Explain why sin x has no inverse on R. Show that it has one on [−π/2, π/2], and find sin⁻¹(−√3/2).

1. 1 · Write what is given and what is to be found. Given: f(x) = sin x. To show: no inverse on R, but an inverse on [−π/2, π/2]. To find: sin⁻¹(−√3/2).
2. 2 · Write the rule. A function has an inverse only if it is one-one and onto.
3. 3 · Draw the figure or sketch. Sketch y = sin x from −2π to 2π. Shade the part from x = −π/2 to x = π/2 and mark (−π/2, −1) and (π/2, 1).
4. 4 · Show that sin x is not one-one on R. sin 0 = sin π = 0, with 0 ≠ π. It is also not onto R, since sin x never equals 2.
5. 5 · Show that the restriction works. On [−π/2, π/2], sin x increases from −1 to 1 and takes every value in [−1, 1] exactly once. So sin: [−π/2, π/2] → [−1, 1] is a bijection, and sin⁻¹: [−1, 1] → [−π/2, π/2] exists.
6. 6 · Find sin⁻¹(−√3/2). We need y in [−π/2, π/2] with sin y = −√3/2. Since sin(π/3) = √3/2, y = −π/3.
7. 7 · Check and box the answer. sin(−π/3) = −sin(π/3) = −√3/2, and −π/3 lies in [−π/2, π/2]. So sin⁻¹(−√3/2) = −π/3.

Scoring tip: Give a numerical example such as sin 0 = sin π when you say 'not one-one'. A bare claim can lose the mark.