Inverse function identities · proof ·what it is and when to use it
Proof: arcsin(−x) = −arcsin x
Inverse sine of a negative — proved by apply sine to both sides, then use the range. Every line below says which rule it uses, so nothing has to be taken on faith.
The proof
Two angles with the same sine are equal once you know both live where sine is one-to-one.
- Let θ = arcsin xθ in [−π/2, π/2], sin θ = x.
- sin(−θ) = −sin θ = −xSine is odd. · Sine is odd
- −θ is also in [−π/2, π/2]The interval is symmetric about 0.
- arcsin(−x) = −θ = −arcsin xUniqueness of the angle in that interval.
Range check twice over: without it, sin(−θ) = −x would only say that −θ is one of infinitely many angles.
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 | 1.570796 | 1.570796 | yes |
| -0.8 | 0.927295 | 0.927295 | yes |
| -0.6 | 0.643501 | 0.643501 | yes |
| -0.4 | 0.411517 | 0.411517 | yes |
| -0.2 | 0.201358 | 0.201358 | yes |
| 0 | 0 | 0 | yes |
Related
- How would I find arcsin(−x) = −arcsin x myself? — the derivation, which is a different question from the proof.
- Inverse sine of a negative: when to use it — the practical side.
- All inverse function identities · proof index