I know the fact that the period of $\sin(x)$ and $\cos(x)$ functions is $2\pi$. My teacher taught me how to find period of $\sin2x$. But I'm interested in how to find the period of nested functions like period of:
- $\sin(\sin(x/3))$
- $\cos(\sin(\sin(x/3)))$
I tried this method:
$\sf Period \,of \,\sin(x/3) = \dfrac{Period \, of \, \sin(x)}{Coefficient\,of\,x \, in\, function}$
$\implies 6\pi$
But how to find period of $\sin(f(x))$ where period of $f(x) = 6\pi$?
Same like this, I stuck on $\sin(\sin(x/3))$.
Please help me, I'm trying to solve it since yesterday but I'm still unable to figure it out.
Best Answer
Let the period of $\cos(\sin(\sin(\frac x3)))$ be $P$, where $P>0$ is the smallest value satisfying the equation $$\cos(\sin(\sin(\frac x3)))=\cos(\sin(\sin(\frac {x+P}{3})))$$ for all values of $x$. $$\implies \sin(\sin(\frac x3))=\pm\sin(\sin(\frac {x+P}{3}))+k\cdot2\pi, k\in\mathbb{Z}$$
Due to the range of $\sin$ this equation has no solution unless $k=0$. So now we solve $$\sin(\sin(\frac x3))=\pm\sin(\sin(\frac {x+P}{3}))=\sin(\pm\sin(\frac {x+P}{3}))$$
So either $$\sin(\frac x3)=\pm\sin(\frac {x+P}{3})+k'\cdot 2\pi,k'\in\mathbb{Z}$$ or $$\sin(\frac x3)=\pi\mp\sin(\frac {x+P}{3})+k'\cdot 2\pi,k'\in\mathbb{Z}$$
Again, due to the range of $\sin$, we must have $k'=0$ in each case, and for the same reason we can ignore the second pair of cases.
So we are reduced to solving $$\sin(\frac x3)=\pm\sin(\frac {x+P}{3})=\sin(\pm\frac {x+P}{3})$$ $$\implies \frac x3=\pm\frac{x+P}{3}+k''\cdot2\pi,k''\in\mathbb{Z}$$ or $$\frac x3=\pi\mp\frac{x+P}{3}+k''\cdot2\pi,k''\in\mathbb{Z}$$
Of these four options, only two can be valid for all values of $x$, i.e. the ones where $x$ is eliminated:
The first one with a $+$ sign gives $$-\frac P3=k''2\pi\implies P=6\pi$$
And the second one with a $-$ sign gives $$0=\pi+\frac P3+k''2\pi\implies P=3\pi$$
So the period is $3\pi$.
You might like to try the first one yourself.