Showing that the function $\sin 2x +\cos 3x $ is periodic

Question

I would like to prove that the function

\begin{equation} \sin 2x +\cos 3x \end{equation}

is periodic and calculate its period. I tried to replace $x+T$ instead of $x$, and I got:

\begin{equation} \sin2x \cos 2T +\cos 2x \sin 2T +\cos 3x \cos 3T -\sin 3x \sin 3T \end{equation}

From this point I do not know how to continue. Any suggestions, please?

Thank you very much

Answer 1

From where you left, you need

\begin{equation} \sin(2x)\cos(2T)+\cos(2x)\sin(2T)+\cos(3x)\cos(3T)-\sin(3x)\sin(3T)=\sin(2x)+\cos(3x) \end{equation} for all $x$. As the functions $\sin(2x)$, $\sin(3x)$, $\cos(2x)$, $\cos(3x)$ are linearly independent, this forces $$ \cos(2T)=1,\ \sin(2T)=0,\ \cos(3T)=1,\ \sin(3T)=0. $$ The equalities $\sin(2T)=\sin(3T)=0$ imply that $T=k\pi$ for some $k\in\mathbb N$ (positive, because $T$ is a period). The $\cos(2k\pi)=1$ for all $k$, but $\cos(3k\pi)=1$, requires $k$ to be even. So the smallest possible value is $T=2\pi$.

Answer 2

Show that this function returns the same values for $x$ and $x+2\pi$ by direct substitution

Answer 3

It is clear that $2\pi$ is a period of $\sin 2x +\cos 3x$. We show that $2\pi$ is the smallest (positive) period.

Let $\alpha$ be a period of our function. Then for all $x$, $$\sin 2x+\cos 3x=\sin 2(x+\alpha) +\cos 3(x+\alpha).\tag{$1$}$$ Differentiate twice, and change signs. (It would be enough to differentiate once, but twice is neater.) We get $$4\sin 2x+9\cos 3x= 4\sin 2(x+\alpha) +9\cos 3(x+\alpha).\tag{$2$}$$ Use Equations $(1)$ and $(2)$ to eliminate the $\cos$ terms. We get $$5\sin 2x =5\sin 2(x+\alpha).$$ Thus $\alpha$ is a period of $\sin 2x$. Similarly, $\alpha$ is a period of $\cos 3x$. But the smallest common positive period of $\sin 2x$ and $\cos 3x$ $\alpha$ is $2\pi$.