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Half angle formulae

Use half angle formulae to find exact sine, cosine and tangent values. Choose signs from the half-angle quadrant and apply the identities to bisectors and triangle geometry.

Before you startDouble angle formulae, exact values and quadrant signs

01 / The half-angle quadrant determines the sign

Halve the interval as well as the angle.

Half-angle formulae often begin with a squared value. Taking its square root gives a magnitude; the quadrant of θ/2 supplies the sign. Knowing the quadrant of θ is useful because it gives an interval that you can halve.

180° < θ < 270° → 90° < θ/2 < 135°

In that example θ/2 is in quadrant two: sine is positive, cosine and tangent are negative. Explore several angles, including negative angles and angles beyond one turn.

Halve the actual angleExplore

Full angle θ

Blue: θ. Gold: θ/2. Compare θ = 60° with θ = 420°: the blue point is the same, but the gold point is opposite. Decimals are rounded to 4 places.

θ = 240°; θ/2 = 120°.

sin(θ/2) = +√((1 − cos θ)/2) = 0.8660.

cos(θ/2) = −√((1 + cos θ)/2) = −0.5000.

tan(θ/2) = −1.7321.

Both rational tangent forms are valid here.

02 / Rearrange the double-angle cosine identities

Put θ/2 into the formula for twice an angle.

Starting with cos 2u = 1 − 2sin² u and cos 2u = 2cos² u − 1, put u = θ/2 and rearrange:

sin²(θ/2) = (1 − cos θ)/2
cos²(θ/2) = (1 + cos θ)/2

These squared identities hold for every real θ. Therefore:

sin(θ/2) = ±√((1 − cos θ)/2)
cos(θ/2) = ±√((1 + cos θ)/2)

Choose each sign independently from the actual half-angle. The ± symbol records the possible choices; it does not mean that sine or cosine has two values at a specified input.

03 / Keep complete angle information

Reducing θ by a full turn changes its half by a half-turn.

Compare θ = 60° and θ = 420°Worked example

Both have cos θ = 1/2 and sin θ = √3/2

The original points coincide on the unit circle.

Their half-angles are 30° and 210°

These are opposite points.

sin 30° = 1/2, but sin 210° = −1/2

The squared half-angle formula gives 1/4 in both cases.

cos 30° = √3/2, but cos 210° = −√3/2

The signs require the actual half-angle interval.

Do not reduce θ modulo 360° before halving it unless you also account for the resulting sign change. The tangent of the half-angle is unchanged by this particular 180° shift, wherever it is defined.

Watch a full turn change the half-angle signs

Pause, replay or seek freely. The notes explain the same idea and stay in view.

04 / Use a given cosine and an interval

Pick the root signs before substituting numbers.

cos θ = −7/25 and 180° < θ < 270°. Find all three half-angle ratios.Worked example

90° < θ/2 < 135°

The half-angle lies in quadrant two.

sin(θ/2) = +√((1 + 7/25)/2) = 4/5

Sine is positive.

cos(θ/2) = −√((1 − 7/25)/2) = −3/5

Cosine is negative.

tan(θ/2) = (4/5)/(−3/5) = −4/3

Divide only after choosing both signs.

If instead 90° < θ < 180° with the same cosine, the half-angle lies in quadrant one. Its sine remains 4/5, but its cosine becomes 3/5 and tangent 4/3.

05 / Find exact half-angles of familiar angles

Keep nested square roots in exact form.

Find sin 22.5° and cos 22.5°Worked example

22.5° = 45°/2 and cos 45° = √2/2

The half-angle is acute, so both roots are positive.

sin 22.5° = √((1 − √2/2)/2) = √(2 − √2)/2

Simplify the denominator inside the root.

cos 22.5° = √(2 + √2)/2

Use the plus inside the cosine formula.

Find cos(7π/8)Worked example

7π/8 is half of 7π/4

The half-angle is in quadrant two.

cos(7π/8) = −√((1 + √2/2)/2)

Choose the negative sign.

cos(7π/8) = −√(2 + √2)/2

The exact magnitude matches cos(π/8), with the opposite sign.

06 / Derive rational forms for tangent

Quotients avoid square roots but still have domains.

Use sin θ = 2sin(θ/2)cos(θ/2), together with the two squared half-angle identities.

tan(θ/2) = sin θ/(1 + cos θ)

This form is valid when cos θ ≠ −1, exactly where tan(θ/2) exists.

tan(θ/2) = (1 − cos θ)/sin θ

This alternative requires sin θ ≠ 0; it excludes some valid zero-tangent inputs.

Find tan 22.5° exactlyWorked example

Use sin 45°/(1 + cos 45°)

The denominator is nonzero.

tan 22.5° = √2/(2 + √2)

Multiply top and bottom by 2.

tan 22.5° = √2 − 1

Rationalise the denominator.

07 / Do not treat a zero denominator as a zero answer

Different algebraic forms can exclude different inputs.

Compare θ = 180° and θ = 360°Worked example

At θ = 180°, tan(θ/2) = tan 90° is undefined

The first rational form has denominator 1 + cos θ = 0.

At θ = 360°, tan(θ/2) = tan 180° = 0

The first rational form is 0/2 = 0.

The second form gives 0/0 at θ = 360°

That form cannot be used there, although the original half-angle tangent exists.

Squaring the tangent gives tan²(θ/2) = (1 − cos θ)/(1 + cos θ), for cos θ ≠ −1. Taking its square root again requires a sign choice; it does not resolve a quadrant ambiguity.

08 / Find the slope of an angle bisector

An acute inclination is halved by the internal bisector.

Find the internal bisector of the acute angle between y = 0 and y = (4/3)x.Worked example

Let the acute inclination be θ, so tan θ = 4/3

A 3–4–5 triangle gives sin θ = 4/5 and cos θ = 3/5.

tan(θ/2) = (4/5)/(1 + 3/5) = 1/2

The bisector inclination is θ/2.

The internal bisector is y = x/2

It passes through the intersection, the origin.

The other angle bisector is y = −2x

The two bisectors are perpendicular.

For lines meeting away from the origin, use the same bisector gradients and then write line equations through their intersection. A vertical line needs an inclination angle rather than an undefined gradient.

09 / Connect a cosine-rule angle with its half

An isosceles triangle gives a geometric check.

An isosceles triangle has equal sides 9 and base 12. Find the sine and cosine of half its apex angle, and its height.Worked example

cos θ = (9² + 9² − 12²)/(2 × 9 × 9) = 1/9

Apply the cosine rule at the apex.

The half-angle is acute

The altitude bisects the apex and the base.

sin(θ/2) = √((1 − 1/9)/2) = 2/3

This also equals the half-base 6 divided by the hypotenuse 9.

cos(θ/2) = √((1 + 1/9)/2) = √5/3

A student claims cos(θ/2) = √((1 + cos θ)/2) for every real θ. Give a counterexample.

Hint

Choose a half-angle where cosine is negative.

Worked solution

At θ = 360°, the left side is cos 180° = −1 while the right side is √1 = 1. The root sign must match the half-angle quadrant.

11 / Recap

The squared formula and the sign are separate steps.

Derive the squared identities from double-angle cosine.
Halve the actual angle interval before choosing root signs.
Do not discard full turns before halving without tracking the sign change.
Use rational tangent forms only on their stated domains.
Connect half-angles to line bisectors, cosine-rule angles and right triangles.

Section 1 of 11 · The half-angle quadrant determines the sign