Which half-angle identities does the explorer cover?

All six standard identities: sin(alpha/2), cos(alpha/2), tan(alpha/2), csc(alpha/2), sec(alpha/2), and cot(alpha/2). Each has its own tab and its own derivation view, either geometric or algebraic.

Why do only sin and cos get geometric proofs?

Sine and cosine are the foundational half-angle identities. Tangent is sin over cos, and cosecant, secant, and cotangent are reciprocals of sin, cos, and tan. The tool limits geometric proofs to sin and cos and treats the other four as algebraic consequences.

How is sin(alpha/2) proved geometrically?

Draw an isosceles triangle with two unit radii meeting at angle alpha. The chord length squared is 2 minus 2 cos alpha by the law of cosines, and also equals 4 sin squared (alpha/2) from the bisected half-chord. Equating gives sin(alpha/2) equal to the square root of (1 minus cos alpha) over 2.

Can I verify an identity at a specific angle?

Yes. Move the alpha slider to the angle you want, then read the two metric cards near the bottom. They show the values of both sides of the identity, computed independently. The formula table at the bottom does the same for all six functions at once.

What does the source column of the formula table mean?

It tells you how each identity was obtained. Sine is labeled geometric. Cosine is via sin (proved using the sin half-angle and Pythagoras). Tangent is via sin and cos. Cosecant, secant, and cotangent are via their reciprocal partners.

Half Angle Identities: Interactive Proofs

Switching Between Functions

Six tabs at the top let you pick which half-angle identity to study:

\sin(\alpha/2)

,

\cos(\alpha/2)

,

\tan(\alpha/2)

,

\csc(\alpha/2)

,

\sec(\alpha/2)

,

\cot(\alpha/2)

.

How selection changes the view:
•

\sin

and

\cos

open the geometric proof scene with a step-by-step animation.
•

\tan

,

\csc

,

\sec

, and

\cot

open the derived identity card with the algebraic chain.
• The active tab is highlighted in deep blue.
• The URL updates with

?halfFn=...

, so links you copy preserve the selected function.

Clicking any row of the formula table at the bottom also jumps to that function.

Adjusting the Angle α

Each view exposes a slider for the base angle $\alpha$ in degrees, between

20°

and

160°

. The half angle is then

\alpha/2

, ranging from

10°

to

80°

.

What changes as you slide:
• On geometric scenes, the SVG triangle resizes and the apex

\alpha

updates immediately.
• The number readout shows the current value of

\alpha

.
• The verification cards recompute both sides of the identity at the new

\alpha/2

.

Sweep through several values to confirm each identity is not a coincidence at one angle but a true equality.

Playing Through a Geometric Proof

When

\sin

or

\cos

is active, an animated proof unfolds in six steps. A toolbar gives you control:

• Reset — return to step 0 with a blank scene.
• Prev / Next — step through one stage at a time.
• Play / Pause — advance automatically.
• Speed selector —

0.5\times

,

1\times

,

1.5\times

,

2\times

.

Each step adds one geometric element (radii, triangle fill, bisector, half-angles, leg labels, final metrics). The right-hand panel logs each step with its name and reasoning.

Reading the Geometric Scene

The SVG shows the unit circle with two radii

OA

and

OB

of length

1

meeting at the center

O

with angle

\alpha

between them.

Elements that appear across the steps:
• Red arc at

O

— the apex angle

\alpha

.
• Indigo chord

AB

— the base of the isosceles triangle.
• Blue segment

OM

— the perpendicular bisector, equal to

\cos(\alpha/2)

.
• Half-angles at

O

— each labeled

\alpha/2

once the bisector is drawn.
• Half-chord labels —

\sin(\alpha/2)

on segments

MA

and

MB

.

A small right-angle mark appears at

M

when the perpendicular bisector becomes visible.

Working with Derived Identities

Selecting

\tan(\alpha/2)

,

\csc(\alpha/2)

,

\sec(\alpha/2)

, or

\cot(\alpha/2)

opens a different card layout. Instead of a triangle, it shows the algebraic derivation as a chain of equations.

Layout of the derived card:
• A short intro explains which earlier identity the current one rests on.
• Jump buttons link directly to the geometric proofs of the source identities.
• A multi-line derivation block shows each manipulation with a brief side note.
• Verification cards confirm both sides match numerically.

