What is the value of \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\)?

Correct answer: \(\frac{\pi}{6}\)

The principal value of arccos is taken in the interval \([0,\pi]\). Since \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\), we have \(\cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}\). The other choices are incorrect: \(\cos\frac{\pi}{3}=\tfrac{1}{2},\;\cos\frac{\pi}{4}=\tfrac{\sqrt{2}}{2}\), and \(\cos\frac{5\pi}{6}=-\tfrac{\sqrt{3}}{2}\). Exam tip: memorize standard trig values and remember arccos range \([0,\pi]\).