Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02
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Half-angle identities with the sign determined by the quadrant

Statement

For every real x, cos⁡(x/2)=εc(1+cos⁡x)/2,sin⁡(x/2)=εs(1−cos⁡x)/2, where εc,εs∈{−1,0,1} are respectively the signs of cos⁡(x/2) and sin⁡(x/2) (so sgn⁡(0)=0). Thus the positive square root is valid only where the relevant half-angle function is nonnegative. The conventions and prerequisite facts used below are recorded in Double-angle and quadratic power-reduction identities, Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Signs, monotonicity intervals, and ranges of sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains.

Facts & Assumptions

Given: A real x.

[L1]

Double-angle and quadratic power-reduction identities gives cos⁡2t=(1+cos⁡2t)/2 and sin⁡2t=(1−cos⁡2t)/2 for every real t.

[L2]

Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0} says that every nonnegative real has a unique nonnegative square root.

Proof

technique · direct
1.1

Apply [L1] with t=x/2. Then cos⁡2(x/2)=(1+cos⁡x)/2 and sin⁡2(x/2)=(1−cos⁡x)/2, so both radicands are nonnegative.

L1
2.1

By [L2], ∣cos⁡(x/2)∣=(1+cos⁡x)/2 and ∣sin⁡(x/2)∣=(1−cos⁡x)/2.

L2step 1.1
3.1

Multiplying each equality of step 2.1 by its sign, with sign 0 when the corresponding value is 0, yields the displayed identities. The sign ranges are therefore exactly {−1,0,1}.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources