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Triple angle formulae

Derive the triple angle formulae for sine, cosine and tangent. Solve multiple-angle equations, connect trigonometry with cubics and keep every interval solution.

Before you startCompound and double angle formulae, factorisation and trig equations

01 / A triple angle is a compound angle

Write 3θ as 2θ + θ, then simplify.

You do not need a separate geometric rule for each multiple angle. Combining compound-angle formulae with double-angle identities produces a cubic expression in a single sine or cosine.

sin 3θ = 3sin θ − 4sin³ θ
cos 3θ = 4cos³ θ − 3cos θ

The circle tracks the original and tripled angles. The lower graph shows the corresponding cubic with input u = sin θ or u = cos θ. Different angles can have the same input u.

Triple an angle, evaluate a cubicExplore

Angle θCubic identity

θ = 30°; 3θ = 90°; u = sin θ = 0.5000.

sin 3θ = 3u − 4u³ = 1.0000.

The cubic uses only inputs −1 ≤ u ≤ 1. Readouts are rounded to four decimal places; the written identities are exact.

02 / Derive the sine triple-angle identity

Replace every squared cosine with 1 − sin² θ.

Expand sin(2θ + θ)Worked example

sin 3θ = sin 2θ cos θ + cos 2θ sin θ

Use the sine addition formula.

= 2sin θ cos² θ + (1 − 2sin² θ)sin θ

Choose a double-angle cosine form in sine.

= 2sin θ(1 − sin² θ) + sin θ − 2sin³ θ

Eliminate the squared cosine.

= 3sin θ − 4sin³ θ

Collect the linear and cubic terms.

The identity holds for every real θ: no division or square root has introduced a restriction.

Watch the angle run at three times the rate

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

03 / Derive the cosine triple-angle identity

Use a cosine-only expression.

Expand cos(2θ + θ)Worked example

cos 3θ = cos 2θ cos θ − sin 2θ sin θ

The subtraction sign belongs to cosine addition.

= (2cos² θ − 1)cos θ − 2sin² θ cos θ

Apply the double-angle formulae.

= 2cos³ θ − cos θ − 2(1 − cos² θ)cos θ

Replace sin² θ.

= 4cos³ θ − 3cos θ

This gives a cubic in cosine.

Useful reverse forms are sin³ θ = (3sin θ − sin 3θ)/4 and cos³ θ = (cos 3θ + 3cos θ)/4.

04 / Derive tangent and state its domain

A quotient requires more care than a polynomial.

Divide sin 3θ by cos 3θ. When cos θ ≠ 0, divide numerator and denominator by cos³ θ and put t = tan θ.

tan 3θ = (3t − t³)/(1 − 3t²)
t = tan θ

The expression needs cos θ ≠ 0 and 1 − 3t² ≠ 0.

Indeed cos 3θ = cos³ θ(1 − 3tan² θ) wherever tan θ exists. At cos θ = 0, cos 3θ is also zero, so the original tangent of the triple angle is undefined too. Here the stated exclusions match the original domain.

Evaluate tan 3θ when tan θ = 1/2Worked example

(3/2 − 1/8)/(1 − 3/4)

The denominator is 1/4, not zero.

tan 3θ = (11/8)/(1/4) = 11/2

Retain the exact fraction.

05 / Compute exact values without finding θ

A polynomial identity uses the supplied signed ratio directly.

Given sin θ = 2/5, find sin 3θWorked example

sin 3θ = 3(2/5) − 4(2/5)³

Substitute the signed value.

= 6/5 − 32/125 = 118/125

No inverse sine or quadrant choice is needed for this result.

Given cos θ = −3/5, find cos 3θWorked example

cos 3θ = 4(−3/5)³ − 3(−3/5)

Keep the minus sign inside the cube.

= −108/125 + 9/5 = 117/125

A negative input cosine can give a positive triple-angle cosine.

06 / Preserve zero factors in equations

Moving all terms to one side reveals extra roots.

Solve sin 3x = sin x for 0° ≤ x ≤ 360°Worked example

3sin x − 4sin³ x = sin x

Replace the triple angle.

2sin x(1 − 2sin² x) = 0

Factor; do not divide by sin x.

sin x = 0 or sin² x = 1/2

The second branch has both positive and negative square roots.

x = 0°, 45°, 135°, 180°, 225°, 315°, 360°

Include both allowed endpoints.

An independent check uses sin A = sin B: either A = B + 360°k or A = 180° − B + 360°k. With A = 3x and B = x, these give x = 180°k or x = 45° + 90°k.

07 / Transform the full interval before solving

Three turns can contain six roots of a simple sine equation.

Solve sin 3x = 1/2 for 0° ≤ x < 360°Worked example

0° ≤ 3x < 1080°

The input angle covers three full turns.

3x = 30°, 150°, 390°, 510°, 750°, 870°

List both sine branches in every turn.

x = 10°, 50°, 130°, 170°, 250°, 290°

Divide every answer by 3.

Solving only one turn of the tripled angle loses four valid roots. The same principle applies in radians: transform both endpoints into the units used by the equation.

08 / Solve a cubic through a cosine identity

Choose an angle interval that represents each cosine input once.

Find all real roots of 8u³ − 6u − 1 = 0Worked example

4u³ − 3u = 1/2

Match the cosine triple-angle expression.

Put u = cos θ with 0 ≤ θ ≤ π

Cosine covers [−1, 1] once on this interval.

cos 3θ = 1/2, with 0 ≤ 3θ ≤ 3π

The allowed tripled-angle interval matters.

3θ = π/3, 5π/3, 7π/3

These are all roots in that interval.

u = cos(π/9), cos(5π/9), cos(7π/9)

The three cosine values are distinct.

A nonzero cubic has at most three real roots

These three roots prove completeness, including ruling out any outside [−1, 1].

The substitution alone does not rule out roots outside [−1, 1]. Here the degree argument completes the proof after three distinct roots have been found.

09 / Do not choose an exact branch from a decimal guess

An algebraic equation may have several legitimate roots.

Explain why cos 20° is a root of 8u³ − 6u − 1 = 0Worked example

cos 60° = 4cos³ 20° − 3cos 20° = 1/2

Use the triple-angle identity.

8cos³ 20° − 6cos 20° − 1 = 0

Multiply by 2 and rearrange.

Rewrite 8sin³ x − 6sin x in terms of a single triple-angle function.

Hint

Factor out −2 from the sine triple-angle formula.

Worked solution

8sin³ x − 6sin x = −2sin 3x.

11 / Recap

Identity, algebra, interval, domain.

Expand 2θ + θ to derive triple-angle formulae.
Use sine-only or cosine-only cubics without unnecessary inverse functions.
Retain tangent exclusions and zero factors.
Expand the input interval before listing multiple-angle roots.
For cubic substitutions, justify completeness and choose the required branch.

Section 1 of 11 · A triple angle is a compound angle