If sin x - sin y = frac{27}{65} and cos x - c | vedclass

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(C) Given: $\sin x - \sin y = \frac{27}{65}$  $(1)$ $\cos x - \cos y = -\frac{21}{65}$  $(2)$ Squaring and adding $(1)$ and $(2)$: $(\sin x - \sin y)^

Vedclass · Accepted Answer

$\frac{63}{65}$

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(C) Given: $\sin x - \sin y = \frac{27}{65}$  $(1)$ $\cos x - \cos y = -\frac{21}{65}$  $(2)$ Squaring and adding $(1)$ and $(2)$: $(\sin x - \sin y)^2 + (\cos x - \cos y)^2 = (\frac{27}{65})^2 + (-\frac{21}{65})^2$ $(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) - 2(\sin x \sin y + \cos x \cos y) = \frac{729 + 441}{4225}$ $1 + 1 - 2 \cos(x - y) = \frac{1170}{4225}$ $2 - 2 \cos(x - y) = \frac{18}{65}$ $2 \cos(x - y) = 2 - \frac{18}{65} = \frac{112}{65} \implies \cos(x - y) = \frac{56}{65}$ Since $\sin^2(x - y) = 1 - (\frac{56}{65})^2 = \frac{4225 - 3136}{4225} = \frac{1089}{4225}$,we have $\sin(x - y) = \pm \frac{33}{65}$. Using the identity $\sin x - \sin y = 2 \sin(\frac{x-y}{2}) \cos(\frac{x+y}{2})$ and $\cos x - \cos y = -2 \sin(\frac{x-y}{2}) \sin(\frac{x+y}{2})$,Dividing the two equations: $\frac{\sin x - \sin y}{\cos x - \cos y} = \frac{2 \sin(\frac{x-y}{2}) \cos(\frac{x+y}{2})}{-2 \sin(\frac{x-y}{2}) \sin(\frac{x+y}{2})} = -\cot(\frac{x+y}{2}) = \frac{27/65}{-21/65} = -\frac{27}{21} = -\frac{9}{7}$. Thus,$	an(\frac{x+y}{2}) = \frac{7}{9}$. Using $\sin(x+y) = \frac{2 	an(\frac{x+y}{2})}{1 + 	an^2(\frac{x+y}{2})} = \frac{2(7/9)}{1 + (49/81)} = \frac{14/9}{130/81} = \frac{14}{9} 	imes \frac{81}{130} = \frac{14 	imes 9}{130} = \frac{126}{130} = \frac{63}{65}$.

If $\sin x - \sin y = \frac{27}{65}$ and $\cos x - \cos y = \frac{-21}{65}$,then find the value of $\sin(x + y)$.

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