Half angle identities without the confusion

Published October 2026.

Half-angle identities go the other way from double-angle ones: they express θ/2 in terms of θ. The formulas are simple; the only real skill is choosing the ± sign, and that comes from knowing which quadrant θ/2 lands in.

The identities

sin(θ/2) = ±√((1 − cos θ) / 2)
cos(θ/2) = ±√((1 + cos θ) / 2)
tan(θ/2) = sin θ / (1 + cos θ) = (1 − cos θ) / sin θ (no ± needed, neat!)

Derivation in one breath: start from the double-angle forms cos 2φ = 1 − 2sin²φ and cos 2φ = 2cos²φ − 1, substitute φ = θ/2, and solve for sin(θ/2) and cos(θ/2). The square root forces the ±.

The sign rule (the part that matters)

Compute θ/2’s quadrant from θ’s range, then take the sign of sine/cosine in that quadrant:

Example: θ = 300° → θ/2 = 150°, which is in quadrant II, where sine is positive and cosine negative. So sin(150°) takes the + root and cos(150°) the − root. If θ = 60°, θ/2 = 30° (quadrant I): both positive.

When θ is unknown and only cos θ is given, the problem must supply θ’s quadrant (or the answer stays ±). No quadrant, no sign, that is not a trick, it is how the math works.

Worked examples

Example 1, exact value. sin(15°): use θ = 30°. sin(15°) = +√((1 − cos 30°)/2) = √((1 − √3/2)/2) = √((2−√3)/4) = √(2−√3)/2 ≈ 0.2588. Positive root: 15° is in quadrant I.

Example 2, with a given cosine. cos θ = 3/5, θ in quadrant IV. Find sin(θ/2). θ/2 is in quadrant II (half of QIV is QII), sine positive there: sin(θ/2) = +√((1 − 3/5)/2) = √(1/5) = √5/5 ≈ 0.4472.

Example 3, tangent shortcut. tan(22.5°): use tan(θ/2) = sin θ/(1 + cos θ) with θ = 45°: (√2/2)/(1 + √2/2) = √2/(2+√2) = √2 − 1 ≈ 0.4142. No sign ambiguity at all, the rationalized form is exact.

Common mistakes

Taking the positive root by default: the ± is load-bearing. sin(195°) is negative, the positive root would be flat wrong.

Halving the quadrant instead of the angle: “θ in QIII so θ/2 in Q-something” is not how it works, halve the angle measure (e.g. θ ∈ (180°, 270°) → θ/2 ∈ (90°, 135°)), then read the quadrant.

Forgetting the /2 inside the root: sin(θ/2) = √((1−cosθ)/2), not √(1−cosθ). The 2 in the denominator is easy to drop under exam pressure.

Exact values worth deriving once

Half-angle identities generate the exact values for 15°, 22.5°, and 75°, the angles that appear in geometry-heavy problems:

sin 15° = √(2−√3)/2 ≈ 0.2588 (from θ = 30°)
cos 15° = √(2+√3)/2 ≈ 0.9659
tan 22.5° = √2 − 1 ≈ 0.4142 (from θ = 45°, via the tangent form)
sin 75° = √(2+√3)/2 ≈ 0.9659 (note: sin 75° = cos 15°, complementary angles)

Derive each once by hand, then keep the table. On a no-calculator exam, these four lines are pure points.

Deriving the tangent half-angle forms

Start from tan(θ/2) = sin(θ/2)/cos(θ/2). Multiply top and bottom by 2cos(θ/2): the numerator becomes 2sin(θ/2)cos(θ/2) = sin θ (double-angle!), and the denominator becomes 2cos²(θ/2) = 1 + cos θ (double-angle again!). So tan(θ/2) = sin θ/(1 + cos θ).

For the second form, multiply by 2sin(θ/2) instead: numerator 2sin²(θ/2) = 1 − cos θ, denominator 2sin(θ/2)cos(θ/2) = sin θ, giving (1 − cos θ)/sin θ. Both are exact, sign-free, and each is convenient in different situations: use the first when sin θ is known, the second when cos θ is known.

Cheat sheet

sin(t/2) = ±√((1−cos t)/2)
cos(t/2) = ±√((1+cos t)/2)
tan(t/2) = sin t/(1+cos t) = (1−cos t)/sin t (no sign needed)
Sign rule: halve the angle range, read the quadrant. When a problem gives cos theta without a quadrant, leave the answer as ±, that is the mathematically complete response. In applied work (physics, engineering), the physical context usually resolves the sign, a length is positive, a measured drop is negative. The identities themselves are just the double-angle formulas solved backwards; the quadrant logic is the only new skill. Practice the quadrant step until it feels automatic through deliberate repetition daily until automatic.

Keep learning

Pair this with our double-angle identities guide, the two are the same formulas read in opposite directions. For the companion angle-unit skill, see degrees vs radians, and compute any trig value with the sin cos tan calculator (degrees or radians, your choice).

Practice: test yourself

1. Find cos(7.5°) using a half-angle identity. (θ = 15°, quadrant I → +√((1+cos 15°)/2); cos 15° = √(2+√3)/2 ≈ 0.9659 → ≈ 0.9914.)
2. If sin θ = −4/5 with θ in QIII, find cos(θ/2). (θ/2 in QII → negative root: −√((1+3/5)/2) = −√(4/5) = −2√5/5. Wait, cos θ for sin θ = −4/5 in QIII is −3/5, so −√((1−3/5)/2) = −√(1/5) = −√5/5.)
3. Simplify √(2+√3)/2. (Recognize cos 15° = √(2+√3)/2 from the half-angle table above.)

Frequently asked questions