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Module 13 · L08 of 12 ~45 min ⚡ +90 XP available

Trigonometric Identities via Complex Numbers

Memorising $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ from a table is a chore. Deriving it from $(\cos\theta + i\sin\theta)^3$ in two lines is elegant — and gives you every multiple-angle identity for free. This lesson uses De Moivre's theorem and the binomial theorem to manufacture $\cos n\theta$ and $\sin n\theta$ identities on demand for $n = 3, 4, 5$ and beyond.

Today's hook — Without looking anything up, expand $(\cos\theta + i\sin\theta)^3$ using the binomial theorem. Then compare with $\cos 3\theta + i\sin 3\theta$ from De Moivre. Equate real and imaginary parts. What identities for $\cos 3\theta$ and $\sin 3\theta$ fall out? Compare your answers after card 05.

0/5QUESTS

Recall — your gut answer first

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Write the binomial expansion of $(a + b)^4$. Before checking — what are the binomial coefficients $\binom{4}{0}, \binom{4}{1}, \ldots, \binom{4}{4}$? Sketch your reasoning below.

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The two moves: De Moivre and Binomial

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Every multiple-angle identity comes from a single trick: express $(\cos\theta + i\sin\theta)^n$ in two different ways and equate. De Moivre gives the compact form $\cos n\theta + i\sin n\theta$. Binomial expansion gives the powers-of-$\cos\theta$ form. Equating real parts produces $\cos n\theta$ as a polynomial in $\cos\theta$ (and $\sin^2\theta$); equating imaginary parts produces $\sin n\theta$.

Strategy: (1) write $(\cos\theta + i\sin\theta)^n$, (2) apply De Moivre on one side and binomial theorem on the other, (3) equate real and imaginary parts, (4) substitute $\sin^2\theta = 1 - \cos^2\theta$ to make $\cos n\theta$ a polynomial in $\cos\theta$ alone (or $\cos^2\theta = 1 - \sin^2\theta$ for $\sin n\theta$).

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

$(\cos\theta + i\sin\theta)^n \;=\; \cos n\theta + i\sin n\theta$

Powers of $i$ cycle

$i^0 = 1,\; i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1$ — they cycle every 4. Track signs carefully when collecting real and imaginary parts.

Real part: even $k$; Imag part: odd $k$

In the binomial sum, terms with $k$ even contribute to $\cos n\theta$ (real); terms with $k$ odd contribute to $\sin n\theta$ (imaginary). Split the sum by parity of $k$.

Convert with Pythagoras

After collecting, use $\sin^2\theta = 1 - \cos^2\theta$ to express $\cos n\theta$ purely as a polynomial in $\cos\theta$ (a Chebyshev polynomial). Symmetrically for $\sin n\theta$.

What you'll master

Know

Key facts

De Moivre: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
Binomial theorem: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ and $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$
$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$

Understand

Concepts

Why equating real parts gives $\cos n\theta$ and imaginary parts gives $\sin n\theta$
How $i^{2k} = (-1)^k$ shapes the sign pattern in the expansion
Why $\cos n\theta$ is always a polynomial in $\cos\theta$ (Chebyshev structure)

Can do

Skills

Derive $\cos n\theta$ and $\sin n\theta$ identities for $n = 3, 4, 5$
Use the resulting identities to solve trig equations like $\cos 3\theta = c$
Express $\sin^n\theta$ or $\cos^n\theta$ as a sum of multiple angles (inverse direction)

Key terms

De Moivre's theorem$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for integer $n$. The bridge between complex powers and multiple angles.

Binomial theorem$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. Used to expand $(\cos\theta + i\sin\theta)^n$ term-by-term.

cis notation$\text{cis}\,\theta = \cos\theta + i\sin\theta$. Convenient shorthand for the polar form of a unit-modulus complex number.

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

Chebyshev polynomialPolynomial $T_n(x)$ defined by $T_n(\cos\theta) = \cos n\theta$. The identity $\cos n\theta = T_n(\cos\theta)$ is exactly what this method produces.

Real / imaginary partsFor $z = a + bi$ (real $a, b$): $\text{Re}(z) = a$, $\text{Im}(z) = b$. Equating real and imaginary parts of an equation between complex numbers gives two real equations.

MEX-N2NESA outcome (Complex Numbers II): uses De Moivre's theorem with complex numbers in polar form, including to derive trigonometric identities.

