QCE Specialist Mathematics - Unit 3 - Mathematical induction and trigonometric proofs

Trigonometric Proofs Using De Moivre's Theorem | QCE Specialist Mathematics

Learn how Specialist Mathematics uses De Moivre's theorem and binomial expansion to prove multi-angle trigonometric identities.

Updated 2026-05-18 - 4 min read

This part of Specialist connects complex numbers with trigonometric identities. The key idea is that the same expression can be expanded in two different ways.

De Moivre's theorem says:

$ (\cos x+i\sin x)^n=\cos(nx)+i\sin(nx). $

The left side can be expanded using the binomial theorem. The right side already separates into a real part and an imaginary part.

Equating real and imaginary parts

For $n=3$:

$ (\cos x+i\sin x)^3=\cos(3x)+i\sin(3x). $

Expand the left side:

$ \cos^3x+3i\cos^2x\sin x+3i^2\cos x\sin^2x+i^3\sin^3x. $

Since $i^2=-1$ and $i^3=-i$:

$ =(\cos^3x-3\cos x\sin^2x)+i(3\cos^2x\sin x-\sin^3x). $

Now match real parts:

$ \cos(3x)=\cos^3x-3\cos x\sin^2x. $

Using $\sin^2x=1-\cos^2x$:

$ \cos(3x)=4\cos^3x-3\cos x. $

Match imaginary parts:

$ \sin(3x)=3\cos^2x\sin x-\sin^3x. $

Using $\cos^2x=1-\sin^2x$:

$ \sin(3x)=3\sin x-4\sin^3x. $

This is the whole trick: the left side gives a binomial expansion involving powers of $\cos x$ and $i\sin x$, while the right side gives $\cos(nx)+i\sin(nx)$. Even powers of $i$ contribute to the real part; odd powers of $i$ contribute to the imaginary part.

For reference:

$ i^0=1,\quad i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1. $

Tracking that cycle prevents sign errors when the expansion becomes longer.

Up to $4x$

QCAA only requires identities up to angles of $4x$. The process is the same, but $n=4$ produces more terms:

$ (\cos x+i\sin x)^4=\cos(4x)+i\sin(4x). $

For a real-part identity, collect the terms without $i$. For an imaginary-part identity, collect the coefficient of $i$.

For $n=4$, the real part comes from the terms with $i^0$, $i^2$ and $i^4$:

$ \cos(4x)=\cos^4x-6\cos^2x\sin^2x+\sin^4x. $

The imaginary part comes from the terms with $i^1$ and $i^3$:

$ \sin(4x)=4\cos^3x\sin x-4\cos x\sin^3x. $

Those are not formulas to memorise blindly; they are outputs of the expansion method. If you forget one, rebuild it from De Moivre's theorem.

Quick check

Exam-style workflow

For triple-angle or quadruple-angle proofs, keep the process mechanical:

1. Write $(\cos x+i\sin x)^n=\cos(nx)+i\sin(nx)$.
2. Expand the left-hand side with the binomial theorem.
3. Replace powers of $i$ using $i^2=-1$, $i^3=-i$ and $i^4=1$.
4. Collect the real terms and the coefficient of $i$.
5. Equate real parts for cosine identities and imaginary parts for sine identities.

Only after that should you simplify with $\sin^2x+\cos^2x=1$. This keeps the proof from turning into a pile of algebra with no clear direction.

Do not skip the phrase "equating real parts" or "equating imaginary parts". It is the justification for moving from a complex equation to a real trigonometric identity. If the final identity is written in powers of only $\cos x$ or only $\sin x$, use $\sin^2x+\cos^2x=1$ at the end to remove the unwanted function.

A complete $4x$ identity build

Start from:

$ (\cos x+i\sin x)^4=\cos(4x)+i\sin(4x). $

The binomial expansion is:

$ \cos^4x+4i\cos^3x\sin x+6i^2\cos^2x\sin^2x+4i^3\cos x\sin^3x+i^4\sin^4x. $

Using $i^2=-1$, $i^3=-i$ and $i^4=1$, the real part is:

$ \cos^4x-6\cos^2x\sin^2x+\sin^4x. $

The imaginary part is:

$ 4\cos^3x\sin x-4\cos x\sin^3x. $

So:

$ \cos(4x)=\cos^4x-6\cos^2x\sin^2x+\sin^4x $

and:

$ \sin(4x)=4\cos^3x\sin x-4\cos x\sin^3x. $

If a question wants $\cos(4x)$ only in terms of $\cos x$, replace $\sin^2x$ with $1-\cos^2x$ and simplify. If it wants only sine, replace $\cos^2x$ instead.

For example:

$ \cos(4x)=\cos^4x-6\cos^2x(1-\cos^2x)+(1-\cos^2x)^2. $

After expanding and collecting:

$ \cos(4x)=8\cos^4x-8\cos^2x+1. $

The same real-part identity can also be written in terms of sine:

$ \cos(4x)=1-8\sin^2x+8\sin^4x. $

For sine, factor the $4x$ expression before simplifying:

$ \sin(4x)=4\sin x\cos x(\cos^2x-\sin^2x). $

Since $\cos^2x-\sin^2x=\cos2x$, this also becomes:

$ \sin(4x)=2\sin2x\cos2x. $

That links the De Moivre proof back to the double-angle identities from Mathematical Methods. In exam work, this is useful because one identity can often be transformed into another acceptable form.

Using the identities after proving them

The identities are not only proof exercises. They can simplify trigonometric equations. For example, from:

$ \cos(3x)=4\cos^3x-3\cos x, $

an equation involving $4u^3-3u$ can be recognised as a triple-angle expression with $u=\cos x$. This can be much cleaner than solving a cubic directly.

Similarly:

$ \sin(3x)=3\sin x-4\sin^3x $

can turn a cubic in $\sin x$ into a single sine of a multiple angle. The important exam habit is to state the substitution clearly. If $u=\cos x$, then the allowed values of $u$ must satisfy $-1\le u\le1$ because it came from a cosine value.

When identities are used inside an equation, do not forget the general-solution step. Solving $\cos(3x)=\frac12$ gives:

$ 3x=2k\pi\pm\frac{\pi}{3}, $

so:

$ x=\frac{2k\pi}{3}\pm\frac{\pi}{9}. $

The De Moivre identity helps compress the expression, but the trigonometric equation still needs all solutions in the requested domain.

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax Algebra and Trigonometry 2e: Polar Form of Complex Numbers
• LibreTexts: Principle of Mathematical Induction