Inverse function identities · proof ·what it is and when to use it
Proof: sin(arccos x) = √(1 − x²)
Sine of an inverse cosine — proved by pythagorean identity on [0, π]. Every line below says which rule it uses, so nothing has to be taken on faith.
The proof
Identical algebra to the arcsin case; the whole difference is which interval supplies the sign.
- Let θ = arccos xSo cos θ = x with θ in [0, π].
- sin²θ = 1 − cos²θ = 1 − x²Pythagorean identity. · Pythagorean identity
- sin θ = +√(1 − x²)sine is ≥ 0 on [0, π].
Both endpoints check out: x = 1 gives 0, x = −1 gives 0, x = 0 gives 1 — exactly what a cosine-known angle predicts.
Where the proof stops applying
x is a number from −1 to 1 — arcsin and arccos are only defined for inputs in this interval — outside it there is no angle to take.
Outside that interval the inverse function returns nothing at all, so the statement is not false there — it has no meaning. The proof above never leaves the interval.
The same claim, checked numerically
A proof is not the same thing as a check, and this page shows both: the reasoning above, and the first few angles fed to the same engine behind the verifier. If they ever disagreed, the data would be broken — the build would fail before publishing (accuracy policy).
| 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 |
Related
- How would I find sin(arccos x) = √(1 − x²) myself? — the derivation, which is a different question from the proof.
- Sine of an inverse cosine: when to use it — the practical side.
- All inverse function identities · proof index