If cos x + cos y - cos (x + y) = frac{3}{2},t | vedclass

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(C) Given: $\cos x + \cos y - \cos (x + y) = \frac{3}{2}$ Using the formula $\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}

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$x = y$

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(C) Given: $\cos x + \cos y - \cos (x + y) = \frac{3}{2}$ Using the formula $\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)$ and $\cos (x+y) = 2 \cos^2 \left(\frac{x+y}{2}ight) - 1$,we get: $2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) - (2 \cos^2 \left(\frac{x+y}{2}ight) - 1) = \frac{3}{2}$ $2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) - 2 \cos^2 \left(\frac{x+y}{2}ight) = \frac{1}{2}$ Multiplying by $2$: $4 \cos^2 \left(\frac{x+y}{2}ight) - 4 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) + 1 = 0$ Let $t = \cos \left(\frac{x+y}{2}ight)$. Then $4t^2 - 4t \cos \left(\frac{x-y}{2}ight) + 1 = 0$. Since $t$ is real,the discriminant $D \geq 0$: $(-4 \cos \left(\frac{x-y}{2}ight))^2 - 4(4)(1) \geq 0$ $16 \cos^2 \left(\frac{x-y}{2}ight) - 16 \geq 0$ $\cos^2 \left(\frac{x-y}{2}ight) \geq 1$. Since the maximum value of $\cos^2 	heta$ is $1$,we must have $\cos^2 \left(\frac{x-y}{2}ight) = 1$. This implies $\frac{x-y}{2} = 0$,so $x = y$.

If $\cos x + \cos y - \cos (x + y) = \frac{3}{2}$,then

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