I'm trying to:
Find period of $a\sin{p_1x\over q_1}+b\cos{p_2x \over q_2}$ where $p_1, p_2, q_1, q_2 \in \mathbb N$ and $p_1q_2\ne p_2q_1$
Let: $$ f(x) = a\sin{p_1x\over q_1} \\ g(x) = b\cos{p_2x \over q_2} $$
Then period of $f(x)$ is $T_f =2\pi {q_1\over p_1}$ and period of $g(x)$ is $T_g = 2\pi {q_2\over p_2}$. Let ${q_1 \over p_1} = n_0$ and ${q_2 \over p_2} = m_0$. So we have:
$$ T_f = 2\pi n_0\\ T_g = 2\pi m_0 $$
Consider $h(x) = f(x) + g(x)$:
$$ h(x) = h(x+2\pi m_0n_0) = f(x+2\pi m_0n_0) + g(x+2\pi m_0n_0) $$
Therefore: $$ T_h = 2\pi m_0n_0 = 2\pi m_0{q_1 \over p_1} = 2\pi n_0{q_2 \over p_2} $$
I'm learning math myself and have nobody to refer to. Could someone tell me whether the above is a correct way of solving that problem?