Question : If $\sin m+\sin n=p, \cos m+\cos n=q$, then find the value of $\sin m \times \sin n+\cos m \times \cos n$.
Option 1: $p^2+q^2-2$
Option 2: $\frac{p^2+q^2-2}{2}$
Option 3: $p+q-p q$
Option 4: $p+q+p q$
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Correct Answer: $\frac{p^2+q^2-2}{2}$
Solution :
Given,
$\sin m+\sin n=p$ and $ \cos m+\cos n=q$
Consider, $\sin m + \sin n = p$
Squaring both sides,
⇒ $(\sin m + \sin n)^2 = p^2$
⇒ $\sin^2m+\sin^2n+2\sin m\sin n=p^2$
We know, $\sin^2x+\cos^2x=1$
⇒ $1 - \cos^2 m + 1 - \cos^2 n + 2 × \sin m × \sin n = p^2$ ................(i)
Now, consider, $\cos m + \cos n = q$
Squaring both sides
⇒ $\cos^2 m + \cos^2 n + 2 × \cos m × \cos n = q^2$ ................(ii)
Add equation (i) and (ii),
⇒ $(1 - \cos^2 m + 1 - cos^2 n + 2 × \sin m × \sin n) +(\cos^2 m + \cos^2 n + 2 × \cos m × \cos n)= p^2+q^2$
⇒ $2 + 2 × \sin m × \sin n + 2 × \cos m × \cos n = p^2 + q^2$
⇒ $2(\sin m × \sin n + \cos m ×\cos n) = p^2 + q^2 - 2$
⇒ $\sin m × \sin n + \cos m × \cos n = \frac{p^2 + q^2 - 2}{2}$
Hence, the correct answer is $\frac{p^2 + q^2 - 2}{2}$.
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