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Triple Angle Identities


How to use
  1. The six tabs across the top switch between sin⁡3θ\sin 3\theta, cos⁡3θ\cos 3\theta, tan⁡3θ\tan 3\theta, csc⁡3θ\csc 3\theta, sec⁡3θ\sec 3\theta and cot⁡3θ\cot 3\theta. Sine and cosine open the step-by-step derivation; the other four open a derived-identity card. Learn more about switching functions
  2. Drag the θ slider between 10°10° and 80°80°. The circle figure redraws, and every numerical check on the page recomputes at the new angle. Learn more about the angle
  3. Press Play to run the derivation one line at a time, or step with ‹ Prev and Next ›. Reset returns to the start and the speed menu sets how fast Play advances. Learn more about playing the derivation
  4. The figure shows 3θ3\theta as 2θ+θ2\theta + \theta on the unit circle at step 1, then the equation chain growing one line per step. Learn more about the figure
  5. The Derivation panel on the right lists every step taken so far with the rule it used. Click any earlier step to jump back to it. Learn more about the derivation panel
  6. The two cards under the controls compare f(3θ)f(3\theta) with the current line of the chain at the chosen angle. They agree on every line, which is the check that no step went wrong. Learn more about the numerical check
  7. On the tan, csc, sec and cot tabs, the card explains where the identity comes from, shows the short derivation, and verifies it at θ. The source buttons jump to the identities it was built from. Learn more about derived identities
  8. The formula table under the tool lists all six identities with their values at θ. Click a row to open that function. Learn more about the formula table

sin(3θ) = 3 sin θ − 4 sin³θ
θ35°
2θθ3θO1sin 3θ = sin(2θ + θ)
Step 0 of 5

sin(3θ)

0.966

sin(2θ + θ)

0.966

Derivation
Press Play to step through the proof.
Function
Identity
Value
Source


Switching Between Functions
Adjusting the Angle
Playing Through the Derivation
Reading the Figure
The Derivation Panel
Checking Each Line Numerically
Working with Derived Identities
Reading the Formula Table
Why Three θ Is Two θ Plus θ
Why Each Result Uses Only One Function
Where Triple-Angle Identities Are Used
Related Concepts and Tools
The Sine Triple-Angle Identity
The Cosine Triple-Angle Identity
The Tangent Triple-Angle Identity
The Cosecant Triple-Angle Identity
The Secant Triple-Angle Identity
The Cotangent Triple-Angle Identity
Sine Step 1: Split the Angle
Sine Step 2: Apply the Angle-Sum Formula
Sine Step 3: Substitute the Double-Angle Identities
Sine Step 4: Apply the Pythagorean Identity
Sine Step 5: Collect Terms
Cosine Step 1: Split the Angle
Cosine Step 2: Apply the Angle-Sum Formula
Cosine Step 3: Substitute the Double-Angle Identities
Cosine Step 4: Apply the Pythagorean Identity
Cosine Step 5: Collect Terms



Switching Between Functions

The tab strip across the top of the tool holds one tab for each of the six functions of 3θ3\theta. The active tab is highlighted and its identity is written out in the bar beneath it.

The tabs fall into two groups. Sine and cosine are proved from scratch: their tabs open a step-by-step derivation that starts from the angle sum and ends at the identity. Tangent, cosecant, secant and cotangent are derived from those two: their tabs open a shorter card that shows how each follows from the sine and cosine results.

Switching tabs resets the derivation to its start, so every function begins from step 0. The angle θ is kept, so all six identities can be compared at the same value. The current function is also written into the page address as `?fn=`, which means a link to this page can open straight onto, say, the cosine derivation.

The formula table under the tool is a second way to switch: clicking any of its rows opens that function.

Adjusting the Angle

The θ slider sits under the identity bar and runs from 10°10° to 80°80° in whole degrees, with the current value printed at its right.

