## Precalculus (6th Edition) Blitzer

The statement talks about the periodicity of the sine function, so the period of the sine function is $2\pi$, or it can be written as $\sin \left( x+2\pi \right)=\sin x$ So, \begin{align} & \sin \frac{7\pi }{3}=\sin \left( 2\pi +\frac{\pi }{3} \right) \\ & =\sin \frac{\pi }{3} \\ & =\frac{\sqrt{3}}{2} \end{align} So, by using the periodic properties of the sine function, the exact value of $\sin \frac{7\pi }{3}$ can be calculated. Hence, the statement is true.