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Triple angle Identities

Triple angle identities express trigonometric functions of 3x3x in terms of functions of xx.

They are useful when a problem contains angles such as 3x3x, 60∘−3x60^\circ-3x, or when equations involve cubic expressions in sine or cosine.

Sine Triple angle Formula

Start with

sin⁡3x=sin⁡(2x+x).\sin 3x=\sin(2x+x).

Using the angle addition formula,

sin⁡3x=sin⁡2xcos⁡x+cos⁡2xsin⁡x.\sin 3x=\sin 2x\cos x+\cos 2x\sin x.

Now substitute

sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x

and

cos⁡2x=1−2sin⁡2x.\cos 2x=1-2\sin^2 x.

This gives

sin⁡3x=2sin⁡xcos⁡2x+sin⁡x(1−2sin⁡2x).\sin 3x = 2\sin x\cos^2 x+\sin x(1-2\sin^2 x).

Since cos⁡2x=1−sin⁡2x\cos^2 x=1-\sin^2 x,

sin⁡3x=2sin⁡x(1−sin⁡2x)+sin⁡x(1−2sin⁡2x).\sin 3x = 2\sin x(1-\sin^2 x)+\sin x(1-2\sin^2 x).

Simplifying,

sin⁡3x=3sin⁡x−4sin⁡3x.\sin 3x=3\sin x-4\sin^3 x.

Cosine Triple angle Formula

Similarly,

cos⁡3x=cos⁡(2x+x).\cos 3x=\cos(2x+x).

Using the angle addition formula,

cos⁡3x=cos⁡2xcos⁡x−sin⁡2xsin⁡x.\cos 3x=\cos 2x\cos x-\sin 2x\sin x.

Substitute

cos⁡2x=2cos⁡2x−1\cos 2x=2\cos^2 x-1

and

sin⁡2x=2sin⁡xcos⁡x.\sin 2x=2\sin x\cos x.

Then

cos⁡3x=(2cos⁡2x−1)cos⁡x−2sin⁡2xcos⁡x.\cos 3x = (2\cos^2 x-1)\cos x-2\sin^2 x\cos x.

Using sin⁡2x=1−cos⁡2x\sin^2 x=1-\cos^2 x,

cos⁡3x=2cos⁡3x−cos⁡x−2(1−cos⁡2x)cos⁡x.\cos 3x = 2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x.

Simplifying,

cos⁡3x=4cos⁡3x−3cos⁡x.\cos 3x=4\cos^3 x-3\cos x.

Tangent Triple angle Formula

Using the tangent addition formula, you get

tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x,\tan 3x = \frac{3\tan x-\tan^3 x}{1-3\tan^2 x},

provided the expressions are defined.

This formula is less commonly memorised, but it can be useful in certain equation solving problems.

Sums of Trigonometric Series

Some problems ask us to evaluate sums such as

sin⁡x+sin⁡2x+sin⁡3x+⋯+sin⁡nx\sin x+\sin 2x+\sin 3x+\cdots+\sin nx

or

cos⁡x+cos⁡2x+cos⁡3x+⋯+cos⁡nx.\cos x+\cos 2x+\cos 3x+\cdots+\cos nx.

There are several ways to approach these sums. A useful method is to multiply by a carefully chosen trigonometric factor so that the sum telescopes.

Sum of Sines

Let

S=sin⁡x+sin⁡2x+⋯+sin⁡nx.S=\sin x+\sin 2x+\cdots+\sin nx.

Multiply both sides by 2sin⁡(x2)2\sin\left(\frac{x}{2}\right).

Using the identity

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B),2\sin A\sin B=\cos(A-B)-\cos(A+B),

we get

2sin⁡(x2)S=∑k=1n2sin⁡(x2)sin⁡(kx).2\sin\left(\frac{x}{2}\right)S = \sum_{k=1}^{n} 2\sin\left(\frac{x}{2}\right)\sin(kx).

Now apply product to sum:

2sin⁡(x2)sin⁡(kx)=cos⁡(kx−x2)−cos⁡(kx+x2).2\sin\left(\frac{x}{2}\right)\sin(kx) = \cos\left(kx-\frac{x}{2}\right) - \cos\left(kx+\frac{x}{2}\right).

So the sum becomes

∑k=1n[cos⁡((k−12)x)−cos⁡((k+12)x)].\sum_{k=1}^{n} \left[ \cos\left(\left(k-\frac{1}{2}\right)x\right) - \cos\left(\left(k+\frac{1}{2}\right)x\right) \right].

Most terms cancel. We are left with

2sin⁡(x2)S=cos⁡(x2)−cos⁡((n+12)x).2\sin\left(\frac{x}{2}\right)S = \cos\left(\frac{x}{2}\right) - \cos\left(\left(n+\frac{1}{2}\right)x\right).

Using sum to product,

cos⁡(x2)−cos⁡((n+12)x)=2sin⁡((n+1)x2)sin⁡(nx2).\cos\left(\frac{x}{2}\right) - \cos\left(\left(n+\frac{1}{2}\right)x\right) = 2\sin\left(\frac{(n+1)x}{2}\right) \sin\left(\frac{nx}{2}\right).