The angle matters in two places. On step 1 of the sine and cosine derivations, the figure draws the unit circle with the 2θ2\theta arc, the extra θ\theta arc and the full 3θ3\theta turn at the chosen angle, so dragging the slider opens and closes the split. Everywhere else, θ is the value at which the identities are checked numerically: the metric cards under the derivation, the verification cards of the derived identities, and the value column of the formula table all recompute as the slider moves.

The range stops short of 90°90° on purpose. At 3θ3\theta near a multiple of 90°90°, tangent, secant or cotangent run off to infinity, and the checks would print ∞\infty instead of numbers. Between 10°10° and 80°80° most values stay finite; where one does not, the card says so rather than showing a meaningless figure.

Playing Through the Derivation

DemoStepping through the proof
Step 0 of 5
    On the sine and cosine tabs, a control bar under the figure drives the derivation. It has five steps, and the counter at the right reads Step n of 5.

  • Play advances one step every few seconds until the end, then changes to Replay. While running it reads Pause.
  • Next › and ‹ Prev move one step at a time and stop Play.
  • Reset returns to step 0, before anything has been drawn.
  • speed menu sets how fast Play advances: 0.5×0.5\times, 1×1\times, 1.5×1.5\times or 2×2\times.

    • Each step does one thing, named in bold in the derivation panel: split the angle, apply the angle-sum formula, substitute the double-angle identities, apply the Pythagorean identity, collect terms. The five steps are the same for sine and cosine, which is why the two derivations are worth running side by side — the moves are identical, only the formulas differ.

Reading the Figure

The figure in the middle of the tool changes character between the first step and the rest.

At step 1 it is a unit circle. A solid radius lies along the positive axis, a dashed indigo radius marks the angle 2θ2\theta, and a second solid radius marks 3θ3\theta. Two short arcs near the centre show the split — 2θ2\theta in light indigo, the extra θ\theta in red — and a longer arc labels the whole 3θ3\theta. A banner at the top states the step in words: sin⁡3θ=sin⁡(2θ+θ)\sin 3\theta = \sin(2\theta + \theta).

From step 2 on, the figure becomes a card headed by the identity being proved, and the equation chain is written into it one line per step. Each line carries the rule that produced it in small grey text at its right, and the final line is set in bold.

This split is deliberate. The geometry does exactly one job — it justifies writing 3θ3\theta as 2θ+θ2\theta + \theta — and the rest of the proof is algebra, so the figure stops pretending otherwise.

The Derivation Panel

The panel headed Derivation on the right of the sine and cosine tabs is a running log of the proof. Before the first step it asks you to press Play; after that, every step taken appears as a numbered entry with its rule in bold and a one- or two-sentence explanation below.

The newest step is outlined in indigo and the panel scrolls to keep it in view. Earlier steps stay listed, so the whole argument can be read top to bottom at any point.

Every entry is clickable. Clicking an earlier step jumps the tool back to it — the figure, the metric cards and the step counter all follow — and stops Play. That makes the panel the fastest way to compare two moments of the proof.

Each explanation ends with two links into this page: one to the section that treats that step in full, with its frozen figure, and one to the section on the identity as a whole.

Checking Each Line Numerically

DemoChecking every line
Step 0 of 5
Two cards under the control bar check the derivation as it runs. The left one always shows f(3θ)f(3\theta) — sin⁡3θ\sin 3\theta or cos⁡3θ\cos 3\theta — at the current angle. The right one shows the current line of the chain, evaluated at the same angle.

Every line of a correct derivation is equal to the thing being derived, so the two cards must agree on every step, to three decimal places. They do. At θ=35°\theta = 35°, for instance, sin⁡105°≈0.966\sin 105° \approx 0.966, and so is sin⁡2θcos⁡θ+cos⁡2θsin⁡θ\sin 2\theta\cos\theta + \cos 2\theta\sin\theta, and so is 3sin⁡θ−4sin⁡3θ3\sin\theta - 4\sin^3\theta.

The check is not a proof — agreement at one angle could be a coincidence — but it catches the common mistakes instantly. A sign dropped in the angle-sum formula, a wrong double-angle form, or a slip in collecting terms would show up as two different numbers. Move the angle slider while a step is showing and the two cards move together, which is the check at every angle at once.

