C2. A function $f(x)$ is periodic with period $T>0$ if $f(x+T)=f(x)$ for all $x$. The smallest such number $T$ is called the least period. For example, the functions $\sin (x)$ and $\cos (x)$ are periodic with least period $2 \pi$.
(a) Let a function $g(x)$ be periodic with the least period $T=\pi$. Determine the least period of $g(x / 3)$.
(b) Determine the least period of $H(x)=\sin (8 x)+\cos (4 x)$.
(c) Determine the least periods of each of $G(x)=\sin (\cos (x))$ and $F(x)=\cos (\sin (x))$.
Solution: A period of $g\left(\frac{x}{3}\right)$ is $3 \pi$ because $g\left(\frac{x+3 \pi}{3}\right)=g\left(\frac{x}{3}+\pi\right)=g\left(\frac{x}{3}\right)$. Note that if $g(x / 3)$ had a period smaller than $3 \pi$, then $g(x)$ would have a period smaller than $\pi$, a contradiction.
The answer is $3 \pi$.
Solution: The least period of $\sin (8 x)$ is $\frac{\pi}{4}$ because $\sin \left(8\left(x+\frac{\pi}{4}\right)\right)=\sin (8 x+2 \pi)=\sin (8 x)$. The least period of $\cos (4 x)$ is $\frac{\pi}{2}$ because $\cos \left(4\left(x+\frac{\pi}{2}\right)\right)=\cos (4 x+2 \pi)=\cos (4 x)$. We pick the larger of the two periods for the sum of the functions. In general, it is the least common multiple of the two periods that must be picked.
Note that if $H(x)$ had a period smaller than $\pi / 2$, then $\cos (4 x)=H(x)-\sin (8 x)$ would also have a period smaller than $\pi / 2$, and so $\cos x$ would have a period smaller than $2 \pi$. Indeed, if $\cos (4 x)=\cos (4(x+T))=$ $\cos (4 x+4 T)$ and $T<\pi / 2$ then $\cos y=\cos (y+4 T)$ and $4 T<4(\pi / 2)=2 \pi$. This is a contradiction.

The answer is $\pi / 2$.
Solution in two parts:
1. $G(x+2 \pi)=\sin (\cos (x+2 \pi))=\sin (\cos (x))=G(x)$.

To prove that $T=2 \pi$ is the smallest positive number consider for example $x=0$. Then $G(0)=\sin (1)$. For $G(T)=\sin (\cos (T))=\sin (1)$, we need either $\cos (T)=1+2 \pi k$ or $\cos (T)=-1+(2 k+1) \pi, k \in \mathbf{Z}$. In the first case we see that the inequality $-1 \leq 1+2 \pi k \leq 1$ is true only for $k=0$. This gives $\cos (T)=1$, and so the smallest positive $T=2 \pi$. In the second case the inequality $-1 \leq-1+(2 k+1) \pi \leq 1$ is true for no integer $k$. Thus, the only solution is $T=2 \pi$.
2. $F(x+\pi)=\cos (\sin (x+\pi))=\cos (-\sin (x))=\cos (\sin (x))=F(x)$.

To prove that $T=\pi$ is the smallest positive number consider for example $x=0$. Then $F(0)=1$. Now, we need $F(0+T)=\cos (\sin (T))=1$. Thus, $\sin T=0$, and so the smallest positive $T=\pi$.
The answers are $2 \pi$ and $\pi$.