Question

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

\\( \sin \frac { 19 \pi } { 3 } \\)

\\( \sin \frac { 19 \pi } { 3 } = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the ex

Explanation:

Step1: Find the period of sine function

The period of \(y = \sin x\) is \(2\pi\). We can use the formula \(\sin(x + 2k\pi)=\sin x\), where \(k\in\mathbb{Z}\).
We want to rewrite \(\frac{19\pi}{3}\) in the form \(x + 2k\pi\). Let's find \(k\) such that \(0\leq x<2\pi\).
We know that \(\frac{19\pi}{3}=6\pi+\frac{\pi}{3}\), and \(6\pi = 3\times2\pi\).

Step2: Apply the periodic property of sine function

Since \(\sin(x + 2k\pi)=\sin x\) with \(x=\frac{\pi}{3}\) and \(k = 3\), we have \(\sin\frac{19\pi}{3}=\sin(6\pi+\frac{\pi}{3})\).
By the periodic property \(\sin(A + 2k\pi)=\sin A\) (\(k\in\mathbb{Z}\)), so \(\sin(6\pi+\frac{\pi}{3})=\sin\frac{\pi}{3}\).

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

We know that for a \(30 - 60-90\) triangle, \(\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\).

Answer:

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