Question

Show that the function f(x)=sin3x + cos5x is periodic and it’s period.

Answer 1

The period of $f(x)$ is $\boxed{2\pi}$.

Recall that $\sin(x)$ and $\cos(x)$ both have periods of $2\pi$. This means

$\sin(x + 2\pi) = \sin(x)$

$\cos(x + 2\pi) = \cos(x)$

Replacing $x$ with $3x$, we have

$\sin(3x + 2\pi) = \sin\left(3 \left(x + \dfrac{2\pi}3\right)\right) = \sin(3x)$

In other words, if we change $x$ by some multiple of $\frac{2\pi}3$, we end up with the same output. So $\sin(3x)$ has period $\frac{2\pi}3$.

Similarly, $\cos(5x)$ has a period of $\frac{2\pi}5$,

$\cos(5x + 2\pi) = \cos\left(5 \left(x + \dfrac{2\pi}5\right)\right) = \cos(5x)$

We want to find the period $p$ of $f(x)$, such that

$f(x + p) = f(x)$

$\implies \sin(3x + p) + \cos(5x + p) = \sin(3x) + \cos(5x)$

On the left side, we have

$\sin(3x + p) = \sin(3x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \sin(3x+2\pi) \cos(p-2\pi) + \cos(3x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \sin(3x) \cos(p-2\pi) + \cos(3x) \sin(p - 2\pi)$

and

$\cos(5x + p) = \cos(5x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \cos(5x+2\pi) \cos(p-2\pi) - \sin(5x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \cos(5x) \cos(p-2\pi) - \sin(5x) \sin(p-2\pi)$

So, in terms of its period, we have

$f(x) = \sin(3x) \cos(p - 2\pi) + \cos(3x) \sin(p - 2\pi) \\\\ ~~~~~~~~ ~~~~+ \cos(5x) \cos(p - 2\pi) - \sin(5x) \sin(p - 2\pi)$

and we need to find the smallest positive $p$ such that

$\begin{cases} \cos(p - 2\pi) = 1 \\ \sin(p - 2\pi) = 0 \end{cases}$

which points to $p=2\pi$, since

$\cos(2\pi-2\pi) = \cos(0) = 1$

$\sin(2\pi - 2\pi) = \sin(0) = 0$