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Module 13 · L09 of 12 ~40 min ⚡ +90 XP available

Binomial Expansion via $(\cos\theta + i\sin\theta)^n$

De Moivre's theorem tells us $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$. The binomial theorem tells us how to expand the left-hand side in powers of $\cos\theta$ and $\sin\theta$. Set the two equal, separate real and imaginary parts, and you have machine-built identities for $\cos n\theta$ and $\sin n\theta$. This is the engine behind every multiple-angle formula in the syllabus.

Today's hook — Without using a double-angle table: expand $(\cos\theta + i\sin\theta)^2$ using the binomial theorem (i.e. $(a+b)^2 = a^2 + 2ab + b^2$). Set it equal to $\cos 2\theta + i\sin 2\theta$. Equate real parts. Equate imaginary parts. What two famous identities have you just derived?

0/5QUESTS

Recall — your gut answer first

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Write out the binomial expansion of $(a+b)^4$ in full (five terms). Then state de Moivre's theorem in one line. Before checking — predict what happens to the real and imaginary parts when you let $a = \cos\theta$ and $b = i\sin\theta$. Where do you expect even powers of $i$ to live?

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The two moves for multiple-angle identities

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Every multiple-angle derivation rewards two habits: expand $(\cos\theta + i\sin\theta)^n$ binomially with $a = \cos\theta$, $b = i\sin\theta$, then collect by powers of $i$ so real terms (even powers of $i$) and imaginary terms (odd powers of $i$) are visible. De Moivre gives the answer on the other side; equating components delivers $\cos n\theta$ and $\sin n\theta$ for free.

The expand-collect-equate flow: (1) write $(\cos\theta + i\sin\theta)^n = \sum_{k=0}^n \binom{n}{k}\cos^{\,n-k}\theta\,(i\sin\theta)^k$, (2) use $i^0=1,\;i^1=i,\;i^2=-1,\;i^3=-i,\;i^4=1$ to split real and imaginary, (3) match against $\cos n\theta + i\sin n\theta$.

$\cos n\theta = \mathrm{Re}\big((\cos\theta + i\sin\theta)^n\big)$ · $\sin n\theta = \mathrm{Im}\big((\cos\theta + i\sin\theta)^n\big)$

$(\cos\theta + i\sin\theta)^n = \sum_{k=0}^{n} \binom{n}{k}\cos^{\,n-k}\theta\,(i\sin\theta)^k$

Track $i^k$ carefully

$i^k$ cycles: $1, i, -1, -i$. Even $k$ contributes to the real part; odd $k$ contributes to the imaginary part. Get the signs right or the whole identity collapses.

De Moivre is the answer key

The LHS expansion is a polynomial in $\cos\theta$ and $\sin\theta$; the RHS is $\cos n\theta + i\sin n\theta$. Equating real and imaginary parts gives both identities in one stroke.

Use $\sin^2 + \cos^2 = 1$ to convert

After equating, you often want $\cos n\theta$ purely in terms of $\cos\theta$. Replace any $\sin^2\theta$ by $1 - \cos^2\theta$ until only cosines remain.

What you'll master

Know

Key facts

De Moivre: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
Binomial theorem: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$
Powers of $i$ cycle: $1,\,i,\,-1,\,-i,\,\ldots$
$\cos n\theta = \mathrm{Re}\big((\cos\theta+i\sin\theta)^n\big)$, $\sin n\theta = \mathrm{Im}(\cdot)$

Understand

Concepts

Why even powers of $i$ contribute to $\cos n\theta$ and odd powers to $\sin n\theta$
Why $\cos n\theta$ is a polynomial in $\cos\theta$ (the Chebyshev polynomial $T_n$)
Why $\sin n\theta = \sin\theta \cdot U_{n-1}(\cos\theta)$ always has a factor of $\sin\theta$

Can do

Skills

Expand $(\cos\theta + i\sin\theta)^n$ binomially for any required $n$
Derive $\cos n\theta$ and $\sin n\theta$ in terms of $\cos\theta$ and $\sin\theta$
Express $\cos n\theta$ purely in powers of $\cos\theta$ via $\sin^2\theta = 1 - \cos^2\theta$

Key terms

De Moivre's theorem$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for any integer $n$. The bridge between powers of a unit-modulus complex number and multiple-angle trig.