Deriving $\cos 3\theta$ and $\sin 3\theta$

core concept

The template: write $(\cos\theta + i\sin\theta)^n$ in two different ways and equate real and imaginary parts.

For $n = 3$: By De Moivre, $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$. By the binomial theorem (writing $c = \cos\theta$, $s = \sin\theta$):

$$(c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = c^3 + 3ic^2 s - 3cs^2 - is^3.$$

Equate real parts: $\cos 3\theta = c^3 - 3cs^2 = \cos^3\theta - 3\cos\theta\sin^2\theta$. Substitute $s^2 = 1 - c^2$:

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

Equate imaginary parts: $\sin 3\theta = 3c^2 s - s^3 = 3\sin\theta\cos^2\theta - \sin^3\theta$. Substitute $c^2 = 1 - s^2$:

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

Pattern. The two identities $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ and $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ are dual. Notice the same coefficients $4$ and $3$ appear, but the cubic and linear terms switch roles — a direct consequence of the symmetry in the binomial expansion.

De Moivre: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ · Expand $(c + is)^n$ using binomial; track powers of $i$: $i^2 = -1$, $i^3 = -i$, $i^4 = 1$ · Real parts $\to \cos n\theta$; imaginary parts $\to \sin n\theta$ · $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$; $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$

Pause — copy De Moivre's theorem, the binomial-expansion procedure (track powers of $i$: $i^2=-1$, $i^3=-i$, $i^4=1$), and $\cos 3\theta = 4\cos^3\theta-3\cos\theta$, $\sin 3\theta = 3\sin\theta-4\sin^3\theta$ into your book.

Quick check: Using $(\cos\theta + i\sin\theta)^2 = \cos 2\theta + i\sin 2\theta$ and equating real parts, which identity for $\cos 2\theta$ follows directly (before any Pythagoras substitution)?

Higher orders: $\cos 4\theta$ and $\cos 5\theta$

core concept

We just saw how to derive $\cos 3\theta = 4\cos^3\theta-3\cos\theta$ and $\sin 3\theta = 3\sin\theta-4\sin^3\theta$ by expanding $(c+is)^3$ and matching real and imaginary parts. That raises a question: what are the analogous results for $n = 4$ and $n = 5$? This card answers it → $\cos 4\theta = 8\cos^4\theta-8\cos^2\theta+1$, $\cos 5\theta = 16\cos^5\theta-20\cos^3\theta+5\cos\theta$, with the sign pattern $(-1)^{k/2}$ tracking even-$k$ terms.

The same template handles any $n$. For $n = 4$, the binomial expansion of $(c + is)^4$ gives:

$$(c + is)^4 = c^4 + 4ic^3 s + 6i^2 c^2 s^2 + 4i^3 c s^3 + i^4 s^4 = c^4 - 6c^2 s^2 + s^4 + i(4c^3 s - 4cs^3).$$

Real parts: $\cos 4\theta = c^4 - 6c^2 s^2 + s^4$. Substitute $s^2 = 1 - c^2$ throughout:

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

Imaginary parts: $\sin 4\theta = 4c^3 s - 4cs^3 = 4\sin\theta\cos\theta(\cos^2\theta - \sin^2\theta) = 2\sin 2\theta \cos 2\theta$.

For $n = 5$, the same method yields $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ and $\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$.

Sign pattern. Real parts use even $k$: $i^{2k} = (-1)^k$, so the real part of $(c+is)^n$ is $\sum_{k \text{ even}} \binom{n}{k} c^{n-k} (-1)^{k/2} s^k$. This is why the signs alternate: $+, -, +, -, \ldots$

$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$ · $\sin 4\theta = 4\sin\theta\cos\theta(\cos^2\theta - \sin^2\theta)$ · $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ · $\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$ · Real part = even-$k$ terms with sign $(-1)^{k/2}$; Imag part = odd-$k$ terms with sign $(-1)^{(k-1)/2}$

Pause — copy $\cos 4\theta = 8c^4-8c^2+1$, $\cos 5\theta = 16c^5-20c^3+5c$ (where $c=\cos\theta$), $\sin 5\theta = 16s^5-20s^3+5s$ (where $s=\sin\theta$), and the sign rule for even/odd $k$ terms into your book.

Did you get this? True or false: when expanding $(\cos\theta + i\sin\theta)^5$ via the binomial theorem, the real part $\cos 5\theta$ comes from terms with $k = 0, 2, 4$ (even $k$).

