Inverse Trig Values | MrWittman.com

Inverse trig asks for an angle.

When you see sin⁻¹, cos⁻¹, or tan⁻¹, the answer is not a ratio — it is the principal angle whose trig value matches the input. This tool shows the allowed ranges, unit-circle location, and common composition traps.

Exact values Principal ranges Unit-circle visuals arcsin(cos θ) style examples

Try a worked example

Pick a problem. Watch the point move on the unit circle, then read why that angle is the inverse trig answer.

Problem

Principal range cheat sheet

These ranges are why inverse trig answers sometimes look “different” from the angle you first imagine.

sin⁻¹(x)

−90° ≤ θ ≤ 90°

Quadrants IV, I, and the y-axis. Output follows the y-value.

cos⁻¹(x)

0° ≤ θ ≤ 180°

Quadrants I and II. Output follows the x-value.

tan⁻¹(x)

−90° < θ < 90°

Quadrants IV and I. Output follows slope y/x.

Do not “cancel” blindly.
sin⁻¹(sin 150°) is not 150° because 150° is outside arcsin’s range. The matching principal angle is 30°.

Reliable process

1Name the output range.

Before doing anything else, decide where the answer is allowed to live.

2Find the reference angle.

Use the familiar special-angle value: 0°, 30°, 45°, 60°, or 90°.

3Choose the allowed quadrant.

Match the sign of the ratio while staying inside the inverse function’s principal range.

Examples students should recognize

sin⁻¹(0)

0°

The angle in [−90°, 90°] with sine 0 is 0°.

cos⁻¹(1)

0°

Cosine is x-coordinate. x = 1 at the point (1, 0).

tan⁻¹(1)

45°

Tangent is slope. The slope is 1 at 45°.

sin⁻¹(−√2/2)

−45°

Reference angle 45°, negative sine, and arcsin must stay from −90° to 90°.

cos⁻¹(−1/2)

120°

Reference angle 60°, negative cosine, and arccos must stay from 0° to 180°.

tan⁻¹(−√3)

−60°

Reference angle 60°, negative tangent, and arctan’s range uses quadrant IV.

Compositions: inverse trig + trig

The key question is: Is the inside angle already in the inverse function’s output range? If yes, it may simplify directly. If no, find the angle inside the allowed range with the same trig value.

sin⁻¹(sin 30°)

30°

30° is already in arcsin’s range, so it comes back unchanged.

sin⁻¹(sin 150°)

30°

150° has sine 1/2, but arcsin can only return −90° to 90°. The matching angle is 30°.

cos⁻¹(cos 240°)

120°

Cos 240° = −1/2. Arccos returns the angle in [0°, 180°], so the answer is 120°.

tan⁻¹(tan 135°)

−45°

Tangent repeats every 180°. The coterminal angle inside arctan’s range is −45°.

sin⁻¹(cos 100°)

−10°

cos 100° = sin(90° − 100°) = sin(−10°). Since −10° is in arcsin’s range, arcsin(cos 100°) = −10°.

cos⁻¹(sin 20°)

70°

sin 20° = cos(90° − 20°) = cos 70°. Since 70° is in arccos’s range, the answer is 70°.