Binomial theorem$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$, where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. The expansion engine for $(\cos\theta + i\sin\theta)^n$.

Real part (Re)For $z = x + iy$, $\mathrm{Re}(z) = x$. Collecting real terms from the binomial expansion gives $\cos n\theta$.

Imaginary part (Im)For $z = x + iy$, $\mathrm{Im}(z) = y$ (note: $y$, not $iy$). Collecting imaginary terms gives $\sin n\theta$.

Multiple-angle identityAn expression for $\cos n\theta$ or $\sin n\theta$ in powers of $\cos\theta$ and $\sin\theta$. Examples: $\cos 2\theta = 2\cos^2\theta - 1$, $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$.

Powers of $i$$i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. Cycles with period 4 — essential for separating real/imaginary parts of binomial terms.

MEX-N2NESA outcome (Complex Numbers II): uses the polar form, de Moivre's theorem and complex roots; applies the binomial theorem to $(\cos\theta+i\sin\theta)^n$ to derive multiple-angle identities.

The method: expand, split by $i$, equate

core concept

Start from the two true statements $$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \quad(\text{de Moivre}),$$ $$(\cos\theta + i\sin\theta)^n = \sum_{k=0}^{n} \binom{n}{k}\cos^{\,n-k}\theta\,(i\sin\theta)^k \quad(\text{binomial}).$$

The right-hand sides must be equal. Split the binomial sum into terms with even $k$ (real, because $i^{2m} = (-1)^m$) and odd $k$ (imaginary, because $i^{2m+1} = (-1)^m i$):

Real part ($k$ even, $k = 2m$): $\displaystyle\cos n\theta = \sum_{m\ge 0} \binom{n}{2m}(-1)^m \cos^{\,n-2m}\theta\,\sin^{2m}\theta$.
Imaginary part ($k$ odd, $k = 2m+1$): $\displaystyle\sin n\theta = \sum_{m\ge 0} \binom{n}{2m+1}(-1)^m \cos^{\,n-2m-1}\theta\,\sin^{2m+1}\theta$.

Hook resolved: for $n = 2$ the expansion gives $$(\cos\theta + i\sin\theta)^2 = \cos^2\theta + 2i\cos\theta\sin\theta + i^2\sin^2\theta = (\cos^2\theta - \sin^2\theta) + i(2\sin\theta\cos\theta).$$ Equating to $\cos 2\theta + i\sin 2\theta$ delivers both double-angle identities at once: $\cos 2\theta = \cos^2\theta - \sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$.

Why this scales. Every multiple-angle identity in the syllabus is one binomial expansion away. You don't memorise $\cos 4\theta$ or $\sin 5\theta$ — you generate them.

The two equalities: de Moivre's LHS = binomial expansion of LHS = $\cos n\theta + i\sin n\theta$ · Even $k$ in $\binom{n}{k}(\cos\theta)^{n-k}(i\sin\theta)^k$ → real part → $\cos n\theta$ · Odd $k$ → imaginary part → $\sin n\theta$ · Sign comes from $i^{2m} = (-1)^m$ — alternates as you step through even $k$ values

Pause — copy the even-$k$/odd-$k$ split rule (even $\to \cos n\theta$, odd $\to \sin n\theta$), the sign from $i^{2m} = (-1)^m$, and the two equalities that make the method work into your book.

Quick check: In the binomial expansion of $(\cos\theta + i\sin\theta)^n$, which terms contribute to $\cos n\theta$?

Expressing $\cos n\theta$ in powers of $\cos\theta$ alone

core concept

We just saw the expand-split-equate method: even-$k$ binomial terms (where $i^k$ is real) give $\cos n\theta$; odd-$k$ terms give $\sin n\theta$, with signs from $(-1)^{k/2}$. That raises a question: $\cos n\theta$ comes out as a polynomial in both $\cos\theta$ and $\sin\theta$ — how do we express it in $\cos\theta$ alone? This card answers it → substitute $\sin^{2m}\theta = (1-\cos^2\theta)^m$ in every even-$k$ term.

