How this instrument works
The half-angle identities find the sine and cosine of half an angle using nothing but the cosine of the full angle — you never need to know θ⁄2 itself, or even θ in degrees, to get an exact numeric answer. They fall out of the double-angle formula cos(2x) = 2cos²x − 1 by substitution: set x = θ⁄2 so that 2x = θ, solve for cos x, and take a square root. Swap in the twin identity cos(2x) = 1 − 2sin²x and sin(θ⁄2) drops out the same way.
Both results carry a square root, so strictly the algebra hands back a ± on each side — the sign is settled by which quadrant θ⁄2 actually falls in, not by the algebra alone. This calculator reports the positive root, which is exactly right whenever θ sits between 0 and π (0° and 180°), so θ⁄2 stays in the first quadrant where sine and cosine are both positive. Push θ past π and the sign of cos(θ⁄2) can flip, a detail every textbook flags with a small ± and a note to check the quadrant by hand.
A neat consequence: because the formula only ever asks for cosθ, two very different angles that share a cosine — say 60° and −60° — hand back identical half-angle results here, even though their halves point opposite ways around the circle. The formula is honestly reporting a magnitude; direction is information cosθ alone never carried in the first place.
- Enter your full angle in the Angle, θ field, in radians — for 60°, that is π⁄3 ≈ 1.0472.
- Read cos(θ⁄2) for the half-angle's cosine, updated the instant you change θ.
- Read sin(θ⁄2) alongside it for the half-angle's sine — both come from the same single input.
- Divide sin(θ⁄2) by cos(θ⁄2) yourself if you need tan(θ⁄2); the sheet reports the two roots, not their ratio.
Worked example — splitting a 60° angle in half
Set θ to 1.0471975511965976 radians, which is exactly π⁄3, or 60° in ordinary degrees. The sheet returns cos(θ⁄2) = 0.8660254037844386 and sin(θ⁄2) = 0.49999999999999994 — the trailing 4 in place of a clean 0.5 is ordinary floating-point rounding from the square root, not a flaw in the formula.
Both figures check out against the familiar 30-60-90 triangle: half of 60° is 30°, and cos 30° = √3⁄2 ≈ 0.8660254, sin 30° = 1⁄2 exactly. The calculator reached those same values using only cos 60° = 0.5 as raw material, never looking up 30° directly — proof that the identity moves information from the full angle to the half angle with nothing lost along the way.
Questions
What is the half-angle formula derived from?
It comes directly from the double-angle identity cos(2x) = 2cos²x − 1. Substitute x = θ⁄2 so that 2x = θ, then solve for cos x and take a square root to get cos(θ⁄2) = √((1 + cosθ)⁄2). Swapping in cos(2x) = 1 − 2sin²x and solving for sin x produces the matching sine formula.
Why does the calculator only give the positive root?
Algebraically each identity carries a ± sign, since a square root has two roots. This sheet reports the positive one, which is correct whenever θ⁄2 lies in the first quadrant — that is, whenever θ itself is between 0 and π (0° and 180°). Outside that range you may need to flip a sign by hand after checking which quadrant θ⁄2 actually lands in.
Is cos(θ⁄2) the same as cos(θ) divided by two?
No — that is the single most common slip with this formula. Trigonometric functions are not linear, so halving the angle does not halve the cosine. For θ = 60°, cosθ divided by two works out to 0.25, but the true cos(30°) is about 0.866, well over three times larger. Always run the angle through the square-root identity, never through simple division.
What angle unit does the Angle field expect?
Radians. The engine evaluates cos(θ) directly, and cosine in code always takes radians, not degrees. To enter 60°, type π⁄3, about 1.0471976; for 90°, type π⁄2, about 1.5707963. Converting a degree figure first means multiplying it by π and dividing by 180.
How do I get tan(θ⁄2) from these two results?
Divide them: tan(θ⁄2) = sin(θ⁄2) ⁄ cos(θ⁄2). A tidier tangent half-angle identity also exists, tan(θ⁄2) = (1 − cosθ) ⁄ sinθ, which skips the square root entirely and is the substitution calculus students meet when turning a rational trigonometric integral into a rational algebraic one.
Where do half-angle identities show up outside a classroom?
Anywhere a full angle is known but a bisected one is needed: finding a triangle's base angles from its apex angle, working out chord lengths from a circle's central angle, or computing the half-width of a camera lens's field of view from its full opening angle.