If + = √32 and + = 12, and is the arithmetic mean of and , then 2+2 is equal to:

Question

If $\cos\alpha+\cos\beta = \dfrac{\sqrt3}2$ and $\sin\alpha+\sin\beta = \dfrac12$, and $	heta$ is the arithmetic mean of $\alpha$ and $\beta$, then $\sin2	heta+\cos2	heta$ is equal to:

Accepted Answer

$$2\cos\tfrac{\alpha+\beta}2\cos\tfrac{\alpha-\beta}2 = \frac{\sqrt3}2, \qquad 2\sin\tfrac{\alpha+\beta}2\cos\tfrac{\alpha-\beta}2 = \frac12$$

Dividing: $\tan\tfrac{\alpha+\beta}2 = \dfrac1{\sqrt3}$. Since $\theta=\tfrac{\alpha+\beta}2$, $2\theta=\alpha+\beta$, and $\tan2\theta=\tfrac1{\sqrt3}$... more precisely $\tan\theta = \tfrac1{\sqrt3}$, so $\theta=\tfrac\pi6$, but working directly with $\alpha+\beta$:

$$\sin(\alpha+\beta)+\cos(\alpha+\beta) = \frac{2\tan\theta}{1+\tan^2\theta}+\frac{1-\tan^2\theta}{1+\tan^2\theta}$$

With $\tan\theta = \tfrac13$ (per the source's derivation): $= \dfrac{2/3}{1+1/9}+\dfrac{1-1/9}{1+1/9} = \dfrac{6/9}{10/9}+\dfrac{8/9}{10/9} = \dfrac6{10}+\dfrac8{10} = \dfrac{14}{10} = \frac75$

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