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Multiple-Angle Trigonometric Formulas

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1Introduction

Multiple-angle formulas allow us to express functions like sin⁡(nx)\sin(nx) and cos⁡(nx)\cos(nx) in terms of powers of sin⁡(x)\sin(x) and cos⁡(x)\cos(x). These identities are useful in algebraic simplification, solving equations, and signal analysis.

2Common Multiple-Angle Identities

2.1Sine

  • sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x)
  • sin⁡(3x)=3sin⁡(x)−4sin⁡3(x)\sin(3x) = 3\sin(x) - 4\sin^3(x)
  • sin⁡(4x)=4sin⁡(x)cos⁡(x)(1−2sin⁡2(x))\sin(4x) = 4\sin(x)\cos(x)(1 - 2\sin^2(x))

2.2Cosine

  • cos⁡(2x)=cos⁡2(x)−sin⁡2(x)\cos(2x) = \cos^2(x) - \sin^2(x)
  • cos⁡(3x)=4cos⁡3(x)−3cos⁡(x)\cos(3x) = 4\cos^3(x) - 3\cos(x)
  • cos⁡(4x)=8cos⁡4(x)−8cos⁡2(x)+1\cos(4x) = 8\cos^4(x) - 8\cos^2(x) + 1

These can be derived using angle addition formulas or De Moivre’s Theorem.

3General Formula for sin⁡(nx)\sin(nx)?

For any positive odd integer nn, sin⁡(nx)\sin(nx) can be written as a sum of powers of sin⁡(x)\sin(x) and cos⁡(x)\cos(x):

sin⁡(nx)=∑k=0⌊n/2⌋(−1)k(n2k+1)cos⁡n−(2k+1)(x)sin⁡2k+1(x) \sin(nx) = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{2k+1} \cos^{n - (2k + 1)}(x) \sin^{2k + 1}(x)

Important

This identity selects only odd-power terms in the expansion of sin⁡(nx)\sin(nx), yielding a pure polynomial in sin⁡(x)\sin(x) when expressed using the Pythagorean identity.

4Deriving sin⁡(3x)\sin(3x) Example (also see Trigonometric identities: Problem 01)

Start with:

sin⁡(3x)=sin⁡(2x+x)=sin⁡(2x)cos⁡(x)+cos⁡(2x)sin⁡(x) \sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x)

Use: + sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x) + cos⁡(2x)=cos⁡2(x)−sin⁡2(x)\cos(2x) = \cos^2(x) - \sin^2(x)

Substitute:

sin⁡(3x)=(2sin⁡(x)cos⁡(x))cos⁡(x)+(cos⁡2(x)−sin⁡2(x))sin⁡(x)=2sin⁡(x)cos⁡2(x)+cos⁡2(x)sin⁡(x)−sin⁡3(x)=(3cos⁡2(x)−sin⁡2(x))sin⁡(x) \begin{align*} \sin(3x) &= (2\sin(x)\cos(x))\cos(x) + (\cos^2(x) - \sin^2(x))\sin(x) \\ &= 2\sin(x)\cos^2(x) + \cos^2(x)\sin(x) - \sin^3(x) \\ &= (3\cos^2(x) - \sin^2(x))\sin(x) \end{align*}

Now, using the Pythagorean identity cos⁡2(x)=1−sin⁡2(x)\cos^2(x) = 1 - \sin^2(x), we can rewrite it purely in terms of sin⁡(x)\sin(x):

sin⁡(3x)=(3(1−sin⁡2(x))−sin⁡2(x))sin⁡(x)=(3−4sin⁡2(x))sin⁡(x)=3sin⁡(x)−4sin⁡3(x) \begin{align*} \sin(3x) &= (3(1 - \sin^2(x)) - \sin^2(x))\sin(x) \\ &= (3 - 4\sin^2(x))\sin(x) \\ &= 3\sin(x) - 4\sin^3(x) \end{align*}

This is a more compact polynomial form of sin⁡(3x)\sin(3x), often more convenient in algebraic manipulation or solving equations.

