Mathematics

Mathematics

Problem

Consider the principal branch of the function y=arcsin⁡xy=\arcsin x. a. (i) Sketch the graph of y=arcsin⁡xy=\arcsin x on the same set of axes as the coordinate system. b. (ii) The graph is then translated 1 unit to the right and 2 units upward. Write the equation of the new curve. c. (iii) Determine the coordinates of the new curve’s endpoints and its yy-intercept. d. (iv) State the domain and range of the translated curve.
Intuitive explanation

The key to understanding arcsin⁡\arcsin is recognizing it as the inverse of sine, but only on a restricted domain where sine is one-to-one. The principal branch locks sine to [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], so arcsin⁡\arcsin outputs angles in that range. This restriction is essential: it ensures every input x∈[−1,1]x \in [-1, 1] maps to exactly one angle, making the inverse well-defined and allowing you to move fluidly between "what angle has this sine value?" and "what's the sine of this angle?"

Intuitive explanation

The key to understanding arcsin⁡\arcsin is recognizing it as the inverse of sine, but only on a restricted domain where sine is one-to-one. The principal branch locks sine to [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], so arcsin⁡\arcsin outputs angles in that range. This restriction is essential: it ensures every input x∈[−1,1]x \in [-1, 1] maps to exactly one angle, making the inverse well-defined and allowing you to move fluidly between "what angle has this sine value?" and "what's the sine of this angle?"

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