The raw real-part formula contains both $\cos\theta$ and $\sin\theta$. A standard HSC ask is to express $\cos n\theta$ purely in $\cos\theta$. The trick is to substitute $\sin^{2m}\theta = (1 - \cos^2\theta)^m$ (every $\sin$ in the real part appears to an even power, so this is always possible).

The resulting polynomial $T_n(\cos\theta) = \cos n\theta$ is the Chebyshev polynomial of the first kind:

$\cos 2\theta = 2\cos^2\theta - 1$
$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$
$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$

For $\sin n\theta$, every term in the imaginary part contains an odd power of $\sin\theta$, so $\sin\theta$ factors out cleanly: $\sin n\theta = \sin\theta \cdot U_{n-1}(\cos\theta)$. You can either leave it mixed or use $\sin^2\theta = 1-\cos^2\theta$ on the leftover even powers.

$$\cos n\theta = \sum_{m\ge 0} \binom{n}{2m}(-1)^m \cos^{\,n-2m}\theta\,(1-\cos^2\theta)^m$$

Common mistake. Don't forget to expand $(1-\cos^2\theta)^m$ all the way out and collect like powers. A half-expanded answer in $\cos 4\theta = \binom{4}{0}\cos^4\theta - \binom{4}{2}\cos^2\theta\sin^2\theta + \binom{4}{4}\sin^4\theta$ form will lose marks if the question asks for "in terms of $\cos\theta$".

Substitute $\sin^{2m}\theta = (1-\cos^2\theta)^m$ to eliminate $\sin\theta$ from $\cos n\theta$ · Standard results: $\cos 2\theta = 2c^2 - 1$, $\cos 3\theta = 4c^3 - 3c$, $\cos 4\theta = 8c^4 - 8c^2 + 1$ (where $c = \cos\theta$) · $\sin n\theta$ always has $\sin\theta$ as a factor (because every imaginary term has odd power of $\sin\theta$)

Pause — copy the substitution $\sin^{2m}\theta = (1-\cos^2\theta)^m$, the standard results $\cos 2\theta = 2c^2-1$, $\cos 3\theta = 4c^3-3c$, $\cos 4\theta = 8c^4-8c^2+1$, and the fact that $\sin\theta$ always factors out of $\sin n\theta$ into your book.

Did you get this? True or false: $\sin n\theta$ always has $\sin\theta$ as a factor when expressed as a polynomial in $\sin\theta$ and $\cos\theta$ via the binomial expansion of $(\cos\theta + i\sin\theta)^n$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · DERIVE $\cos 4\theta$ AND $\sin 4\theta$ (MIXED FORM)

By expanding $(\cos\theta + i\sin\theta)^4$ using the binomial theorem and equating real and imaginary parts with de Moivre, derive expressions for $\cos 4\theta$ and $\sin 4\theta$ in terms of $\cos\theta$ and $\sin\theta$.

Let $c = \cos\theta$, $s = \sin\theta$. Binomial: $(c + is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4$. Use $i^2 = -1$, $i^3 = -i$, $i^4 = 1$: $(c+is)^4 = c^4 + 4ic^3 s - 6c^2 s^2 - 4ic s^3 + s^4$.

Write the binomial coefficients $\binom{4}{k} = 1, 4, 6, 4, 1$ before substituting. Then replace each $(is)^k$ using the $i^k$ cycle so the signs are explicit.

PROBLEM 2 · DERIVE $\cos 5\theta$ AND $\sin 5\theta$

Use the binomial expansion of $(\cos\theta + i\sin\theta)^5$ to find $\cos 5\theta$ and $\sin 5\theta$ in terms of $\cos\theta$ and $\sin\theta$.

With $c = \cos\theta$, $s = \sin\theta$, binomial coefficients $\binom{5}{k} = 1, 5, 10, 10, 5, 1$: $(c+is)^5 = c^5 + 5c^4(is) + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5$.

