Question

find one positive and one negative angle coterminal with an angle of $\frac{pi}{6}$.
$\frac{11pi}{6},\frac{-7pi}{6}$
$\frac{3pi}{6},\frac{-21pi}{6}$
$\frac{21pi}{6},\frac{-3pi}{6}$
$\frac{7pi}{6},\frac{-11pi}{6}$
question 7 (5 points)
find the exact value of $sec(-660)^{circ}$.
2
$\frac{sqrt{3}}{2}$
$\frac{1}{2}$
$\frac{1}{sqrt{2}}$

Explanation:

Step1: Recall coterminal - angle formula

Coterminal angles of an angle $\theta$ are given by $\theta + 2k\pi$ ($k\in\mathbb{Z}$). For $\theta=\frac{\pi}{6}$, when $k = 1$, the positive - coterminal angle is $\frac{\pi}{6}+2\pi=\frac{\pi + 12\pi}{6}=\frac{13\pi}{6}$. When $k=- 1$, the negative - coterminal angle is $\frac{\pi}{6}-2\pi=\frac{\pi - 12\pi}{6}=-\frac{11\pi}{6}$.

Step2: Recall secant function property

We know that $\sec\theta=\frac{1}{\cos\theta}$, and $\cos(-\alpha)=\cos\alpha$. So, $\sec(-660^{\circ})=\sec(660^{\circ})$. Also, $660^{\circ}=360^{\circ}+300^{\circ}$, and $\cos$ has a period of $360^{\circ}$, so $\cos(660^{\circ})=\cos(300^{\circ})$. Since $\cos(300^{\circ})=\cos(360^{\circ}-60^{\circ})=\cos60^{\circ}=\frac{1}{2}$, then $\sec(660^{\circ})=\frac{1}{\cos(660^{\circ})}=\frac{1}{\cos(300^{\circ})}=\frac{1}{\cos60^{\circ}} = 2$.

Answer:

For the first question: $\frac{7\pi}{6},-\frac{11\pi}{6}$
For the second question: $2$