If 0

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(B) Given $\cos x + \cos y - \cos(x + y) = \frac{3}{2}$.Using the identity $\cos x + \cos y = 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2})$ and $\cos(x+y

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$\frac{1+\sqrt{3}}{2}$

Vedclass · Answer

(B) Given $\cos x + \cos y - \cos(x + y) = \frac{3}{2}$.Using the identity $\cos x + \cos y = 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2})$ and $\cos(x+y) = 2 \cos^2(\frac{x+y}{2}) - 1$,we get:$2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) - (2 \cos^2(\frac{x+y}{2}) - 1) = \frac{3}{2}$$-2 \cos^2(\frac{x+y}{2}) + 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + 1 = \frac{3}{2}$$2 \cos^2(\frac{x+y}{2}) - 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + \frac{1}{2} = 0$Multiply by $2$:$4 \cos^2(\frac{x+y}{2}) - 4 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + 1 = 0$This is a quadratic in $\cos(\frac{x+y}{2})$. Completing the square:$(2 \cos(\frac{x+y}{2}) - \cos(\frac{x-y}{2}))^2 + 1 - \cos^2(\frac{x-y}{2}) = 0$$(2 \cos(\frac{x+y}{2}) - \cos(\frac{x-y}{2}))^2 + \sin^2(\frac{x-y}{2}) = 0$Since both terms are squares,they must be zero:$\sin(\frac{x-y}{2}) = 0 \Rightarrow x = y$$2 \cos(\frac{x+y}{2}) - \cos(0) = 0$ $\Rightarrow 2 \cos x = 1$ $\Rightarrow \cos x = \frac{1}{2}$Since $0 Then $\sin x = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos y = \cos(\frac{\pi}{3}) = \frac{1}{2}$.Therefore,$\sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1+\sqrt{3}}{2}$.

If $0 < x, y < \pi$ and $\cos x + \cos y - \cos(x + y) = \frac{3}{2}$,then $\sin x + \cos y$ is equal to ...... .

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