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Question

If $0<x,y<\pi$ and $\cos x+\cos y-\cos(x+y) = \dfrac32$, then $\sin x+\cos y$ is equal to:

Accepted Answer

Using sum-to-product and $\cos(x+y)=2\cos^2\tfrac{x+y}2-1$:

$$2\cos\tfrac{x+y}2\cos\tfrac{x-y}2 - 2\cos^2\tfrac{x+y}2 = \tfrac12$$

This rearranges (completing the square in $\cos\tfrac{x-y}2$) to:

$$\left[\cos\tfrac{x-y}2-2\cos\tfrac{x+y}2\right]^2 + \sin^2\tfrac{x-y}2 = 0$$

Both terms are non-negative, so both vanish: $\sin\tfrac{x-y}2=0 \implies x=y$, and $\cos0-2\cos x=0 \implies \cos x=\tfrac12$.

So $\cos x=\cos y=\tfrac12$, and $\sin x=\sqrt{1-\tfrac14}=\tfrac{\sqrt3}2$.

$$\sin x+\cos y = \frac{\sqrt3}2+\frac12 = \frac{1+\sqrt3}2$$

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