Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
Question
Find the principal value of $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
✓
Answer
Let cos-1$\left(\frac{\sqrt{3}}{2}\right)$ = x Then, cos x = $\frac{\sqrt{3}}{2}$ = cos $\left(\frac{\pi}{6}\right)$ We know that the principle value range of cos-1 is [0, $\pi$] and cos $\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$ Therefore, the principal value of cos-1 $\left(\frac{\sqrt{3}}{2}\right) \text { is } \frac{\pi}{6}$
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