Question

find the exact value of the expression \\( \tan \left( \arccos \left( \frac { 1 } { 2 } \
ight) \
ight) \\).

\\( \tan \left( \arccos \left( \frac { 1 } { 2 } \
ight) \
ight) = \square \\)
(type an exact answer, using \\( \pi \\) and radicals as needed.)

Explanation:

Step1: Let $\theta=\arccos(\frac{1}{2})$

By the definition of the inverse cosine function, if $\theta = \arccos(x)$, then $\cos\theta=x$ and $\theta\in[0,\pi]$. So, if $\theta=\arccos(\frac{1}{2})$, then $\cos\theta=\frac{1}{2}$ and $\theta\in[0,\pi]$. We know that $\theta=\frac{\pi}{3}$ since $\cos(\frac{\pi}{3})=\frac{1}{2}$.

Step2: Find $\tan\theta$

We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Since $\theta = \frac{\pi}{3}$, $\sin\theta=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$ and $\cos\theta=\frac{1}{2}$. Then $\tan\theta=\frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$.

Answer:

$\sqrt{3}$