\theta-\phi)=1 / 2$ ...(i)
$2 \cos (\theta+\phi) \cos (\theta-\phi)=3 / 2$ ...(ii)
Squaring (1) and (2) and then adding $\cos ^{2}(\theta-\phi)=\frac{1}{4}+\frac{9}{4}=\frac{5}{2} \Rightarrow \cos ^{2}(\theta-\phi)=\frac{5}{8}$
$2 \sin (\theta+\phi) \cos (\theta-\phi)=1 / 2$ ...(i)
$2 \cos (\theta+\phi) \cos (\theta-\phi)=3 / 2$ ...(ii)
Squaring (1) and (2) and then adding $\cos ^{2}(\theta-\phi)=\frac{1}{4}+\frac{9}{4}=\frac{5}{2} \Rightarrow \cos ^{2}(\theta-\phi)=\frac{5}{8}$
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