Working with Derived Identities

DemoDerived identities and the table
Step 0 of 5
The tangent, cosecant, secant and cotangent tabs open a card instead of a step-by-step derivation, because none of the four needs one. Each is built from identities already proved.

The card has three parts. How this identity follows gives the one-sentence reason — tangent is sine over cosine, cosecant is the reciprocal of sine, and so on — followed by source buttons naming the identities it depends on. Clicking a source button switches to that tab, so the dependency can be followed back to its proof. Derivation lays out the short algebra, two or three lines, each with a note on what was done. Verify at θ compares the two sides numerically at the current angle.

Tangent is the only one with a real step: after dividing sin⁡3θ\sin 3\theta by cos⁡3θ\cos 3\theta, top and bottom are divided by cos⁡3θ\cos^3\theta so that every term becomes a power of tan⁡θ\tan\theta. The other three are reciprocals, and their derivations are a definition and a substitution.

Reading the Formula Table

Under the tool, a table lists all six triple-angle identities at once, one row per function. Its columns are the function, the identity, its value at the current θ, and its source — angle sum for sine and cosine, or the functions it was derived from for the other four.

The active function's row is marked with an indigo bar on its left. Clicking any row switches the tool to that function, the same as clicking its tab.

The table is useful in two ways. As a reference, it holds the six results in one place without scrolling through the derivations. As a check, its value column shows all six functions of 3θ3\theta at the same angle, so relationships between them can be read off directly: the tangent value is the sine value divided by the cosine value, and the three reciprocals are one over their partners. Moving the angle slider updates the whole column.

Why Three θ Is Two θ Plus θ

There is no new identity behind the triple-angle formulas. They are what the angle-sum identities give when one of the two angles is itself a double angle.

The angle-sum identities express sin⁡(α+β)\sin(\alpha + \beta) and cos⁡(α+β)\cos(\alpha + \beta) in terms of the sines and cosines of α\alpha and β\beta. Taking α=β=θ\alpha = \beta = \theta gives the double-angle identities. Taking α=2θ\alpha = 2\theta and β=θ\beta = \theta gives an expression for sin⁡3θ\sin 3\theta that still contains sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta, and substituting the double-angle identities removes them.

The same move repeats: 4θ=3θ+θ4\theta = 3\theta + \theta, 5θ=4θ+θ5\theta = 4\theta + \theta, and so on, each built from the one before. The triple angle is the first case where the build takes more than one application, which is why its derivation is worth following line by line. The sum and difference and double-angle identities are both set out on the identities lesson.

Why Each Result Uses Only One Function

Step 4 of both derivations uses the Pythagorean identity, sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, to remove one of the two functions. Without it, the sine result would be correct but mixed:

sin⁡3θ=3sin⁡θcos⁡2θ−sin⁡3θ\sin 3\theta = 3\sin\theta\cos^2\theta - \sin^3\theta


Replacing cos⁡2θ\cos^2\theta by 1−sin⁡2θ1 - \sin^2\theta turns that into a polynomial in sin⁡θ\sin\theta alone, 3sin⁡θ−4sin⁡3θ3\sin\theta - 4\sin^3\theta. The cosine derivation does the same in the other direction and ends at 4cos⁡3θ−3cos⁡θ4\cos^3\theta - 3\cos\theta.

The one-function form is what makes the identities useful. It says that cos⁡3θ\cos 3\theta is a cubic polynomial in cos⁡θ\cos\theta, which lets a triple-angle equation be solved as a cubic equation, and it is the start of a pattern: cos⁡nθ\cos n\theta is always a polynomial of degree nn in cos⁡θ\cos\theta. The removal works because each derivation happens to leave only even powers of the unwanted function, and even powers can always be rewritten with the Pythagorean identity.

Where Triple-Angle Identities Are Used

The triple-angle identities turn up wherever an angle and its triple have to be related.

