2007 AMC 12A Problem 17

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.