section iii: calculator - inactive free - response\ndirections: answer the following problems completely and…

Explanation:

Step1: Recall tangent - angle formula

We know that $\tan\left(\frac{11\pi}{6}\right)=\tan\left(2\pi-\frac{\pi}{6}\right)$. Since $\tan(2k\pi - \alpha)=-\tan\alpha$ for integer $k$, then $\tan\left(2\pi-\frac{\pi}{6}\right)=-\tan\frac{\pi}{6}$. And $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$, so $\tan\left(\frac{11\pi}{6}\right)=-\frac{\sqrt{3}}{3}$.

Step2: Recall arccosine definition

The function $y = \arccos x$ gives the angle $\theta$ such that $\cos\theta=x$ and $0\leq\theta\leq\pi$. If $x = \frac{\sqrt{2}}{2}$, and we know that $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ and $\frac{\pi}{4}\in(0,\pi)$, so $\arccos\frac{\sqrt{2}}{2}=\frac{\pi}{4}$.

Step3: Recall tangent function properties

We know that $\tan\theta=-\sqrt{3}$. The tangent function $y = \tan\theta$ has a period of $\pi$. The principal - value solution of $\tan\theta=-\sqrt{3}$ in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is $\theta=-\frac{\pi}{3}$. In the general form, the solutions of $\tan\theta = -\sqrt{3}$ are $\theta=-\frac{\pi}{3}+k\pi,k\in\mathbb{Z}$. For the domain of all real - valued $\theta$, the solutions are $\theta=\frac{2\pi}{3}+k\pi,k\in\mathbb{Z}$ (when $k = 0$, $\theta=\frac{2\pi}{3}$; when $k = 1$, $\theta=\frac{2\pi}{3}+\pi=\frac{5\pi}{3}$, etc.).

Answer:

a) $-\frac{\sqrt{3}}{3}$ b) $\frac{\pi}{4}$ c) $\theta=\frac{2\pi}{3}+k\pi,k\in\mathbb{Z}$