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 + \frac{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 + \frac{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`