Exact values. Setting θ=20°\theta = 20° in the cosine identity gives cos⁡60°=4cos⁡320°−3cos⁡20°\cos 60° = 4\cos^3 20° - 3\cos 20°, so cos⁡20°\cos 20° is a root of the cubic 4x3−3x−12=04x^3 - 3x - \frac{1}{2} = 0. This is the classical proof that a 60°60° angle cannot be trisected with straightedge and compass: the cubic has no solution built from square roots alone.

Cubic equations. Run the other way, the identity solves cubics. An equation of the form 4x3−3x=c4x^3 - 3x = c with ∣c∣≤1|c| \le 1 has the solution x=cos⁡(13arccos⁡c)x = \cos\left(\frac{1}{3}\arccos c\right), which is the trigonometric method for cubics with three real roots.

Integration and series. Solving the sine identity for sin⁡3θ\sin^3\theta gives sin⁡3θ=3sin⁡θ−sin⁡3θ4\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}, which reduces a cube to first powers — the same power-reduction move the power-reducing identities make for squares.

Related Concepts and Tools

The triple-angle identities sit at the end of a chain of identities, and the tools for the earlier links are the natural neighbours.

• Double Angle Identities — the identities substituted at step 3 of both derivations, each with its own geometric proof.
• Half Angle Identities — the double-angle identities solved the other way, for θ/2\theta/2.
• Pythagorean Identities — the identity used at step 4 to write each result in one function.
• Basic Trigonometric Identities — the reciprocal and quotient identities behind the derived tan, csc, sec and cot forms.
• Unit Circle — the setting of the step-1 figure, where 3θ3\theta is split into 2θ+θ2\theta + \theta.

For the theory in prose, the trigonometric identities lesson covers every family in one place, including the triple-angle identities themselves.

The Sine Triple-Angle Identity

The identity sin⁡(3θ)=3sin⁡θ−4sin⁡3θ\sin(3\theta) = 3\sin\theta - 4\sin^3\theta is derived in the tool in five steps, from the angle sum to a cubic in sin⁡θ\sin\theta.
sin(3θ) = 3 sin θ − 4 sin³θsin 3θ =sin(2θ + θ)split the angle=sin 2θ · cos θ + cos 2θ · sin θangle-sum formula=2 sin θ cos²θ + (1 − 2 sin²θ) sin θsubstitute both double-angle identities=2 sin θ (1 − sin²θ) + sin θ − 2 sin³θcos²θ = 1 − sin²θ=3 sin θ − 4 sin³θcollect terms
The complete sine derivation, frozen

Five lines from sin(2θ + θ) to 3 sin θ − 4 sin³θ: the angle sum, both double-angle identities, and one Pythagorean substitution.

The derivation runs: split the angle, apply the angle-sum formula, substitute the double-angle identities, apply the Pythagorean identity, and collect terms.

The result is odd in sin⁡θ\sin\theta, as it must be: sin⁡3θ\sin 3\theta changes sign with θ\theta. It is also bounded as it should be — at sin⁡θ=1\sin\theta = 1 it gives 3−4=−13 - 4 = -1, which is sin⁡270°\sin 270°.

The Cosine Triple-Angle Identity

The identity cos⁡(3θ)=4cos⁡3θ−3cos⁡θ\cos(3\theta) = 4\cos^3\theta - 3\cos\theta is derived in the same five steps as the sine identity, ending in a cubic in cos⁡θ\cos\theta.
cos(3θ) = 4 cos³θ − 3 cos θcos 3θ =cos(2θ + θ)split the angle=cos 2θ · cos θ − sin 2θ · sin θangle-sum formula=(2 cos²θ − 1) cos θ − 2 sin²θ cos θsubstitute both double-angle identities=2 cos³θ − cos θ − 2 (1 − cos²θ) cos θsin²θ = 1 − cos²θ=4 cos³θ − 3 cos θcollect terms
The complete cosine derivation, frozen

The same five moves for cosine, ending at 4 cos³θ − 3 cos θ, written in cos θ alone.

