We need to find the exact values of three trigonometric functions without using a calculator: 12) $\sin{\frac{5\pi}{3}}$ 13) $\tan{\frac{-5\pi}{6}}$ 14) $\csc{\frac{5\pi}{3}}$

1. Problem Description

We need to find the exact values of three trigonometric functions without using a calculator:

12) $\sin{\frac{5\pi}{3}}$

13) $\tan{\frac{-5\pi}{6}}$

14) $\csc{\frac{5\pi}{3}}$

2. Solution Steps

12) $\sin{\frac{5\pi}{3}}$

The angle $\frac{5\pi}{3}$ is in the fourth quadrant. The reference angle is $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$.

Since sine is negative in the fourth quadrant, we have

$\sin{\frac{5\pi}{3}} = -\sin{\frac{\pi}{3}}$.

We know that $\sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2}$.

Therefore, $\sin{\frac{5\pi}{3}} = -\frac{\sqrt{3}}{2}$.

13) $\tan{\frac{-5\pi}{6}}$

Since tangent is an odd function, $\tan(-x) = -\tan(x)$. Therefore, $\tan{\frac{-5\pi}{6}} = -\tan{\frac{5\pi}{6}}$.

The angle $\frac{5\pi}{6}$ is in the second quadrant. The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.

Since tangent is negative in the second quadrant, $\tan{\frac{5\pi}{6}} = -\tan{\frac{\pi}{6}}$.

Therefore, $\tan{\frac{-5\pi}{6}} = -\tan{\frac{5\pi}{6}} = -(-\tan{\frac{\pi}{6}}) = \tan{\frac{\pi}{6}}$.

We know that $\tan{\frac{\pi}{6}} = \frac{\sin{\frac{\pi}{6}}}{\cos{\frac{\pi}{6}}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

Therefore, $\tan{\frac{-5\pi}{6}} = \frac{\sqrt{3}}{3}$.

14) $\csc{\frac{5\pi}{3}}$

We know that $\csc(x) = \frac{1}{\sin(x)}$.

From the first question, we already found that $\sin{\frac{5\pi}{3}} = -\frac{\sqrt{3}}{2}$.

Therefore, $\csc{\frac{5\pi}{3}} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$.

3. Final Answer

12) $-\frac{\sqrt{3}}{2}$

13) $\frac{\sqrt{3}}{3}$

14) $-\frac{2\sqrt{3}}{3}$