Always write all six terms before simplifying. Missing a middle term is the most common error in expansions of odd $n$.

PROBLEM 3 · EXPRESS $\cos 4\theta$ PURELY IN $\cos\theta$

Starting from the mixed expression $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$ derived in Problem 1, express $\cos 4\theta$ as a polynomial in $\cos\theta$ only.

Substitute $\sin^2\theta = 1 - \cos^2\theta$. Let $c = \cos\theta$ so $\sin^2\theta = 1 - c^2$ and $\sin^4\theta = (1 - c^2)^2 = 1 - 2c^2 + c^4$.

Substituting in $\sin^2$ once is cleaner than juggling $\sin^4$ alone. Square the result rather than expanding $\sin^4$ from scratch.

Fill the gap: Expressed as a polynomial in $\cos\theta$ alone, $\cos 3\theta = $ $\cos^3\theta - $ $\cos\theta$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to evaluate $i^k$

Writing $(\cos\theta + i\sin\theta)^4 = \cos^4\theta + 4\cos^3\theta(i\sin\theta) + 6\cos^2\theta(i\sin\theta)^2 + \ldots$ and leaving $(i\sin\theta)^k$ unsimplified makes it impossible to split real and imaginary parts. Always reduce $(i\sin\theta)^k = i^k \sin^k\theta$ and apply the $i^k$ cycle before grouping.

Trap 02

Wrong signs from $i^2$ and $i^3$

$i^2 = -1$ flips the sign of every $k = 2$ term; $i^3 = -i$ flips every $k = 3$ term. A missed sign in the $k = 2$ or $k = 3$ term wrecks both $\cos n\theta$ and $\sin n\theta$. Write the $i^k$ values above each term as you expand.

Trap 03

Stopping at the mixed form when the question asks for "$\cos\theta$ only"

"Express $\cos 4\theta$ in terms of $\cos\theta$" requires substituting $\sin^2\theta = 1 - \cos^2\theta$ and collecting. Leaving $\sin^2\theta$ or $\sin^4\theta$ in the final answer typically loses the last mark.

Did you get this? True or false: in the binomial expansion of $(\cos\theta + i\sin\theta)^6$, the $k = 3$ term contributes to the real part because $\binom{6}{3} = 20$ is even.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

By expanding $(\cos\theta + i\sin\theta)^3$ binomially and equating real and imaginary parts, derive $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos\theta$ and $\sin\theta$.

Use the result from Q1 plus $\sin^2\theta = 1 - \cos^2\theta$ to show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.

Use the result from Q1 to show that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ (this time eliminate $\cos\theta$ via $\cos^2\theta = 1 - \sin^2\theta$).

Use the $\sin 5\theta$ expansion derived in Worked Example 2 to express $\sin 5\theta$ as a polynomial in $\sin\theta$ alone.

Without expanding $(\cos\theta + i\sin\theta)^6$ in full, state which values of $k$ (in the binomial expansion) contribute to $\cos 6\theta$ and what sign each carries.

Odd one out: Three of these identities are correct multiple-angle expansions. Which one is NOT?

Revisit your thinking

Earlier you expanded $(\cos\theta + i\sin\theta)^2$ and predicted where even and odd powers of $i$ would land.

Even powers of $i$ ($i^0 = 1$, $i^2 = -1$) sit in the real part and assemble $\cos n\theta$. Odd powers ($i^1 = i$, $i^3 = -i$) sit in the imaginary part and assemble $\sin n\theta$. The whole derivation is a marriage between de Moivre (one equation, two pieces of information) and the binomial theorem (the expansion machinery). Once you see this, every multiple-angle identity from $\cos 2\theta$ to $\sin 7\theta$ is mechanical — no memorisation required.

