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TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Half-angle identities with the sign determined by the quadrant

Statement

For every real xx, cos⁡(x/2)=εc(1+cos⁡x)/2,sin⁡(x/2)=εs(1−cos⁡x)/2,\cos(x/2)=\varepsilon_c\sqrt{(1+\cos x)/2},\qquad \sin(x/2)=\varepsilon_s\sqrt{(1-\cos x)/2}, where εc,εs∈{−1,0,1}\varepsilon_c,\varepsilon_s\in\{-1,0,1\} are respectively the signs of cos⁡(x/2)\cos(x/2) and sin⁡(x/2)\sin(x/2) (so sgn⁡(0)=0\operatorname{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\sqrt{a} \ge 0 with (a)2=a(\sqrt{a})^2 = a; the positives are {x2:x≠0}\{x^2 : x \neq 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 xx.

[L1]

Double-angle and quadratic power-reduction identities gives cos⁡2t=(1+cos⁡2t)/2\cos^2t=(1+\cos2t)/2 and sin⁡2t=(1−cos⁡2t)/2\sin^2t=(1-\cos2t)/2 for every real tt.

[L2]

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

Proof

technique · direct
1.1

Apply [L1] with t=x/2t=x/2. Then cos⁡2(x/2)=(1+cos⁡x)/2\cos^2(x/2)=(1+\cos x)/2 and sin⁡2(x/2)=(1−cos⁡x)/2\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|\cos(x/2)|=\sqrt{(1+\cos x)/2} and ∣sin⁡(x/2)∣=(1−cos⁡x)/2|\sin(x/2)|=\sqrt{(1-\cos x)/2}.

L2step 1.1
3.1

Multiplying each equality of step 2.1 by its sign, with sign 00 when the corresponding value is 00, yields the displayed identities. The sign ranges are therefore exactly {−1,0,1}\{-1,0,1\}.

step 2.1∎

Depends on

Used by

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Sources