Applications to Trig Identities & Series
The substitution $z=\cos\theta+i\sin\theta$ turns De Moivre's theorem into a machine for generating multiple-angle formulae, rewriting powers of sine and cosine, and summing trigonometric series in closed form.
The key substitution
Let $z=\cos\theta+i\sin\theta$, so $|z|=1$. Since $\dfrac{1}{z}=z^{-1}=\cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta$ (the reciprocal of a modulus-$1$ number is its conjugate), adding and subtracting $z$ and $\dfrac1z$ isolates $\cos\theta$ and $\sin\theta$ exactly:
Applying De Moivre's theorem to $z^n$ and $z^{-n}$ gives the same pair of identities one level up, at angle $n\theta$:
These four identities are the entire toolkit for this lesson: every technique below is just the binomial theorem applied to one of the first pair, then reorganised using the second pair.
Multiple angles in terms of powers
To express $\cos n\theta$ or $\sin n\theta$ in terms of powers of $\cos\theta$ and $\sin\theta$, expand $(\cos\theta+i\sin\theta)^n$ using the binomial theorem, then compare with $\cos n\theta+i\sin n\theta$ (De Moivre) — the real part of the expansion must equal $\cos n\theta$, and the imaginary part must equal $\sin n\theta$.
By De Moivre, $\cos3\theta+i\sin3\theta=(\cos\theta+i\sin\theta)^3$. Expand the right-hand side by the binomial theorem, writing $c=\cos\theta,\,s=\sin\theta$ for brevity:
Grouping real and imaginary parts: $(c^3-3cs^2) + i(3c^2s-s^3)$.
Comparing with $\cos3\theta+i\sin3\theta$:
$\cos3\theta = c^3-3cs^2 = \cos^3\theta - 3\cos\theta(1-\cos^2\theta) = 4\cos^3\theta-3\cos\theta$ (using $\sin^2\theta=1-\cos^2\theta$)
$\sin3\theta = 3c^2s-s^3 = 3(1-\sin^2\theta)\sin\theta-\sin^3\theta = 3\sin\theta-4\sin^3\theta$ (using $\cos^2\theta=1-\sin^2\theta$)
After the binomial expansion, always substitute $i^2=-1$ (giving real terms a sign flip) and $i^3=-i$ before separating real and imaginary parts — the odd powers of $i$ (from odd powers of $\sin\theta$) contribute to the imaginary part, the even powers of $i$ contribute to the real part. Getting this parity right is the crux of the whole method.
Powers in terms of multiple angles (the reverse direction)
To go the other way — expressing $\cos^n\theta$ or $\sin^n\theta$ in terms of $\cos k\theta$ or $\sin k\theta$ — raise $z+\dfrac1z=2\cos\theta$ (or $z-\dfrac1z=2i\sin\theta$) to the $n$th power using the binomial theorem, then regroup the resulting terms into conjugate pairs $z^k+\dfrac{1}{z^k}=2\cos k\theta$.
Start from $z+\dfrac1z=2\cos\theta$ and raise both sides to the power $4$:
Expanding the left-hand side by the binomial theorem:
Using $z^4+\dfrac{1}{z^4}=2\cos4\theta$ and $z^2+\dfrac{1}{z^2}=2\cos2\theta$:
For $\sin^n\theta$ with $n$ odd, raising $z-\dfrac1z=2i\sin\theta$ to an odd power produces a leftover factor of $i$ on the left that must be divided out. For instance, $(z-\tfrac1z)^3=8i^3\sin^3\theta=-8i\sin^3\theta$, while expanding the left directly gives $\left(z^3-\dfrac{1}{z^3}\right)-3\left(z-\dfrac1z\right)=2i\sin3\theta-6i\sin\theta$. Equating and dividing every term by $2i$ gives $\sin^3\theta=\dfrac{3\sin\theta-\sin3\theta}{4}$ — always check the power of $i$ on each side matches before dividing it out.
Summing trigonometric series
Sums like $\cos\theta+\cos2\theta+\cdots+\cos n\theta$ can be evaluated exactly by recognising them as the real (or imaginary) part of a geometric series in $z=\cos\theta+i\sin\theta$, which can then be summed using the standard geometric series formula.
Let $z=\cos\theta+i\sin\theta$. By De Moivre, $z^k=\cos k\theta+i\sin k\theta$, so the required sum is the imaginary part of $S=z+z^2+\cdots+z^n$, a geometric series with first term $z$, ratio $z$, and $n$ terms:
Factor top and bottom symmetrically to expose the modulus-argument structure: $z^n-1 = z^{n/2}\left(z^{n/2}-z^{-n/2}\right) = z^{n/2}\cdot2i\sin\dfrac{n\theta}{2}$, and $z-1=z^{1/2}\left(z^{1/2}-z^{-1/2}\right)=z^{1/2}\cdot2i\sin\dfrac{\theta}{2}$.
Since $z^{\frac{n+1}{2}}=\cos\dfrac{(n+1)\theta}{2}+i\sin\dfrac{(n+1)\theta}{2}$, taking the imaginary part of $S$ gives exactly the claimed result.
Taking the real part of $S$ instead of the imaginary part gives the corresponding formula for $\cos\theta+\cos2\theta+\cdots+\cos n\theta$ — the two are easy to mix up. Always state clearly at the start which part of $S$ you need before finishing the algebra.
Exam-style questions
Question 1 6 marks
Use De Moivre's theorem to show that $\sin3\theta=3\sin\theta-4\sin^3\theta$.
$\cos3\theta+i\sin3\theta=(\cos\theta+i\sin\theta)^3$ (B1)
$=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta$ (M1 binomial expansion)
Imaginary part: $\sin3\theta = 3\cos^2\theta\sin\theta-\sin^3\theta$ (M1 compare imaginary parts)
$= 3(1-\sin^2\theta)\sin\theta-\sin^3\theta = 3\sin\theta-3\sin^3\theta-\sin^3\theta = 3\sin\theta-4\sin^3\theta$ (M1 use cos²θ = 1 − sin²θ, A1 correct simplification, A1 final form)
Question 2 7 marks
(a) Show that $\cos^4\theta=\dfrac18(\cos4\theta+4\cos2\theta+3)$. (b) Hence find the exact value of $\displaystyle\int_0^{\pi/2}\cos^4\theta\,\mathrm d\theta$.
(a) As derived in the lesson: $\left(z+\tfrac1z\right)^4=16\cos^4\theta$, expand to get $z^4+4z^2+6+\tfrac{4}{z^2}+\tfrac1{z^4}$, group into $2\cos4\theta+8\cos2\theta+6$, divide by $16$ (M1, M1, A1).
(b) $\displaystyle\int_0^{\pi/2}\cos^4\theta\,d\theta = \frac18\int_0^{\pi/2}(\cos4\theta+4\cos2\theta+3)\,d\theta = \frac18\left[\frac{\sin4\theta}{4}+2\sin2\theta+3\theta\right]_0^{\pi/2}$ (M1, A1 integration)
At $\theta=\dfrac{\pi}{2}$: $\sin2\pi=0$, $\sin\pi=0$, so the bracket $=\dfrac{3\pi}{2}$; at $\theta=0$ the bracket is $0$ (M1).
$\displaystyle\int_0^{\pi/2}\cos^4\theta\,d\theta = \frac18\times\frac{3\pi}{2} = \frac{3\pi}{16}$ (A1)