Academy
Formulas/maths/M3 De Moivre/Expanding cos nθ and sin nθ

Expanding cos nθ and sin nθ

Expand the RHS using the binomial theorem, then separate real and imaginary parts. Real part gives cos nθ as a polynomial in cos θ; imaginary part gives sin nθ. For example, cos 3θ = 4cos³θ − 3cos θ and sin 3θ = 3sin θ − 4sin³θ both fall out of (cis θ)³.
Derivation

De Moivre's theorem turns the problem of expanding cos⁡nθ\cos n\theta and sin⁡nθ\sin n\theta into a binomial expansion. The powers of ii separate the real and imaginary parts automatically.

The Method

By De Moivre's theorem:

cos⁡nθ+isin⁡nθ=(cos⁡θ+isin⁡θ)n\cos n\theta + i\sin n\theta = (\cos\theta + i\sin\theta)^n

Expand the right side using the binomial theorem, use the cycle i2=−1i^2=-1, i3=−ii^3=-i, i4=1i^4=1, then separate real and imaginary parts.

Case n=3n = 3

(cos⁡θ+isin⁡θ)3=cos⁡3θ+3cos⁡2θ(isin⁡θ)+3cos⁡θ(isin⁡θ)2+(isin⁡θ)3(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3 =cos⁡3θ+3icos⁡2θsin⁡θ−3cos⁡θsin⁡2θ−isin⁡3θ= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta

Real part:

cos⁡3θ=cos⁡3θ−3cos⁡θsin⁡2θ=cos⁡3θ−3cos⁡θ(1−cos⁡2θ)\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = \cos^3\theta - 3\cos\theta(1-\cos^2\theta) cos⁡3θ=4cos⁡3θ−3cos⁡θ\boxed{\cos 3\theta = 4\cos^3\theta - 3\cos\theta}

Imaginary part:

sin⁡3θ=3cos⁡2θsin⁡θ−sin⁡3θ=3(1−sin⁡2θ)sin⁡θ−sin⁡3θ\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta = 3(1-\sin^2\theta)\sin\theta - \sin^3\theta sin⁡3θ=3sin⁡θ−4sin⁡3θ\boxed{\sin 3\theta = 3\sin\theta - 4\sin^3\theta}

Case n=4n = 4

(cos⁡θ+isin⁡θ)4=cos⁡4θ+4icos⁡3θsin⁡θ−6cos⁡2θsin⁡2θ−4icos⁡θsin⁡3θ+sin⁡4θ(\cos\theta + i\sin\theta)^4 = \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta

Real part:

cos⁡4θ=cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta:

=cos⁡4θ−6cos⁡2θ(1−cos⁡2θ)+(1−cos⁡2θ)2=8cos⁡4θ−8cos⁡2θ+1= \cos^4\theta - 6\cos^2\theta(1-\cos^2\theta) + (1-\cos^2\theta)^2 = 8\cos^4\theta - 8\cos^2\theta + 1

Imaginary part:

sin⁡4θ=4cos⁡3θsin⁡θ−4cos⁡θsin⁡3θ=4sin⁡θcos⁡θ(cos⁡2θ−sin⁡2θ)=2sin⁡2θcos⁡2θ\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta = 4\sin\theta\cos\theta(\cos^2\theta - \sin^2\theta) = 2\sin 2\theta\cos 2\theta

(Consistent with sin⁡4θ=sin⁡2(2θ)=2sin⁡2θcos⁡2θ\sin 4\theta = \sin 2(2\theta) = 2\sin 2\theta\cos 2\theta.) ✓\checkmark

The General Pattern

For general nn, the binomial expansion of (cos⁡θ+isin⁡θ)n(\cos\theta + i\sin\theta)^n gives:

  • cos⁡nθ\cos n\theta = sum of even-powered terms (i0=1i^0=1, i2=−1i^2=-1, i4=1i^4=1, ...) — a polynomial in cos⁡θ\cos\theta of degree nn
  • sin⁡nθ\sin n\theta = sum of odd-powered terms (i1=ii^1=i, i3=−ii^3=-i, ...) — always has a sin⁡θ\sin\theta factor, multiplied by a polynomial in cos⁡θ\cos\theta of degree n−1n-1

This gives the Chebyshev-type expansions: cos⁡nθ\cos n\theta as a degree-nn polynomial in cos⁡θ\cos\theta, which is the foundation of Chebyshev polynomials Tn(cos⁡θ)=cos⁡nθT_n(\cos\theta) = \cos n\theta.