Assertion (A): sin^-1(sin 5π/6) = π/6. Reason (R): sin^-1(sin θ) = θ only for θ in [-π/2, π/2].

Question 108

HardOriginal question

Assertion (A): $\sin^{-1}\left(\sin \frac{5\pi}{6}\right) = \frac{\pi}{6}$ .
Reason (R): $\sin^{-1}(\sin \theta) = \theta$ only for $\theta$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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CorrectNot quiteAnswer A

Correct answer: A

A is true: $\sin \frac{5\pi}{6} = \frac{1}{2}$ , and the angle on the branch with sine $\frac{1}{2}$ is $\frac{\pi}{6}$ .
R is true and explains why the answer is not $\frac{5\pi}{6}$ itself.

Concept: When an inverse undoes a function
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsValues of expressions

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