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

PROBLEM 1 · DERIVE $\cos 4\theta$ IN $\cos\theta$ ONLY

Use De Moivre and the binomial theorem to express $\cos 4\theta$ as a polynomial in $\cos\theta$.

By De Moivre: $(\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta$. By the binomial theorem with $c = \cos\theta$, $s = \sin\theta$: $(c + is)^4 = c^4 + 4ic^3 s + 6i^2 c^2 s^2 + 4i^3 c s^3 + i^4 s^4 = c^4 - 6c^2 s^2 + s^4 + i(4c^3 s - 4cs^3)$.

Use $i^2 = -1$, $i^3 = -i$, $i^4 = 1$ to simplify each term. Group real and imaginary parts before equating.

PROBLEM 2 · DERIVE $\sin 5\theta$ IN $\sin\theta$ ONLY

Show that $\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$.

Expand $(c + is)^5$ by binomial: $c^5 + 5ic^4 s + 10i^2 c^3 s^2 + 10i^3 c^2 s^3 + 5i^4 c s^4 + i^5 s^5 = c^5 - 10c^3 s^2 + 5cs^4 + i(5c^4 s - 10c^2 s^3 + s^5)$. Pascal row 5: $1, 5, 10, 10, 5, 1$.

Carefully apply $i^2 = -1$, $i^4 = 1$ to the even-$k$ terms (real part), and $i^3 = -i$, $i^5 = i$ to the odd-$k$ terms (imaginary part).

PROBLEM 3 · USE $\cos 3\theta$ TO SOLVE A CUBIC

Show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$, then use this identity to solve $8x^3 - 6x - 1 = 0$ by setting $x = \cos\theta$.

From $(c + is)^3 = \cos 3\theta + i\sin 3\theta$, equate real parts: $\cos 3\theta = c^3 - 3cs^2 = c^3 - 3c(1 - c^2) = 4c^3 - 3c$. So $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.

This is the identity we want as a tool. Memorise its derivation, not just the result.

Fill the gap: $\cos 3\theta = $ $\cos^3\theta - $$\cos\theta$, and $\sin 3\theta = $$\sin\theta - $$\sin^3\theta$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to swap to a single trig function

After equating real parts you get $\cos n\theta$ in mixed $\cos, \sin$ form. To match the standard identity you MUST substitute $\sin^2\theta = 1 - \cos^2\theta$ (or vice versa). Stopping early loses marks for "not fully simplified".

Trap 02

Mishandling powers of $i$

$i^2 = -1$, $i^3 = -i$, $i^4 = 1$. A common slip is treating $i^3$ as $-1$ or $i^4$ as $i$. Write the cycle down beside your working and tick off each term.

Trap 03

Using the wrong Pythagoras substitution

For $\cos n\theta$ in $\cos\theta$ only, replace $\sin^2\theta$ with $1 - \cos^2\theta$. For $\sin n\theta$ in $\sin\theta$ only, replace $\cos^2\theta$ with $1 - \sin^2\theta$. Mixing them produces a non-polynomial answer.

Did you get this? True or false: $\cos 4\theta$ cannot be expressed as a polynomial in $\cos\theta$ alone because the expansion of $(\cos\theta + i\sin\theta)^4$ contains $\sin^4\theta$ terms.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Derive $\cos 2\theta$ and $\sin 2\theta$ from $(\cos\theta + i\sin\theta)^2$. Show both the mixed form and the single-trig-function form using Pythagoras.

Use $(\cos\theta + i\sin\theta)^5$ and the binomial theorem to derive $\cos 5\theta$ as a polynomial in $\cos\theta$.

Use the identity $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ to solve $4x^3 - 3x + \tfrac12 = 0$ by setting $x = \sin\theta$.

Verify the identity $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$ at three specific angles: $\theta = 0, \pi/4, \pi/2$.

Derive a formula for $\tan 3\theta$ in terms of $\tan\theta$. (Hint: divide $\sin 3\theta$ by $\cos 3\theta$, then divide numerator and denominator by $\cos^3\theta$.)

Odd one out: Three of these multiple-angle identities are correct. Which one is WRONG?

Revisit your thinking

Earlier you expanded $(\cos\theta + i\sin\theta)^3$ using the binomial theorem and equated with $\cos 3\theta + i\sin 3\theta$ from De Moivre.