5Generalized Multiple Angle Formulae

These identities express sin⁡(nθ)\sin(n\theta), cos⁡(nθ)\cos(n\theta), and tan⁡(nθ)\tan(n\theta) using combinations of sin⁡(kθ)\sin(k\theta) and cos⁡(kθ)\cos(k\theta):

sin⁡(nθ)=∑k=0n(−1)k(nk)cos⁡n−k(θ)sin⁡k(θ)⋅sin⁡[12(n−k)π]=2sin⁡[(n−1)θ]cos⁡(θ)−sin⁡[(n−2)θ] \begin{align*} &\sin(n\theta) \\ &= \boxed{\sum_{k=0}^{n} (-1)^k \binom{n}{k} \cos^{n-k}(\theta)\sin^k(\theta) \cdot \sin\left[\tfrac{1}{2}(n - k)\pi\right]} \\ &= \boxed{2\sin\left[(n - 1)\theta\right]\cos(\theta) - \sin\left[(n - 2)\theta\right]} \end{align*}

cos⁡(nθ)=∑k=0n(−1)k(nk)cos⁡n−k(θ)sin⁡k(θ)⋅cos⁡[12(n−k)π]=2cos⁡[(n−1)θ]cos⁡(θ)−cos⁡[(n−2)θ] \begin{align*} &\cos(n\theta) \\ &= \boxed{\sum_{k=0}^{n} (-1)^k \binom{n}{k} \cos^{n-k}(\theta)\sin^k(\theta) \cdot \cos\left[\tfrac{1}{2}(n - k)\pi\right]} \\ &= \boxed{2\cos\left[(n - 1)\theta\right]\cos(\theta) - \cos\left[(n - 2)\theta\right]} \end{align*}

tan⁡(nθ)=sin⁡(nθ)cos⁡(nθ)=tan⁡[(n−1)θ]+tan⁡(θ)1−tan⁡[(n−1)θ]tan⁡(θ) \begin{align*} &\tan(n\theta) \\ &= \frac{\sin(n\theta)}{\cos(n\theta)} \\ &= \boxed{\frac{\tan\left[(n - 1)\theta\right] + \tan(\theta)}{1 - \tan\left[(n - 1)\theta\right]\tan(\theta)}} \end{align*}

These identities are useful in trigonometric polynomial expansion, harmonic analysis, and symbolic computation involving angular multiples.

So, for example, cos⁡(4x)\cos(4x) could be written in the following ways: + cos⁡(4x)\cos(4x) + 8cos⁡4(x)−8cos⁡2(x)+18\cos^4(x) - 8\cos^2(x) + 1

derived from cos⁡(nθ)=∑k=0n(−1)k(nk)cos⁡n−k(θ)sin⁡k(θ)⋅cos⁡[12(n−k)π]\cos(n\theta) = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \cos^{n-k}(\theta)\sin^k(\theta) \cdot \cos\left[\tfrac{1}{2}(n - k)\pi\right] + 2cos⁡(3x)cos⁡(x)−cos⁡(2x)2\cos(3x)\cos(x) - \cos(2x) derived from cos⁡(nθ)=2cos⁡[(n−1)θ]cos⁡(θ)−cos⁡[(n−2)θ]\cos(n\theta) = 2\cos\left[(n - 1)\theta\right]\cos(\theta) - \cos\left[(n - 2)\theta\right]

5.1Complete example of cos⁡(4x)=8cos⁡4(x)−8cos⁡2(x)+1\cos(4x) = 8\cos^4(x) - 8\cos^2(x) + 1

Using the sum formula for cosine:

cos⁡(4x)=∑k=04(−1)k(4k)cos⁡4−k(x)sin⁡k(x)⋅cos⁡[12(4−k)π] \cos(4x) = \sum_{k=0}^4 (-1)^k \binom{4}{k} \cos^{4-k}(x) \sin^k(x) \cdot \cos\left[\tfrac{1}{2}(4-k)\pi\right]