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Multiple choice

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Short answer

ApplyBand 33 marks

Q1. By expanding $(\cos\theta + i\sin\theta)^3$ using the binomial theorem and applying de Moivre's theorem, show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. (3 marks)

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ApplyBand 44 marks

Q2. Using the binomial expansion of $(\cos\theta + i\sin\theta)^5$, show that $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$. (4 marks)

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AnalyseBand 54 marks

Q3. (a) Use the binomial expansion of $(\cos\theta + i\sin\theta)^4$ to write $\sin 4\theta$ in terms of $\cos\theta$ and $\sin\theta$. (b) Hence, or otherwise, show that $\dfrac{\sin 4\theta}{\sin\theta} = 4\cos\theta(2\cos^2\theta - 1)$ (assume $\sin\theta \neq 0$). (4 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $(c+is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = (c^3 - 3cs^2) + i(3c^2 s - s^3)$. Equating with $\cos 3\theta + i\sin 3\theta$: $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$ and $\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta$.

2. $\cos 3\theta = c^3 - 3c(1 - c^2) = c^3 - 3c + 3c^3 = 4c^3 - 3c = 4\cos^3\theta - 3\cos\theta$. ✓

3. $\sin 3\theta = 3c^2 s - s^3 = 3(1 - s^2)s - s^3 = 3s - 3s^3 - s^3 = 3\sin\theta - 4\sin^3\theta$. ✓

4. $\sin 5\theta = 5c^4 s - 10c^2 s^3 + s^5$. Use $c^2 = 1 - s^2$: $c^4 = (1 - s^2)^2 = 1 - 2s^2 + s^4$. So $5(1 - 2s^2 + s^4)s - 10(1 - s^2)s^3 + s^5 = 5s - 10s^3 + 5s^5 - 10s^3 + 10s^5 + s^5 = 16s^5 - 20s^3 + 5s$.

5. For $(\cos\theta + i\sin\theta)^6$, even $k \in \{0, 2, 4, 6\}$ contribute to the real part. Signs from $i^{2m} = (-1)^m$: $k = 0$ (+), $k = 2$ ($-$), $k = 4$ (+), $k = 6$ ($-$). So $\cos 6\theta = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta$.

Q1 (3 marks): Binomial expansion $(c+is)^3 = (c^3 - 3cs^2) + i(3c^2 s - s^3)$ [1]. De Moivre gives $\cos 3\theta + i\sin 3\theta$; equating real parts: $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$ [1]. Substitute $\sin^2\theta = 1 - \cos^2\theta$: $\cos 3\theta = c^3 - 3c(1 - c^2) = 4\cos^3\theta - 3\cos\theta$ [1].

Q2 (4 marks): $(c+is)^5 = c^5 + 5ic^4 s - 10c^3 s^2 - 10ic^2 s^3 + 5c s^4 + is^5$ [1]. Real part: $\cos 5\theta = c^5 - 10c^3 s^2 + 5c s^4$ [1]. Substitute $s^2 = 1 - c^2$, $s^4 = (1-c^2)^2 = 1 - 2c^2 + c^4$: $\cos 5\theta = c^5 - 10c^3(1 - c^2) + 5c(1 - 2c^2 + c^4)$ [1] $= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5 = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ [1].

Q3 (4 marks): (a) Imaginary part of $(c+is)^4 = c^4 + 4ic^3 s - 6c^2 s^2 - 4ic s^3 + s^4$: $\sin 4\theta = 4c^3 s - 4c s^3$ [1] $= 4\sin\theta\cos\theta(\cos^2\theta - \sin^2\theta)$ [1]. (b) $\dfrac{\sin 4\theta}{\sin\theta} = 4\cos\theta(\cos^2\theta - \sin^2\theta)$ [1]. Replace $\sin^2\theta = 1 - \cos^2\theta$: $4\cos\theta(\cos^2\theta - (1 - \cos^2\theta)) = 4\cos\theta(2\cos^2\theta - 1)$ [1].

Boss battle · The Multiple-Angle Master

earn bronze · silver · gold

Five timed questions on binomial expansions of $(\cos\theta + i\sin\theta)^n$, equating real/imaginary parts and converting to polynomials in $\cos\theta$. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering quick multiple-angle identity questions. Lighter alternative to the boss.

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