Is $\cos(4x)+\sin(2x)$ periodic and how do I calculate the primitive period?

Question

My first attempt is under this, i can work out the primitive period of both of the $\cos(4x)$ and $\sin(2x)$ but how do I calculate the primitive period of $\cos(4x)+\sin(2x)$?

My attempt:

Let $u=4x$ then $x=\frac{u}{4}$ and $\cos(u)$ is $2\pi$ periodic thus $T=\frac{2\pi}{4}$ hence $\cos(4x)$ is periodic with primitive period $T=\frac{\pi}{4}$.

Now Let $H=2x$ and thus $x=\frac{H}{2}$ and $\sin(H)$ is also $2\pi$ periodic Thus $\sin(2x)$ is periodic with primitive period $T=\frac{2\pi}{2}=\pi$ but i dont know how to combine these results to calculate the primitive period of the sum of both $\cos(4x)$ and $\sin(2x)$

Answer 1

Your attempt seems fine. Choose $T = \pi$, because that is a multiple of $T_1 = \pi /4$ and $T_2 = \pi$. Then, \begin{align} f(x + T) & = \cos (4(x+T)) + \sin (2(x+T)) \\ & = \cos (4x + 4T) + \sin (2x + 2T) \\ & = \cos(4x+4\pi) + \sin (2x + 2 \pi) \\ & = \cos (4x) + \sin (2x) \\ & = f(x). \end{align}

Answer 2

If the period of $f(x)$ is $2T$, and the period of $g(x)$ is $T$, then their sum, $h(x) = f(x)+g(x)$, is periodic with period no shorter than $2T$, as you have essentially shown: For any $x$,

$$ h(x+2T) = f(x+2T)+g(x+2T) = f(x)+g(x) = h(x) $$

It then remains to show that $h(x)$ is not periodic with period equal to $T$. Find two values, separated by $T$, for which $h(x)$ is not equal: that is, find $x_0$ such that $h(x_0) \not= h(x_0+T)$.

Answer 3

Your function is $$f(x)=\cos4x+\sin2x=1-2\sin^22x+\sin2x$$

Since it is a function of $\sin2x$ which has period $\pi$, the period of $f$ is also $\pi$