We are asked to find the exact values of the following 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 are asked to find the exact values of the following 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) To find $sin(\frac{5\pi}{3})$, we first find the reference angle. $\frac{5\pi}{3}$ is in the fourth quadrant, and its reference angle is $2\pi - \frac{5\pi}{3} = \frac{6\pi - 5\pi}{3} = \frac{\pi}{3}$.

Since sine is negative in the fourth quadrant, $sin(\frac{5\pi}{3}) = -sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$.

13) To find $tan(\frac{-5\pi}{6})$, we first find the reference angle. $\frac{-5\pi}{6}$ is in the third quadrant. The reference angle is $\frac{-5\pi}{6} + \pi = \frac{\pi}{6}$.

Since tangent is positive in the third quadrant, $tan(\frac{-5\pi}{6}) = tan(\frac{\pi}{6}) = \frac{sin(\frac{\pi}{6})}{cos(\frac{\pi}{6})} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

14) To find $csc(\frac{5\pi}{3})$, we can use the result from problem

1

2. $csc(\theta) = \frac{1}{sin(\theta)}$.

Since we found $sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2}$, we have $csc(\frac{5\pi}{3}) = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$.

3. Final Answer

12) $sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2}$

13) $tan(\frac{-5\pi}{6}) = \frac{\sqrt{3}}{3}$

14) $csc(\frac{5\pi}{3}) = -\frac{2\sqrt{3}}{3}$