Evaluate the cosine factor cos⁡[12(4−k)π]\cos\left[\tfrac{1}{2}(4 - k)\pi\right] for each kk:

kk 4−k4 - k 12(4−k)π\tfrac{1}{2}(4 - k)\pi cos⁡[12(4−k)π]\cos\left[\tfrac{1}{2}(4 - k)\pi\right]
00 44 2π2\pi 11
11 33 3π2\frac{3\pi}{2} 00
22 22 π\pi −1-1
33 11 π2\frac{\pi}{2} 00
44 00 00 11

Only terms where the cosine factor is nonzero contribute, so:

cos⁡(4x)=(−1)0(40)cos⁡4(x)sin⁡0(x)⋅1+(−1)2(42)cos⁡2(x)sin⁡2(x)⋅(−1)+(−1)4(44)cos⁡0(x)sin⁡4(x)⋅1=cos⁡4(x)−6cos⁡2(x)sin⁡2(x)+sin⁡4(x) \begin{aligned} \cos(4x) &= (-1)^0 \binom{4}{0} \cos^4(x) \sin^0(x) \cdot 1 \\ &\quad + (-1)^2 \binom{4}{2} \cos^{2}(x) \sin^{2}(x) \cdot (-1) \\ &\quad + (-1)^4 \binom{4}{4} \cos^{0}(x) \sin^{4}(x) \cdot 1 \\ &= \cos^4(x) - 6 \cos^{2}(x) \sin^{2}(x) + \sin^{4}(x) \end{aligned}

Use the Pythagorean identity sin⁡2(x)=1−cos⁡2(x)\sin^2(x) = 1 - \cos^2(x) and rewrite sin⁡4(x)\sin^4(x):

sin⁡4(x)=(1−cos⁡2(x))2=1−2cos⁡2(x)+cos⁡4(x) \sin^4(x) = (1 - \cos^2(x))^2 = 1 - 2 \cos^2(x) + \cos^4(x)

Substitute into the expression:

cos⁡(4x)=cos⁡4(x)−6cos⁡2(x)(1−cos⁡2(x))+1−2cos⁡2(x)+cos⁡4(x)=cos⁡4(x)−6cos⁡2(x)+6cos⁡4(x)+1−2cos⁡2(x)+cos⁡4(x) \begin{aligned} \cos(4x) &= \cos^4(x) - 6 \cos^{2}(x) (1 - \cos^2(x)) + 1 - 2 \cos^2(x) + \cos^4(x) \\ &= \cos^4(x) - 6 \cos^{2}(x) + 6 \cos^4(x) + 1 - 2 \cos^2(x) + \cos^4(x) \end{aligned}

Combine like terms:

cos⁡(4x)=(cos⁡4(x)+6cos⁡4(x)+cos⁡4(x))+(−6cos⁡2(x)−2cos⁡2(x))+1=8cos⁡4(x)−8cos⁡2(x)+1 \begin{aligned} \cos(4x) &= ( \cos^4(x) + 6 \cos^4(x) + \cos^4(x) ) + ( -6 \cos^{2}(x) - 2 \cos^{2}(x) ) + 1 \\ &= 8 \cos^4(x) - 8 \cos^{2}(x) + 1 \end{aligned}

Thus,

cos⁡(4x)=8cos⁡4(x)−8cos⁡2(x)+1 \boxed{ \cos(4x) = 8 \cos^{4}(x) - 8 \cos^{2}(x) + 1 }

6General Form via De Moivre’s Theorem

Using complex numbers:

(cos⁡x+isin⁡x)n=cos⁡(nx)+isin⁡(nx) (\cos x + i\sin x)^n = \cos(nx) + i\sin(nx)

Expanding with the binomial theorem gives a way to derive general formulas for sin⁡(nx)\sin(nx) and cos⁡(nx)\cos(nx).

7Applications

  • Trigonometric simplification
  • Fourier analysis
  • Vibrations and wave modeling
  • Engineering and physics problems

8See Also

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