The derivation runs: split the angle, apply the angle-sum formula, substitute the double-angle identities, apply the Pythagorean identity, and collect terms.

The polynomial 4x3−3x4x^3 - 3x is the third Chebyshev polynomial, and it is the identity behind the uses in exact values and cubic equations.

The Tangent Triple-Angle Identity

The identity tan⁡(3θ)=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan(3\theta) = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} is not proved from scratch; it is the sine identity divided by the cosine identity.
tan(3θ) = (3 tan θ − tan³θ) / (1 − 3 tan²θ)tan(3θ) =sin(3θ) / cos(3θ)definition=(3 sin θ − 4 sin³θ) / (4 cos³θ − 3 cos θ)substitute the two identities=(3 tan θ − tan³θ) / (1 − 3 tan²θ)divide top and bottom by cos³θ
tan(3θ), derived

Sine over cosine, then top and bottom divided by cos³θ, which turns every term into a power of tan θ.

After the division, top and bottom are divided by cos⁡3θ\cos^3\theta. Each term of 3sin⁡θ−4sin⁡3θ3\sin\theta - 4\sin^3\theta and of 4cos⁡3θ−3cos⁡θ4\cos^3\theta - 3\cos\theta has degree three in sine and cosine, so each becomes a power of tan⁡θ\tan\theta once the sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta relation is used. The denominator vanishes at tan⁡2θ=13\tan^2\theta = \frac{1}{3}, which is θ=30°\theta = 30°, where 3θ=90°3\theta = 90° and tangent is undefined.

The Cosecant Triple-Angle Identity

The identity csc⁡(3θ)=13sin⁡θ−4sin⁡3θ\csc(3\theta) = \dfrac{1}{3\sin\theta - 4\sin^3\theta} is the reciprocal of the sine identity.
csc(3θ) = 1 / (3 sin θ − 4 sin³θ)csc(3θ) =1 / sin(3θ)definition=1 / (3 sin θ − 4 sin³θ)substitute sin(3θ)
csc(3θ), derived

The reciprocal of the sine identity: two lines, with no new algebra.

Its derivation is two lines: the definition csc⁡3θ=1sin⁡3θ\csc 3\theta = \frac{1}{\sin 3\theta} and a substitution. It is undefined wherever sin⁡3θ=0\sin 3\theta = 0, which in the tool's range happens at θ=60°\theta = 60°.

The Secant Triple-Angle Identity

The identity sec⁡(3θ)=14cos⁡3θ−3cos⁡θ\sec(3\theta) = \dfrac{1}{4\cos^3\theta - 3\cos\theta} is the reciprocal of the cosine identity.
sec(3θ) = 1 / (4 cos³θ − 3 cos θ)sec(3θ) =1 / cos(3θ)definition=1 / (4 cos³θ − 3 cos θ)substitute cos(3θ)
sec(3θ), derived

The reciprocal of the cosine identity: two lines, with no new algebra.

As with cosecant, the derivation is a definition and a substitution. It is undefined wherever cos⁡3θ=0\cos 3\theta = 0, which in the tool's range happens at θ=30°\theta = 30°.

The Cotangent Triple-Angle Identity

The identity cot⁡(3θ)=1−3tan⁡2θ3tan⁡θ−tan⁡3θ\cot(3\theta) = \dfrac{1 - 3\tan^2\theta}{3\tan\theta - \tan^3\theta} is the reciprocal of the tangent identity, with numerator and denominator exchanged.
cot(3θ) = (1 − 3 tan²θ) / (3 tan θ − tan³θ)cot(3θ) =1 / tan(3θ)definition=(1 − 3 tan²θ) / (3 tan θ − tan³θ)substitute tan(3θ)
cot(3θ), derived

The reciprocal of the tangent identity, with numerator and denominator swapped.

It could equally be written in cot⁡θ\cot\theta by dividing top and bottom by tan⁡3θ\tan^3\theta, but the tool keeps it in tan⁡θ\tan\theta so that it reads directly as the tangent identity turned over. It is undefined wherever sin⁡3θ=0\sin 3\theta = 0.

