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Inverse Trigonometric Functions · Question 88 of 120

Question 88

MediumOriginal question

What is sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin \frac{2 \pi}{3}\right)?

Options for question 88

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. sin⁡2π3=32\sin \frac{2 \pi}{3} = \frac{\sqrt{3}}{2}.

  2. The principal value with that sin is π3\frac{\pi}{3}.

Worth noting

2π3\frac{2\pi}{3} is outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so sin⁡−1\sin^{-1} cannot return it.

Concept
An inverse function applied to a function
Typical time
90 s
Your time
—
The rule this testsValues of expressions
Practise this chapter →

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

Practise this chapter →

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