This split keeps the geometric ideas isolated to two functions and treats the other four as algebraic consequences.

Reading the Formula Table

A reference table beneath every scene lists all six identities at once:

• Function column — the name of the trig function with

\alpha/2

argument.
• Identity column — the right-hand side of the formula.
• Value column — the numeric value at the current

\alpha/2

.
• Source column — labels each identity as geometric (

\sin

,

\cos

) or via X for derived ones.

Click any row to make that function active. The current row gets a deep-blue left border and tinted background.

Verifying Identities Numerically

Every scene includes two metric cards that compute both sides of the active identity at the current

\alpha

.

Example for

\sin(\alpha/2)

:
• Left card shows

\sin(\alpha/2)

.
• Right card shows

\sqrt{(1 - \cos\alpha)/2}

.

The two numbers always match (within rounding to three decimals). Sweeping the slider while watching the cards is a fast empirical check that the identity holds for every

\alpha

, not just the one in the picture. The formula table mirrors this across all six functions simultaneously.

Geometric Proofs: sin(α/2) and cos(α/2)

The two foundational identities are proved by drawing an isosceles triangle with two unit radii.

sin(α/2) — apply the law of cosines and equate with the squared chord:

|AB|^2 = 2 - 2\cos\alpha = (2\sin(\alpha/2))^2

Solving gives

\sin(\alpha/2) = \sqrt{(1 - \cos\alpha)/2}

.

cos(α/2) — from Pythagoras in the half-triangle,

\cos^2(\alpha/2) = 1 - \sin^2(\alpha/2)

. Substituting the sin half-angle identity:

\cos(\alpha/2) = \sqrt{(1 + \cos\alpha)/2}

For full coverage and equivalent forms (including sign choices by quadrant), see the half angle identities theory page.

Derived Identities: tan(α/2), csc(α/2), sec(α/2), cot(α/2)

The four remaining identities follow directly from the first two:

tan(α/2) — from

\tan = \sin/\cos

applied to the half angle:

\tan(\alpha/2) = \sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}}

csc(α/2) — reciprocal of

\sin(\alpha/2)

:

\csc(\alpha/2) = \sqrt{\frac{2}{1 - \cos\alpha}}

sec(α/2) — reciprocal of

\cos(\alpha/2)

:

\sec(\alpha/2) = \sqrt{\frac{2}{1 + \cos\alpha}}

cot(α/2) — reciprocal of

\tan(\alpha/2)

:

\cot(\alpha/2) = \sqrt{\frac{1 + \cos\alpha}{1 - \cos\alpha}}

For step-by-step derivations and alternative forms, see the trigonometric identities page and the reciprocal identities page.

Why Half-Angle Identities Matter

Half-angle identities are essential whenever a problem asks for a trig function at an angle that is not on the unit circle but is half of one that is.

• Exact-value computation — find

\sin 15°

,

\cos 22.5°

,

\tan 75°

from

\sin 30°

,

\cos 45°

,

\cos 150°

.
• Integration — the Weierstrass substitution

t = \tan(\alpha/2)

converts rational trig integrals into rational functions of

t

.
• Equation solving — reduce equations mixing

\sin\alpha

and

\sin(\alpha/2)

to single-argument form.
• Geometry — apothems, chord lengths, and inscribed-polygon side lengths all use half-angle formulas.

For applications and worked examples, see the trigonometric identities applications page.

Related Concepts and Tools

Continue exploring with these connected resources:

• Double Angle Identities — companion formulas for

\sin(2\theta)

,

\cos(2\theta)

, and beyond.
• Pythagorean Identities —

\sin^2\theta + \cos^2\theta = 1

and its companions, used in the cos half-angle proof.
• Sum and Difference Identities — base relations from which double- and half-angle identities are derived.
• Unit Circle — geometric setup for every identity in this tool.
• Trigonometric Functions Graphs — see how

\sin

,

\cos

,

\tan

and their reciprocals evolve as the angle varies.
• Triangle Explorer — interactive triangles with built-in law of sines and law of cosines.

The Sine Half-Angle Identity

The identity

\sin(\alpha/2) = \sqrt{(1 - \cos\alpha)/2}

is proved geometrically by measuring the chord of an isosceles triangle two ways — the same figure as the double-angle proofs, read in the other direction.

