What is the principal value of cos^-1((√(3))/2)?

Inverse Trigonometric Functions · Question 42 of 120

Question 42

EasyOriginal question

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

- \frac{\pi}{6}

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\frac{\pi}{6}

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\frac{\pi}{3}

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\frac{11 \pi}{6}

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Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Find the angle $\theta$ in the branch $[0, \pi]$ with $\cos \theta = \frac{\sqrt{3}}{2}$ .
That angle is $\frac{\pi}{6}$ .

Worth noting

$\frac{\pi}{3}$ has cosine $\frac{1}{2}$ ; $\frac{\sqrt{3}}{2}$ belongs to $\frac{\pi}{6}$ .

Ten questions from Inverse Trigonometric Functions, at your own pace.

Ten questions from Inverse Trigonometric Functions, at your own pace.