In which interval does this α = sin^-1(1/3) lie?

Question 57

MediumOriginal question

Let $\alpha = \sin^{-1}\left(\frac{1}{3}\right)$ .

In which interval does this $\alpha = \sin^{-1}\left(\frac{1}{3}\right)$ lie?

\left(0, \frac{\pi}{6}\right)

Correct answerYour answer

\left(\frac{\pi}{3}, \frac{\pi}{2}\right)

Correct answerYour answer

\left(\frac{\pi}{6}, \frac{\pi}{4}\right)

Correct answerYour answer

\left(\frac{\pi}{4}, \frac{\pi}{3}\right)

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$\sin^{-1}$ is increasing, and $0 < \frac{1}{3} < \frac{1}{2}$ .
So $\sin^{-1} 0 < \alpha < \sin^{-1} \frac{1}{2}$ , that is $0 < \alpha < \frac{\pi}{6}$ .

Worth noting

$\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ is where $\cos^{-1}\left(\frac{1}{3}\right)$ lies; $\cos^{-1}$ is decreasing, $\sin^{-1}$ increasing.

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

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