Therefore,

sin⁡x+sin⁡2x+⋯+sin⁡nx=sin⁡(nx2)sin⁡((n+1)x2)sin⁡(x2).\sin x+\sin 2x+\cdots+\sin nx = \frac{ \sin\left(\frac{nx}{2}\right) \sin\left(\frac{(n+1)x}{2}\right) }{ \sin\left(\frac{x}{2}\right) }.

This formula is valid when sin⁡(x2)≠0\sin\left(\frac{x}{2}\right)\neq 0. Special cases where the denominator is zero should be handled separately.

Sum of Cosines

A similar telescoping argument gives

cos⁡x+cos⁡2x+⋯+cos⁡nx=sin⁡(nx2)cos⁡((n+1)x2)sin⁡(x2),\cos x+\cos 2x+\cdots+\cos nx = \frac{ \sin\left(\frac{nx}{2}\right) \cos\left(\frac{(n+1)x}{2}\right) }{ \sin\left(\frac{x}{2}\right) },

again assuming sin⁡(x2)≠0\sin\left(\frac{x}{2}\right)\neq 0.

These two formulas are especially useful when evaluating long trigonometric sums.

Worked Examples

Example 1

Simplify sin⁡5xcos⁡2x\sin 5x\cos 2x.

Use the product to sum identity

sin⁡Acos⁡B=12(sin⁡(A+B)+sin⁡(A−B)).\sin A\cos B = \frac{1}{2}(\sin(A+B)+\sin(A-B)).

With A=5xA=5x and B=2xB=2x,

sin⁡5xcos⁡2x=12(sin⁡7x+sin⁡3x).\sin 5x\cos 2x = \frac{1}{2}(\sin 7x+\sin 3x).

Example 2

Express sin⁡2x\sin^2 x in terms of cos⁡2x\cos 2x.

From the double angle identity,

cos⁡2x=1−2sin⁡2x.\cos 2x=1-2\sin^2 x.

Rearranging,

sin⁡2x=1−cos⁡2x2.\sin^2 x=\frac{1-\cos 2x}{2}.

This identity is useful when reducing powers of trigonometric functions.

Example 3

Solve 4cos⁡3x−3cos⁡x=124\cos^3 x - 3\cos x = \frac{1}{2} for 0∘≤x<360∘0^\circ \le x < 360^\circ.

Recognise the left hand side as the cosine triple angle formula:

4cos⁡3x−3cos⁡x=cos⁡3x.4\cos^3 x - 3\cos x = \cos 3x.

So the equation becomes

cos⁡3x=12.\cos 3x = \frac{1}{2}.

We know that cos⁡θ=12\cos \theta = \frac{1}{2} when θ=60∘+360∘k\theta = 60^\circ + 360^\circ k or θ=300∘+360∘k\theta = 300^\circ + 360^\circ k.

So

3x=60∘+360∘kor3x=300∘+360∘k.3x = 60^\circ + 360^\circ k \quad\text{or}\quad 3x = 300^\circ + 360^\circ k.

From the first family: x=20∘,140∘,260∘x = 20^\circ, 140^\circ, 260^\circ.

From the second family: x=100∘,220∘,340∘x = 100^\circ, 220^\circ, 340^\circ.

The six solutions in [0∘,360∘)[0^\circ, 360^\circ) are

x=20∘,100∘,140∘,220∘,260∘,340∘.x = 20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ, 340^\circ.

Example 4

Evaluate

sin⁡10∘+sin⁡20∘+⋯+sin⁡80∘.\sin 10^\circ+\sin 20^\circ+\cdots+\sin 80^\circ.

This is a sine series with first angle 10∘10^\circ, common step 10∘10^\circ, and last angle 80∘80^\circ.

Using the sum formula with x=10∘x=10^\circ and n=8n=8,

sin⁡10∘+sin⁡20∘+⋯+sin⁡80∘=sin⁡40∘sin⁡45∘sin⁡5∘.\sin 10^\circ+\sin 20^\circ+\cdots+\sin 80^\circ = \frac{ \sin 40^\circ\sin 45^\circ }{ \sin 5^\circ }.

This is already exact form.

Example 5

Show that

sin⁡x+sin⁡3x+sin⁡5x+⋯+sin⁡(2n−1)x=sin⁡2(nx)sin⁡x.\sin x+\sin 3x+\sin 5x+\cdots+\sin(2n-1)x = \frac{\sin^2(nx)}{\sin x}.

Let

S=sin⁡x+sin⁡3x+⋯+sin⁡(2n−1)x.S=\sin x+\sin 3x+\cdots+\sin(2n-1)x.

Multiply by 2sin⁡x2\sin x:

2sin⁡xS=∑k=1n2sin⁡xsin⁡((2k−1)x).2\sin x S = \sum_{k=1}^{n}2\sin x\sin((2k-1)x).

