chapter 6 quiz\n170 points possible answered: 1/17\nquestion 2\nevaluate the following expressions.\n\\(…

Explanation:

Step1: Evaluate (\cos^{-1}\left(\frac{\sqrt{2}}{2}\right))

Let (\theta=\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)). By the definition of the inverse - cosine function, (\cos\theta=\frac{\sqrt{2}}{2}) and (0\leq\theta\leq\pi). We know that (\theta = \frac{\pi}{4}) since (\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}). Then (\sin\left(\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)=\sin\left(\frac{\pi}{4}\right)). Using the unit - circle or the sine function for special angles, (\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}).

Step2: Evaluate (\cos^{-1}\left(\frac{\sqrt{3}}{2}\right))

Let (\alpha=\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)). By the definition of the inverse - cosine function, (\cos\alpha=\frac{\sqrt{3}}{2}) and (0\leq\alpha\leq\pi). We know that (\alpha=\frac{\pi}{6}) since (\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}). Then (\tan\left(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)=\tan\left(\frac{\pi}{6}\right)). Using the unit - circle or the tangent function for special angles, (\tan\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{3}).

Answer:

(\frac{\sqrt{2}}{2}), (\frac{\sqrt{3}}{3})