Theorem. The period of simple harmonic motion is $T=\frac{2\pi}{\omega}.$

Proof. (Guevara) The period of cosine is the smallest number $p$ such that $\cos{t}=\cos{\left(t+p\right)}$ for all $t$ in the domain of cosine, so $p=2\pi.$ Let $\nabla$ denote the domain of cosine. Therefore, for all $t\in\nabla, p=2\pi$ is the smallest number such that: \begin{align} x\left(t\right) &=\cos{\left(\omega t+\phi\right)}\\ &=\cos{\left(\omega t+\phi+2\pi\right)}\\ &=\cos{\left(\omega t+2\pi+\phi\right)}\\ &=\cos{ \left[ \omega \left( t+\frac{2\pi}{\omega} \right) +\phi \right]}\\ &=x\left(t+\frac{2\pi}{\omega}\right) \end{align} This says that $p=2\pi$ is the smallest number such that $x\left(t\right)=x\left(t+\frac{2\pi}{\omega}\right)$ for all $t\in\nabla=\mathrm{dom}x$ in the domain of $x.$ Since, by definition, the period of $x$ is the smallest number $T$ such that $x\left(t\right)=x\left(t+T\right).$ Therefore, $T=\frac{2\pi}{\omega}.$ \[\tag*{$\blacksquare$}\]

Proof. (Serway 391) The phase increases by $2\pi\ \mathrm{rad}$ in time $T.$ Therefore, \begin{align} \omega t+\phi+2\pi &= \omega\left(t+T\right)+\phi\\ t+\frac{2\pi}{\omega} &= t+T\\ T &= \frac{2\pi}{\omega} \end{align} \[\tag*{$\blacksquare$}\]

Proof. (Rice 190) By definition, one period of $\cos{\left(\omega t-\phi\right)}$ is contained in the interval $0\leq\omega t-\phi\leq2\pi.$ Therefore, \begin{equation} \frac{\phi}{\omega}\le t\le\frac{2\pi}{\omega}+\frac{\phi}{\omega} \end{equation} Thus, the period is the length of this interval: \begin{equation} T=\left(\frac{2\pi}{\omega}+\frac{\phi}{\omega}\right) - \frac{\phi}{\omega}=\frac{2\pi}{\omega} \end{equation} \[\tag*{$\blacksquare$}\]

Works Cited

Serway, Raymond A. and Robert J. Beichner. Physics for Scientists and Engineers, Fifth Edition. Brooks/Cole, 2000. Print.

Rice, Bernard J. and Jerry D. Strange. Plane Trigonometry, Seventh Edition. Boston: PWS Publishing Company, 1996. Print