## Calculus (3rd Edition)

(a) $a=\pi/3,\ \ b=\pi/6$ (b) $a= \pi$ (other answers are possible.)
(a) Consider the values: $a=\pi/3,\ \ b=\pi/6$ Then we have \begin{align*} \cos a&= \frac{1}{2}\\ \cos b&= \frac{\sqrt{3}}{2} \end{align*} But $$\cos (a+b)=\cos (\pi/2)=0$$ Thus, we see that $$\cos (a+b) \neq \cos a+\cos b$$ (b) Consider $a= \pi$: \begin{align*} \cos(a/2)&=0\\ \frac{\cos a}{2}&=-0.5 \end{align*} Thus $$\cos \frac{a}{2} \neq \frac{\cos a}{2}$$