evaluate the following expressions.\n$\\sin\\left(\\cos^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)\\right)$\n$\\…

Explanation:

Step1: 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\theta), and (\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2})

Step2: 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\alpha), and (\tan\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{3})

Answer:

The value of (\sin\left(\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)) is (\frac{\sqrt{2}}{2}) and the value of (\tan\left(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)) is (\frac{\sqrt{3}}{3})