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Our post on inverse trigonometric functions shows how to simplify single values such as sin⁻¹(sin 2). This one goes a step further: it draws the whole graph of y = sin⁻¹(sin x), turns it into one formula that works for any x, and then uses the graph for equations, derivatives and an area. Once you can sketch it, these questions take a few lines.
What does the syllabus say?
NCERT's Class 12 chapter Inverse Trigonometric Functions (2026-27 reprint) states that sin(sin⁻¹ x) = x if −1 ≤ x ≤ 1 and sin⁻¹(sin x) = x if −π/2 ≤ x ≤ π/2. It does not draw the graph beyond that interval. The JEE (Main) 2026 syllabus lists "inverse trigonometrical functions their properties" in Unit 14 and "graphs of simple functions" in Unit 7. The JEE (Advanced) 2026 brochure lists inverse trigonometric functions (principal value only) and their elementary properties.
How do you build the graph?
The output of sin⁻¹ always lies in [−π/2, π/2]. For each x, you need the angle in that range with the same sine.
On [−π/2, π/2]: y = x.
On [π/2, 3π/2]: sin(π − x) = sin x and π − x lies in range, so y = π − x.
On [3π/2, 5π/2]: sin(x − 2π) = sin x, so y = x − 2π.
On [−3π/2, −π/2]: y = −π − x.
Each piece is a straight line of slope +1 or −1, so the graph is a zigzag of triangles between −π/2 and π/2, with peaks at x = π/2 + 2kπ and valleys at x = −π/2 + 2kπ. It repeats every 2π, like sin x, but with straight sides.
Worked example 1: sin⁻¹(sin 10)
10 ÷ π ≈ 3.18, so k = 3 (3π ≈ 9.425).
x − kπ = 10 − 3π ≈ 0.575, which lies in [−π/2, π/2].
k is odd, so y = −(10 − 3π) = 3π − 10 ≈ −0.575.
Check: sin 10 ≈ −0.544 and sin(−0.575) ≈ −0.544.
Worked example 2: sin⁻¹(sin(−4))
−4 ÷ π ≈ −1.27, so k = −1.
x − kπ = −4 + π ≈ −0.858.
k is odd, so y = 4 − π ≈ 0.858.
Check: sin(−4) ≈ 0.757 and sin(0.858) ≈ 0.757.
Worked example 3: sin⁻¹(sin 7)
7 ÷ π ≈ 2.23, so k = 2, which is even. y = 7 − 2π ≈ 0.717.
What else does the graph give you?
Derivative. On every piece the slope is +1 or −1. In fact d/dx sin⁻¹(sin x) = cos x ÷ |cos x|, which is undefined at the corners x = π/2 + kπ, where cos x = 0.
An area. From 0 to π the graph is a triangle with base π and height π/2, so the integral from 0 to π of sin⁻¹(sin x) dx = (1/2)(π)(π/2) = π²/4 ≈ 2.467. From 0 to 2π the area above and below the axis cancels, and the integral is 0.
An equation. sin⁻¹(sin x) = x − 2π holds exactly when x lies in [3π/2, 5π/2], the piece where the graph is that line.
The cos⁻¹ cousin. cos⁻¹(cos x) lives in [0, π]: it equals x on [0, π] and 2π − x on [π, 2π], another zigzag, but between 0 and π.
Which slips cost marks?
Treating 10 as degrees. Without a degree sign, 10 means 10 radians.
Using the nearest multiple of 2π. Use the nearest multiple of π, then the sign (−1)ᵏ.
Losing the sign on odd pieces. For k odd, the answer is kπ − x, not x − kπ.
Confusing the two compositions. sin(sin⁻¹ x) = x for every x in [−1, 1]. sin⁻¹(sin x) = x only on [−π/2, π/2].
If the zigzag is hard to picture, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson, so you can revisit principal values before drawing again.
A practice routine
Sketch y = sin⁻¹(sin x) from −2π to 3π, marking peaks, valleys and slopes.
Use the formula for x = 4, 6 and 20, and check on a calculator. (Answers: π − 4, 6 − 2π and 20 − 6π.)
Sketch cos⁻¹(cos x) on the same axes and find where the two graphs meet between 0 and π.
Find the integral of sin⁻¹(sin x) from 0 to 3π/2 using triangle areas.
For how graphs of inverses come from reflecting in y = x, see our post on inverse function graphs. Tonight, draw step 1 on graph paper before you use the formula on anything.
