Inverse function identities · proof ·what it is and when to use it
Proof: arccos(−x) = π − arccos x
Inverse cosine of a negative — proved by cosine of a supplementary angle. Every line below says which rule it uses, so nothing has to be taken on faith.
The proof
One identity plus the range of arccos; the range is what makes the answer unique.
- Let θ = arccos xθ in [0, π], cos θ = x.
- cos(π − θ) = −cos θ = −xSupplementary-angle rule. · Cosine of a difference
- π − θ is in [0, π]The interval is closed under θ → π − θ.
- arccos(−x) = π − θ = π − arccos xarccos is the unique angle in [0, π].
Read it as: a negative cosine means an obtuse angle, and obtuse angles are π minus their reference angle.
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 | 0.643501 | 0.643501 | yes |
| -0.6 | 0.927295 | 0.927295 | yes |
| -0.4 | 1.159279 | 1.159279 | yes |
| -0.2 | 1.369438 | 1.369438 | yes |
| 0 | 1.570796 | 1.570796 | yes |
Related
- How would I find arccos(−x) = π − arccos x myself? — the derivation, which is a different question from the proof.
- Inverse cosine of a negative: when to use it — the practical side.
- All inverse function identities · proof index