Sine Step 1: Split the Angle

The derivation begins with geometry: 3θ3\theta is written as 2θ+θ2\theta + \theta.
2θθ3θO1sin 3θ = sin(2θ + θ)
Step 1: 3θ as 2θ + θ on the unit circle, θ = 35°

The 2θ arc and one more θ arc make up the whole 3θ turn. Everything after this step is algebra on that split.

On the unit circle, turning through 3θ3\theta is the same as turning through 2θ2\theta and then one more θ\theta, so sin⁡3θ=sin⁡(2θ+θ)\sin 3\theta = \sin(2\theta + \theta). This is the only geometric step; it is what lets the angle-sum identities take over.

Sine Step 2: Apply the Angle-Sum Formula

The angle-sum formula sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta is applied with α=2θ\alpha = 2\theta and β=θ\beta = \theta.
sin(3θ) = 3 sin θ − 4 sin³θsin 3θ =sin(2θ + θ)split the angle=sin 2θ · cos θ + cos 2θ · sin θangle-sum formula
Step 2: the angle-sum formula applied

sin(2θ + θ) expands into two products, each still containing a double angle.

sin⁡3θ=sin⁡2θcos⁡θ+cos⁡2θsin⁡θ\sin 3\theta = \sin 2\theta\cos\theta + \cos 2\theta\sin\theta
The expression is correct but still contains double angles, which the next step removes.

Sine Step 3: Substitute the Double-Angle Identities

Both double-angle identities are substituted: sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta and cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta.
sin(3θ) = 3 sin θ − 4 sin³θsin 3θ =sin(2θ + θ)split the angle=sin 2θ · cos θ + cos 2θ · sin θangle-sum formula=2 sin θ cos²θ + (1 − 2 sin²θ) sin θsubstitute both double-angle identities
Step 3: both double-angle identities substituted

sin 2θ and cos 2θ are replaced, so only functions of θ itself remain — but both sin and cos.

sin⁡3θ=2sin⁡θcos⁡2θ+(1−2sin⁡2θ)sin⁡θ\sin 3\theta = 2\sin\theta\cos^2\theta + (1 - 2\sin^2\theta)\sin\theta
The form of cos⁡2θ\cos 2\theta was chosen to be in sin⁡θ\sin\theta, so only one cos⁡2θ\cos^2\theta is left to deal with.

Sine Step 4: Apply the Pythagorean Identity

The Pythagorean identity replaces cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta.
sin(3θ) = 3 sin θ − 4 sin³θsin 3θ =sin(2θ + θ)split the angle=sin 2θ · cos θ + cos 2θ · sin θangle-sum formula=2 sin θ cos²θ + (1 − 2 sin²θ) sin θsubstitute both double-angle identities=2 sin θ (1 − sin²θ) + sin θ − 2 sin³θcos²θ = 1 − sin²θ
Step 4: the Pythagorean identity applied

cos²θ becomes 1 − sin²θ, and the expression is now written in sin θ alone.

sin⁡3θ=2sin⁡θ(1−sin⁡2θ)+sin⁡θ−2sin⁡3θ\sin 3\theta = 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta
Every term is now in sin⁡θ\sin\theta alone, the point of writing each result in one function.

Sine Step 5: Collect Terms

Expanding and collecting finishes the chain.
sin(3θ) = 3 sin θ − 4 sin³θsin 3θ =sin(2θ + θ)split the angle=sin 2θ · cos θ + cos 2θ · sin θangle-sum formula=2 sin θ cos²θ + (1 − 2 sin²θ) sin θsubstitute both double-angle identities=2 sin θ (1 − sin²θ) + sin θ − 2 sin³θcos²θ = 1 − sin²θ=3 sin θ − 4 sin³θcollect terms
Step 5: terms collected

The chain closes at 3 sin θ − 4 sin³θ, a cubic in sin θ.

