Range of f(x) = sin(x) from R to R:
AR
BRandom
C[0, 1]
D[-1, 1] (sine bounded)
Answer & Solution
Correct answer: D. [-1, 1] (sine bounded)
sin x ∈ [-1, 1] for all real x. f is many-one (periodic), so not bijection from R to R. Restricting domain to [-π/2, π/2] makes it bijection [-π/2, π/2] → [-1, 1], inverse = arcsin.
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