Question

example 6: use reference angles to find the exact value of each of the following trigonometric functions. be sure to sketch the angle and its reference angle and label the reference triangle formed by the reference angle and the x - axis.
a. sin 300°
b. tan(5π/4)
c. sec(-π/6)

Explanation:

Step1: Find reference - angle for $\sin300^{\circ}$

The angle $300^{\circ}$ is in the fourth - quadrant. The reference angle $\theta_{r}=360^{\circ}-300^{\circ}=60^{\circ}$. In the fourth - quadrant, $\sin$ is negative. So, $\sin300^{\circ}=-\sin60^{\circ}=-\frac{\sqrt{3}}{2}$.

Step2: Find reference - angle for $\tan\frac{5\pi}{4}$

The angle $\frac{5\pi}{4}$ is in the third - quadrant. $\frac{5\pi}{4}=\pi+\frac{\pi}{4}$, so the reference angle $\theta_{r}=\frac{\pi}{4}$. In the third - quadrant, $\tan$ is positive. So, $\tan\frac{5\pi}{4}=\tan\frac{\pi}{4}=1$.

Step3: Find reference - angle for $\sec(-\frac{\pi}{6})$

Since $\sec(-\alpha)=\sec\alpha$, the angle $\frac{\pi}{6}$ is in the first - quadrant. The reference angle for $\frac{\pi}{6}$ is $\frac{\pi}{6}$ itself. And $\sec\alpha=\frac{1}{\cos\alpha}$, so $\sec\frac{\pi}{6}=\frac{1}{\cos\frac{\pi}{6}}=\frac{2\sqrt{3}}{3}$.

Answer:

a. $\sin300^{\circ}=-\frac{\sqrt{3}}{2}$
b. $\tan\frac{5\pi}{4}=1$
c. $\sec(-\frac{\pi}{6})=\frac{2\sqrt{3}}{3}$