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What condition on x makes sin⁻¹x real?

Answer and explanation

Correct answer: −1 ≤ x ≤ 1

For a real number y, sin y always lies between −1 and 1 inclusive. Therefore the inverse sine equation sin y = x has a real solution only when x belongs to the interval [−1, 1]. The endpoints are included because sin(−π/2) = −1 and sin(π/2) = 1. Hence the required condition is −1 ≤ x ≤ 1, making option A correct. Values greater than 1 or less than −1 cannot be outputs of the real sine function, so options B and C are impossible. Option D is false because inverse sine is not real for every real input; for example, sin⁻¹2 has no real value. This interval is the real domain of the principal inverse sine function.

Related tags

Inverse-TrigonometryReal-DomainArcsineIntroduction To Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMathematicsClass 12 Mcq

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