ections: evaluate the following expressions w...

# Explanation: ## Step1: Recall inverse - cosine range The range of \(y = \cos^{-1}(x)\) is \([0,\pi]\). We know that \(\cos(\frac{\pi}{3})=\frac{1}{2}\), so \(\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}\). ## Step2: Recall inverse - sine range The range of \(y=\sin^{-1}(x)\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\). Since \(\sin(-\frac{\pi}{4}) =-\frac{\sqrt{2}}{2}\), then \(\sin^{-1}(-\frac{\sqrt{2}}{2})=-\frac{\pi}{4}\). ## Step3: Recall inverse - tangent range The range of \(y = \tan^{-1}(x)\) is \((-\frac{\pi}{2},\frac{\pi}{2})\). As \(\tan(\frac{\pi}{4}) = 1\), so \(\arctan(1)=\frac{\pi}{4}\). ## Step4: For \(\arccos(-\frac{\sqrt{3}}{2})\) Since the range of \(y=\cos^{-1}(x)\) is \([0,\pi]\) and \(\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}\), then \(\arccos(-\frac{\sqrt{3}}{2})=\frac{5\pi}{6}\). ## Step5: For \(\arcsin(-\frac{1}{2})\) With the range of \(y = \sin^{-1}(x)\) being \([-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\sin(-\frac{\pi}{6})=-\frac{1}{2}\), we have \(\arcsin(-\frac{1}{2})=-\frac{\pi}{6}\). ## Step6: For \(\tan^{-1}(-\frac{\sqrt{3}}{3})\) Given the range of \(y=\tan^{-1}(x)\) is \((-\frac{\pi}{2},\frac{\pi}{2})\) and \(\tan(-\frac{\pi}{6})=-\frac{\sqrt{3}}{3}\), so \(\tan^{-1}(-\frac{\sqrt{3}}{3})=-\frac{\pi}{6}\). ## Step7: For \(\tan^{-1}(-1)\) Since the range of \(y = \tan^{-1}(x)\) is \((-\frac{\pi}{2},\frac{\pi}{2})\) and \(\tan(-\frac{\pi}{4})=-1\), then \(\tan^{-1}(-1)=-\frac{\pi}{4}\). ## Step8: For \(\cos^{-1}(\frac{\sqrt{3}}{2})\) As the range of \(y=\cos^{-1}(x)\) is \([0,\pi]\) and \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\), we get \(\cos^{-1}(\frac{\sqrt{3}}{2})=\frac{\pi}{6}\). ## Step9: For \(\arcsin(0)\) With the range of \(y = \sin^{-1}(x)\) being \([-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\sin(0) = 0\), so \(\arcsin(0)=0\). ## Step10: For \(\cos^{-1}(\frac{\sqrt{2}}{2})\) Since the range of \(y=\cos^{-1}(x)\) is \([0,\pi]\) and \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\), then \(\cos^{-1}(\frac{\sqrt{2}}{2})=\frac{\pi}{4}\). ## Step11: For \(\arccos(-\frac{\sqrt{3}}{2})\) (again) The range of \(y=\cos^{-1}(x)\) is \([0,\pi]\) and \(\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}\), so \(\arccos(-\frac{\sqrt{3}}{2})=\frac{5\pi}{6}\). ## Step12: For \(\arctan(\sqrt{3})\) Given the range of \(y=\tan^{-1}(x)\) is \((-\frac{\pi}{2},\frac{\pi}{2})\) and \(\tan(\frac{\pi}{3})=\sqrt{3}\), so \(\arctan(\sqrt{3})=\frac{\pi}{3}\). # Answer: 1. \(\frac{\pi}{3}\) 2. \(-\frac{\pi}{4}\) 3. \(\frac{\pi}{4}\) 4. \(\frac{5\pi}{6}\) 5. \(-\frac{\pi}{6}\) 6. \(-\frac{\pi}{6}\) 7. \(-\frac{\pi}{4}\) 8. \(\frac{\pi}{6}\) 9. \(0\) 10. \(\frac{\pi}{4}\) 11. \(\frac{5\pi}{6}\) 12. \(\frac{\pi}{3}\)