The expansion gives $\cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$. Real parts: $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta$ (after $\sin^2\theta = 1 - \cos^2\theta$). Imaginary parts: $\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta$ (after $\cos^2\theta = 1 - \sin^2\theta$). The whole method is just binomial + De Moivre + Pythagoras — once you see this pattern, every multiple-angle identity becomes a 2-line derivation.

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. Use $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$ and the binomial theorem to derive $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. (2 marks)

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ApplyBand 43 marks

Q2. By expanding $(\cos\theta + i\sin\theta)^4$, show that $\sin 4\theta = 4\sin\theta\cos\theta(2\cos^2\theta - 1)$. (3 marks)

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AnalyseBand 53 marks

Q3. (a) Derive $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. (b) Hence find the three real solutions of $8x^3 - 6x - \sqrt{2} = 0$, expressing them in terms of cosines of specific angles. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $(c + is)^2 = c^2 + 2ics - s^2$. Real: $\cos 2\theta = c^2 - s^2 = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$. Imag: $\sin 2\theta = 2cs = 2\sin\theta\cos\theta$.

2. Real part of $(c+is)^5$: $c^5 - 10c^3 s^2 + 5cs^4$. Substitute $s^2 = 1 - c^2$, $s^4 = (1 - c^2)^2$: $\cos 5\theta = c^5 - 10c^3(1 - c^2) + 5c(1 - 2c^2 + c^4) = 16c^5 - 20c^3 + 5c$.

3. $4\sin^3\theta - 3\sin\theta + \tfrac12 = 0 \Rightarrow -(3\sin\theta - 4\sin^3\theta) + \tfrac12 = 0 \Rightarrow \sin 3\theta = \tfrac12$. $3\theta = \tfrac{\pi}{6} + 2k\pi$ or $\tfrac{5\pi}{6} + 2k\pi$, giving three distinct $\sin\theta$ values: $\sin\tfrac{\pi}{18},\; \sin\tfrac{5\pi}{18},\; \sin\tfrac{13\pi}{18}$.

4. $\theta = 0$: LHS $= 1$, RHS $= 8 - 8 + 1 = 1$ ✓. $\theta = \pi/4$: LHS $= \cos\pi = -1$, RHS $= 8(\tfrac14) - 8(\tfrac12) + 1 = -1$ ✓. $\theta = \pi/2$: LHS $= \cos 2\pi = 1$, RHS $= 0 - 0 + 1 = 1$ ✓.

5. $\tan 3\theta = \dfrac{\sin 3\theta}{\cos 3\theta} = \dfrac{3c^2 s - s^3}{c^3 - 3cs^2}$. Divide top and bottom by $c^3$: $\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$.

Q1 (2 marks): Expansion: $(c + is)^3 = c^3 + 3ic^2 s - 3cs^2 - is^3$ [1]. Real part: $\cos 3\theta = c^3 - 3cs^2 = c^3 - 3c(1 - c^2) = 4c^3 - 3c$ [1]. Therefore $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.

Q2 (3 marks): $(c + is)^4 = c^4 + 4ic^3 s - 6c^2 s^2 - 4ics^3 + s^4$ [1]. Imag part: $\sin 4\theta = 4c^3 s - 4cs^3 = 4cs(c^2 - s^2)$ [1]. Using $s^2 = 1 - c^2$: $c^2 - s^2 = 2c^2 - 1$, so $\sin 4\theta = 4\sin\theta\cos\theta(2\cos^2\theta - 1)$ [1].

Q3 (3 marks): (a) Standard derivation gives $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ [1]. (b) Let $x = \cos\theta$: $8\cos^3\theta - 6\cos\theta = \sqrt{2}$, so $2(4\cos^3\theta - 3\cos\theta) = \sqrt{2}$, giving $\cos 3\theta = \tfrac{1}{\sqrt{2}}$ [1]. $3\theta = \pm\tfrac{\pi}{4} + 2k\pi$; three distinct $\theta \in [0, \pi]$: $\theta = \tfrac{\pi}{12}, \tfrac{7\pi}{12}, \tfrac{3\pi}{4}$. Three roots: $x = \cos\tfrac{\pi}{12},\; \cos\tfrac{7\pi}{12},\; \cos\tfrac{3\pi}{4}$ [1].

Boss battle · The Identity Forger

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Module overview · Extension 2