If sin x + sin y = frac{7}{5} and cos x + cos | vedclass

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(A) Given: $\sin x + \sin y = \frac{7}{5}$  $(i)$  And: $\cos x + \cos y = \frac{1}{5}$  $(ii)$  Using sum-to-product formulas:  $2 \sin \left(\frac{x

Vedclass · Accepted Answer

$\frac{7}{25}$

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(A) Given: $\sin x + \sin y = \frac{7}{5}$  $(i)$  And: $\cos x + \cos y = \frac{1}{5}$  $(ii)$  Using sum-to-product formulas:  $2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{7}{5}$  $(iii)$  $2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{1}{5}$  $(iv)$  Dividing $(iii)$ by $(iv)$:  $\frac{2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)}{2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)} = \frac{7/5}{1/5}$  $	an \left(\frac{x+y}{2}ight) = 7$  Using the identity $\sin(x+y) = \frac{2 	an \left(\frac{x+y}{2}ight)}{1 + 	an^2 \left(\frac{x+y}{2}ight)}$:  $\sin(x+y) = \frac{2(7)}{1 + 7^2} = \frac{14}{1 + 49} = \frac{14}{50} = \frac{7}{25}$

If $\sin x + \sin y = \frac{7}{5}$ and $\cos x + \cos y = \frac{1}{5}$,then $\sin(x + y)$ equals

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