2sin⁡θ−2sin⁡3θ+sin⁡θ−2sin⁡3θ=3sin⁡θ−4sin⁡3θ2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta
This is the sine triple-angle identity.

Cosine Step 1: Split the Angle

The cosine derivation starts from the same split: cos⁡3θ=cos⁡(2θ+θ)\cos 3\theta = \cos(2\theta + \theta).
2θθ3θO1cos 3θ = cos(2θ + θ)
Step 1: 3θ as 2θ + θ on the unit circle, θ = 35°

The same split as for sine: 3θ is the 2θ turn followed by one more θ.

The geometry is identical to the sine case — the 2θ2\theta turn followed by one more θ\theta — and again it is the only geometric step.

Cosine Step 2: Apply the Angle-Sum Formula

The angle-sum formula for cosine, cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta, is applied with α=2θ\alpha = 2\theta, β=θ\beta = \theta.
cos(3θ) = 4 cos³θ − 3 cos θcos 3θ =cos(2θ + θ)split the angle=cos 2θ · cos θ − sin 2θ · sin θangle-sum formula
Step 2: the angle-sum formula applied

cos(2θ + θ) expands into a difference of two products.

cos⁡3θ=cos⁡2θcos⁡θ−sin⁡2θsin⁡θ\cos 3\theta = \cos 2\theta\cos\theta - \sin 2\theta\sin\theta
The minus sign is the one place the cosine derivation differs in structure from the sine one.

Cosine Step 3: Substitute the Double-Angle Identities

The double-angle identities are substituted, this time with cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1.
cos(3θ) = 4 cos³θ − 3 cos θcos 3θ =cos(2θ + θ)split the angle=cos 2θ · cos θ − sin 2θ · sin θangle-sum formula=(2 cos²θ − 1) cos θ − 2 sin²θ cos θsubstitute both double-angle identities
Step 3: both double-angle identities substituted

cos 2θ and sin 2θ are replaced by their double-angle forms in θ.

cos⁡3θ=(2cos⁡2θ−1)cos⁡θ−2sin⁡2θcos⁡θ\cos 3\theta = (2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta
Choosing the form of cos⁡2θ\cos 2\theta in cos⁡θ\cos\theta leaves a single sin⁡2θ\sin^2\theta to remove.

Cosine Step 4: Apply the Pythagorean Identity

The Pythagorean identity replaces sin⁡2θ\sin^2\theta with 1−cos⁡2θ1 - \cos^2\theta.
cos(3θ) = 4 cos³θ − 3 cos θcos 3θ =cos(2θ + θ)split the angle=cos 2θ · cos θ − sin 2θ · sin θangle-sum formula=(2 cos²θ − 1) cos θ − 2 sin²θ cos θsubstitute both double-angle identities=2 cos³θ − cos θ − 2 (1 − cos²θ) cos θsin²θ = 1 − cos²θ
Step 4: the Pythagorean identity applied

sin²θ becomes 1 − cos²θ, so every term is in cos θ.

cos⁡3θ=2cos⁡3θ−cos⁡θ−2(1−cos⁡2θ)cos⁡θ\cos 3\theta = 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta
Every term is now in cos⁡θ\cos\theta alone.

Cosine Step 5: Collect Terms

Expanding and collecting finishes the chain.
cos(3θ) = 4 cos³θ − 3 cos θcos 3θ =cos(2θ + θ)split the angle=cos 2θ · cos θ − sin 2θ · sin θangle-sum formula=(2 cos²θ − 1) cos θ − 2 sin²θ cos θsubstitute both double-angle identities=2 cos³θ − cos θ − 2 (1 − cos²θ) cos θsin²θ = 1 − cos²θ=4 cos³θ − 3 cos θcollect terms
Step 5: terms collected

The chain closes at 4 cos³θ − 3 cos θ, a cubic in cos θ.

2cos⁡3θ−cos⁡θ−2cos⁡θ+2cos⁡3θ=4cos⁡3θ−3cos⁡θ2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta = 4\cos^3\theta - 3\cos\theta
This is the cosine triple-angle identity.