The complete sine derivation, frozen

The chord of angle α measured two ways — law of cosines against the half-chord — solved for sin(α/2) = √((1 − cos α)/2).

The proof's six stages: setup, law of cosines, bisect, read off the half-chord, square the chord, and equate and solve.

Within the tool's range (

\alpha

from

20°

to

160°

) the half angle stays acute, so the positive square root is always the right choice. This identity feeds cosecant's formula directly and supplies half of tangent's.

The Cosine Half-Angle Identity

The identity

\cos(\alpha/2) = \sqrt{(1 + \cos\alpha)/2}

needs no second measurement: it rests on the sine half-angle identity plus one application of Pythagoras inside the right triangle the bisector creates.

The complete cosine derivation, frozen

Pythagoras inside the bisected triangle, with the sine result substituted, yields cos(α/2) = √((1 + cos α)/2).

Its six stages: setup, bisect, identify the legs, apply Pythagoras, substitute the sin half-angle, and take the root.

The sign flip between the two formulas —

1 - \cos\alpha

for sine,

1 + \cos\alpha

for cosine — comes straight from that Pythagorean complement. Secant's identity inverts this one.

The Tangent Half-Angle Identity

Tangent of the half angle divides the two geometric results:

tan(α/2), derived

Sine over cosine, both half-angle roots substituted, combined under a single radical.

\tan(\alpha/2) = \frac{\sin(\alpha/2)}{\cos(\alpha/2)} = \sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}}

Combining the two square roots into one is legitimate because both radicands are positive in the tool's range. Equivalent forms

\sin\alpha/(1 + \cos\alpha)

and

(1 - \cos\alpha)/\sin\alpha

avoid the root entirely — that last form is the heart of the Weierstrass substitution. Sources: sine and cosine; its own reciprocal gives cotangent.

The Cosecant Half-Angle Identity

Cosecant inverts the sine result:

csc(α/2), derived

The sine half-angle identity inverted under its root.

\csc(\alpha/2) = \frac{1}{\sin(\alpha/2)} = \sqrt{\frac{2}{1 - \cos\alpha}}

Inverting under the root flips the fraction — the

2

moves to the numerator. The formula diverges as

\alpha \to 0

, where the half angle's sine vanishes; inside the tool's

20°

–

160°

range both verification cards stay finite and equal. Source: the sine half-angle identity.

The Secant Half-Angle Identity

Secant inverts the cosine result:

sec(α/2), derived

The cosine half-angle identity inverted under its root.

\sec(\alpha/2) = \frac{1}{\cos(\alpha/2)} = \sqrt{\frac{2}{1 + \cos\alpha}}

Like its partner cosecant, it satisfies

|\sec(\alpha/2)| \ge 1

wherever defined. It would diverge only as

\alpha \to 360°

, far outside the slider's range. Source: the cosine half-angle identity.

The Cotangent Half-Angle Identity

Cotangent flips the tangent result upside down:

cot(α/2), derived

The tangent formula flipped: the fraction under the radical turns upside down.

\cot(\alpha/2) = \frac{1}{\tan(\alpha/2)} = \sqrt{\frac{1 + \cos\alpha}{1 - \cos\alpha}}

The inversion swaps numerator and denominator under the root. Because tangent was itself derived, cotangent stands two steps from the geometry, resting ultimately on sine and cosine.

Sine Half-Angle, Step 1: Setup

The sine half-angle proof opens with two unit radii

OA

and

OB

meeting at the center with apex angle

\alpha

— and a goal: express

\sin(\alpha/2)

using only

\alpha

.

Step 1: the setup

Two unit radii meeting at the full angle α, joined by the chord AB.

The half angle does not exist in the figure yet; it will appear the moment the apex is bisected.

Sine Half-Angle, Step 2: Law of Cosines on Triangle OAB

Measure the chord first: the law of cosines with two unit sides and included angle

\alpha

gives

Step 2: law of cosines

|AB|² = 2 − 2 cos α, read straight off triangle OAB.

|AB|^2 = 1 + 1 - 2\cos\alpha = 2 - 2\cos\alpha

One expression for the squared chord, in terms of the full angle only.

