Inverse function identities · derivation ·proof
Where cos(arcsin x) = √(1 − x²) comes from
Draw the right triangle. The fastest route in an exam: build a triangle from the given ratio and read the missing side off Pythagoras.
Building it step by step
- Opposite = x, hypotenuse = 1Because sin θ = x/1.
- Adjacent = √(1 − x²)Pythagoras on the triangle.
- cos θ = adjacent / hypotenuse = √(1 − x²)Definition of cosine.
The triangle silently assumes x ≥ 0; the range argument in the proof is what makes the formula valid for negative x too.
Where this route is used
The workhorse of calculus substitution: an angle is known only through its sine and you need its cosine. It also appears whenever a right triangle is drawn from sin θ = x.
- simplify
- integrate
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 | 0 | 0 | yes |
| -0.8 | 0.6 | 0.6 | yes |
| -0.6 | 0.8 | 0.8 | yes |
| -0.4 | 0.916515 | 0.916515 | yes |
| -0.2 | 0.979796 | 0.979796 | yes |
| 0 | 1 | 1 | yes |
Neighbouring derivations
- sin(arcsin x) = x — Sine of an inverse sine
- tan(arcsin x) = x / √(1 − x²) — Tangent of an inverse sine
- sin(arccos x) = √(1 − x²) — Sine of an inverse cosine
How to recall it under pressure
Re-derive it from the first line rather than searching memory: start from cos(arcsin(x)) and do the one algebraic move the derivation above makes. If you need the statement itself, it is on the cosine of an inverse sine page.