trigidentity.com

Half-angle identities · proof ·what it is and when to use it

Proof: sin(x/2) = ±√((1 − cos x)/2)

Half-angle identity for sine — proved by solve cos 2u for sin u. Every line below says which rule it uses, so nothing has to be taken on faith.

sin(x/2) = ±√((1 − cos x)/2)

The proof

Rearranging the double-angle cosine, then a change of name.

  1. cos 2u = 1 − 2sin²u
  2. sin²u = (1 − cos 2u)/2
    Solve for sin²u.
  3. sin u = ±√((1 − cos 2u)/2)
    Take the root; sign unknown until the quadrant of u is known.
  4. u = x/2 → sin(x/2) = ±√((1 − cos x)/2)
    Rename 2u as x.

With 0° < x < 360°, the half angle x/2 lies between 0° and 180° where sine is positive — so the + root is correct on that interval and only there.

Where the proof stops applying

Requires x between 0° and 360° — the + root is the valid one for 0° < x < 360° (x/2 in the upper half-plane); outside that interval sin(x/2) is negative and the identity needs the − root.

This is not a technicality: outside the interval, the statement as written is false, and the sign has to be chosen by hand.

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).

Both sides evaluated at the same angles — left side sin(t/2), right side ± sqrt((1 - cos(t))/2).
AngleLeft sideRight sideAgree
7.5°0.0654030.065403yes
15°0.1305260.130526yes
18°0.1564340.156434yes
22.5°0.195090.19509yes
30°0.2588190.258819yes
37°0.3173050.317305yes

Related

sampled at 800 valid angles inside its stated intervalconditional: x between 0° and 360°Source: OpenStax Precalculus, Ch. 7.3 · Paul's Online Math Notes · revised 2026-09-26 ·how we check ·accuracy policy ·report an error