Sine Half-Angle, Step 3: Bisect

Drop

OM

perpendicular to

AB

: it splits the apex into two halves of

\alpha/2

and lands on the chord's midpoint

M

.

Step 3: the bisector

OM halves the apex into two α/2 angles and lands on the chord midpoint.

This is where the half angle enters the picture — as a genuine geometric angle, not an algebraic trick.

Sine Half-Angle, Step 4: Read Off the Half-Chord

Right triangle

OMA

has hypotenuse

1

and angle

\alpha/2

at

O

, so its opposite leg is exactly the half-chord:

Step 4: the half-chord

MA = sin(α/2), so the full chord is 2 sin(α/2).

MA = \sin(\alpha/2) \qquad AB = 2\sin(\alpha/2)

A second expression for the chord — this one in terms of the half angle.

Sine Half-Angle, Step 5: Square the Chord

Square the half-chord expression to match the law-of-cosines form:

Step 5: square the chord

(2 sin(α/2))² = 4 sin²(α/2) — ready to meet the law-of-cosines expression.

|AB|^2 = (2\sin(\alpha/2))^2 = 4\sin^2(\alpha/2)

Both measurements of

|AB|^2

are now on the table.

Sine Half-Angle, Step 6: Equate and Solve

Set the two chord measurements equal and solve for the half-angle sine:

Step 6: equate and solve

Setting the two chord expressions equal isolates sin(α/2) under the root.

2 - 2\cos\alpha = 4\sin^2(\alpha/2) \;\Longrightarrow\; \sin(\alpha/2) = \sqrt{\frac{1 - \cos\alpha}{2}}

The verification cards keep both sides numerically equal across the whole slider range.

Cosine Half-Angle, Step 1: Setup

The cosine half-angle proof starts from the same figure: unit radii

OA

and

OB

with apex

\alpha

, aiming for

\cos(\alpha/2)

in terms of

\alpha

.

Step 1: the setup

The same unit triangle at angle α — this time the target is the adjacent leg.

Unlike sine's proof, this one will finish without ever measuring the chord by law of cosines.

Cosine Half-Angle, Step 2: Bisect

Drop the perpendicular bisector

OM

immediately: two half-angles of

\alpha/2

at the center, and a right angle at

M

.

Step 2: the bisector

OM splits the apex; right triangle OMA now carries the half-angle α/2.

Right triangle

OMA

is the whole stage for the rest of this proof.

Cosine Half-Angle, Step 3: Identify the Legs

In right triangle

OMA

, the hypotenuse is

1

and the angle at

O

is

\alpha/2

, so both legs are the basic ratios:

Step 3: the legs

Against the unit hypotenuse, the legs are sin(α/2) and cos(α/2) themselves.

MA = \sin(\alpha/2) \qquad OM = \cos(\alpha/2)

The bisector segment

OM

is the quantity the proof is after.

Cosine Half-Angle, Step 4: Apply Pythagoras

The legs of a right triangle with unit hypotenuse obey the Pythagorean identity:

Step 4: Pythagoras

cos²(α/2) = 1 − sin²(α/2) inside the right triangle.

\sin^2(\alpha/2) + \cos^2(\alpha/2) = 1 \;\Longrightarrow\; \cos^2(\alpha/2) = 1 - \sin^2(\alpha/2)

Cosine is now expressed through the half-angle sine, which step 5 already knows how to replace.

Cosine Half-Angle, Step 5: Substitute the Sin Half-Angle

Insert the result of the sine proof,

\sin^2(\alpha/2) = (1 - \cos\alpha)/2

:

Step 5: substitute

The sine half-angle result replaces sin²(α/2), leaving only cos α inside.

\cos^2(\alpha/2) = 1 - \frac{1 - \cos\alpha}{2} = \frac{1 + \cos\alpha}{2}

The sign inside flips from minus to plus — the fingerprint of the Pythagorean complement.

Cosine Half-Angle, Step 6: Take the Root

Take the positive square root (the half angle is acute throughout the tool's range):

Step 6: take the root

The square root delivers cos(α/2) = √((1 + cos α)/2).

\cos(\alpha/2) = \sqrt{\frac{1 + \cos\alpha}{2}}

Both verification cards agree at every slider position — the numerical seal on the proof.

Half Angle Trigonometric Identities