Suppose that \sin a+\sin b=√\tfrac53 and \cos a+\cos b=1. What is \cos(a-b)?

2007 AMC 12A Problem 17 — Solution

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17.

Suppose that $\sin a+\sin b=\sqrt{\tfrac53}$ and $\cos a+\cos b=1.$ What is $\cos(a-b)?$

$\sqrt{\dfrac53}-1$

$\dfrac13$

$\dfrac12$

$\dfrac23$

$1$

Squaring both equations gives $\sin^2 a+2\sin a\sin b+\sin^2 b=\tfrac53$ and $\cos^2 a+2\cos a\cos b$ $+\cos^2 b=1.$

Adding and using $\sin^2\theta+\cos^2\theta=1$ twice, $\begin{aligned} &2+2(\sin a\sin b+\cos a\cos b) \\ &=\tfrac83. \end{aligned}$

So $\cos(a-b)=\sin a\sin b$ $+\cos a\cos b$ $=\tfrac13.$

Thus, the correct answer is B.