use the fact that the trigonometric functions are periodic to find the exact value of the given exp\n\n\\(…

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})