Inverse function identities · derivation ·proof
Where arcsin x + arccos x = π/2 comes from
Two acute angles of one right triangle. Draw the triangle once and the identity is visible before it is written down.
Building it step by step
- Right triangle, opposite side x, hypotenuse 1Sets sin of one acute angle to x.
- That acute angle = arcsin xDefinition.
- The other acute angle = arccos xIts cosine is the same ratio x.
- arcsin x + arccos x = π/2The two acute angles of a right triangle sum to 90°.
This is the version to keep in your head; the algebraic proof above is what extends it to negative x.
Where this route is used
Whenever both inverse functions appear in one expression — usually to swap one for the other, to simplify a sum of angles, or to check that two answers in a textbook are the same number.
- simplify
- verify
- solve
The result, checked at real values
Deriving a formula and testing it are different things; this table is the test. The numbers come from the same engine as the verifier, computed when the site was built.
| Angle | Left side | Right side | Agree |
|---|---|---|---|
| -1 | 1.570796 | 1.570796 | yes |
| -0.8 | 1.570796 | 1.570796 | yes |
| -0.6 | 1.570796 | 1.570796 | yes |
| -0.4 | 1.570796 | 1.570796 | yes |
| -0.2 | 1.570796 | 1.570796 | yes |
| 0 | 1.570796 | 1.570796 | yes |
Neighbouring derivations
- cos(arcsin x) = √(1 − x²) — Cosine of an inverse sine
- sin(arccos x) = √(1 − x²) — Sine of an inverse cosine
- sin θ = cos(90° − θ) — Cofunction identity for sine
How to recall it under pressure
Re-derive it from the first line rather than searching memory: start from arcsin(x) + arccos(x) and do the one algebraic move the derivation above makes. If you need the statement itself, it is on the inverse sine plus inverse cosine page.