Question

use the fact that the trigonometric functions are periodic to find the exact value of the given expressior

\\( \sin \frac { 13 \pi } { 3 } \\)

\\( \sin \frac { 13 \pi } { 3 } = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expressi

Explanation:

Step1: Find the period of the sine function

The period of the sine function \(y = \sin(x)\) is \(2\pi\). We can write \(\frac{13\pi}{3}=4\pi+\frac{\pi}{3}\). Since \(\sin(x + 2k\pi)=\sin(x)\) for any integer \(k\), and here \(k = 2\) (because \(4\pi=2\times2\pi\)), we have \(\sin(\frac{13\pi}{3})=\sin(4\pi+\frac{\pi}{3})\).

Step2: Simplify using the periodicity

By the periodicity property \(\sin(x + 2k\pi)=\sin(x)\), when \(x=\frac{\pi}{3}\) and \(k = 2\), we get \(\sin(4\pi+\frac{\pi}{3})=\sin(\frac{\pi}{3})\).

Step3: Evaluate \(\sin(\frac{\pi}{3})\)

We know that \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\) from the unit - circle definition of the sine function (for a \(30 - 60-90\) triangle with hypotenuse \(r = 1\), opposite side \(y=\frac{\sqrt{3}}{2}\) when the angle is \(\frac{\pi}{3}\)).

Answer:

\(\frac{\sqrt{3}}{2}\)