Using

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B),2\sin A\sin B=\cos(A-B)-\cos(A+B),

we get

2sin⁡xsin⁡((2k−1)x)=cos⁡((2k−2)x)−cos⁡(2kx).2\sin x\sin((2k-1)x) = \cos((2k-2)x)-\cos(2kx).

Therefore the sum telescopes:

2sin⁡xS=(1−cos⁡2x)+(cos⁡2x−cos⁡4x)+⋯+(cos⁡(2n−2)x−cos⁡2nx).2\sin xS = (1-\cos 2x)+(\cos 2x-\cos 4x)+\cdots+ (\cos(2n-2)x-\cos 2nx).

All middle terms cancel, leaving

2sin⁡xS=1−cos⁡2nx.2\sin xS=1-\cos 2nx.

Since

1−cos⁡2nx=2sin⁡2(nx),1-\cos 2nx=2\sin^2(nx),

we obtain

S=sin⁡2(nx)sin⁡x.S= \frac{\sin^2(nx)}{\sin x}.

Key Contest Insight

Most trigonometric problems become easier after changing the form of the expression.

Use:

  • Product to sum when products appear
  • Sum to product when sums need to be factored
  • Double angle identities when squares appear
  • Auxiliary angle when an expression has the form asin⁡x+bcos⁡xa\sin x+b\cos x
  • Triple angle identities when cubes or 3x3x appear
  • Telescoping when a long trigonometric series appears

The main challenge is recognising which identity makes the expression simpler.

Strategy Tips

  • Do not expand blindly. Look for structure first.
  • Products such as sin⁡Acos⁡B\sin A\cos B often suggest product to sum.
  • Squares such as sin⁡2x\sin^2 x and cos⁡2x\cos^2 x often suggest double angle identities.
  • For expressions like asin⁡x+bcos⁡xa\sin x+b\cos x, use the auxiliary angle method to find maximum and minimum values quickly.
  • Cubes such as sin⁡3x\sin^3 x and cos⁡3x\cos^3 x often suggest triple angle identities.
  • Long sums with evenly spaced angles often telescope after multiplying by a suitable sine factor.
  • Always check whether special cases make denominators zero.
  • In degree notation, keep track of whether angles are in degrees or radians.

Common Pitfalls

  • Mixing degrees and radians.
  • Forgetting the negative sign in
cos⁡X−cos⁡Y=−2sin⁡(X+Y2)sin⁡(X−Y2).\cos X-\cos Y = -2\sin\left(\frac{X+Y}{2}\right) \sin\left(\frac{X-Y}{2}\right).
  • Using a formula without checking whether division by zero occurs.
  • Applying a product to sum identity with the wrong signs.
  • Forgetting that triple angle identities contain cubic terms.
  • Expanding trigonometric expressions when a direct identity would simplify them.

Practice Problems

StatusSourceProblem NameDifficultyTags
CustomNormal
Show TagsProduct-to-Sum, Trigonometry
CustomHard
Show TagsSum-to-Product, Trigonometry
CustomHard
Show TagsTelescoping Series, Trigonometry
Berkeley Math Circle: Monthly Contest 8Hard
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MathNetHard
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Harvard-MIT Mathematics TournamentHard
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Harvard-MIT Mathematics TournamentHard
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Harvard-MIT Mathematics TournamentHard
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Harvard-MIT Math TournamentHard
Show TagsSingle-variable, Trigonometric functions
Harvard-MIT Math TournamentHard
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Harvard-MIT Math TournamentHard
Show TagsApplications, Single-variable, Trigonometric functions
Harvard-MIT Math TournamentHard
Show TagsApplications, Derivatives, Trigonometric functions
Harvard-MIT Math TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsODEs, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsApplications, Single-variable, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsSingle-variable, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, ODEs, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsDerivatives, Trigonometric functions
Harvard-MIT November TournamentHard
Show TagsFunctions, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsDerivatives, Limits, Trigonometric functions
Berkeley Math Circle: Monthly Contest 8Hard
Show TagsTrigonometric functions
MathNetHard
Show TagsFunctions, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Math TournamentHard
Show TagsSingle-variable, Trigonometric functions
Harvard-MIT Math TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Math TournamentHard
Show TagsApplications, Single-variable, Trigonometric functions
Harvard-MIT Math TournamentHard
Show TagsApplications, Derivatives, Trigonometric functions
Harvard-MIT Math TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsODEs, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsApplications, Single-variable, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsTrigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsSingle-variable, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsLimits, ODEs, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsDerivatives, Trigonometric functions
Harvard-MIT November TournamentHard
Show TagsFunctions, Trigonometric functions
Harvard-MIT Mathematics TournamentHard
Show TagsDerivatives, Limits, Trigonometric functions
HMMT November 2012Hard
Show TagsTrigonometric functions
HMMT FebruaryHard
Show TagsFunctions, Trigonometric functions
HMMT February 2016Hard
Show TagsSingle-variable, Trigonometric functions

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