AoPS - Problem

Source:

Tags: Comc, 2017 COMC

lion11202

12.10.2018 19:22

Source: 2017 Canadian Open Math Challenge, Problem 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 $2\pi$. $\qquad$(a) Let a function $g(x)$ be periodic with the least period $T = \pi$. Determine the least period of $g(x/3)$. $\qquad$(b) Determine the least period of $H(x) = sin(8x) + cos(4x)$ $\qquad$(c) Determine the least periods of each of $G(x) = sin(cos(x))$ and $F(x) = cos(sin(x))$.

b) If $H(x)$ has period $p$, then so do $H(x)+H(-x)=2\cos(4x)$ and $H(x)-H(-x)=2\sin 8x$. Hence $\sin x$ has period $8p$ and $\cos x$ has period $4p$. This implies $8p\equiv 0(2\pi),4p\equiv 0(2\pi)$ hence $p\equiv 0(\pi/2)$ and $p={\pi\over 2}$ is a period, hence the smallest one. c) If $p>0$ is a period for $G$ then $0=G(\pi/2)=G(\pi/2+p)=\sin (\cos (\pi/2+p))=-\sin (\sin p)\Rightarrow \sin p=0$ hence $p=k\pi$ for some integer $k\ge 1$. Now $G(x+\pi)=-G(x)$ hence the least period is $2\pi$. If $p>0$ is a period for $F$ then $1=F(0)=F(p)=\cos(\sin p)$ hence $\sin p=0$ and $p=k\pi$ where $k\ge 1$. The least period is $\pi$.