Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 1 · Complex familiar

Question 16

[6 marks] Technology-free
The question

Given $\sin(\theta)\neq0$, use De Moivre’s theorem to prove the identity

$$\frac{\sin(4\theta)}{\sin(\theta)}=8\cos^{3}(\theta)-4\cos(\theta)$$

[6 marks]
Watch the situation first

Turn the angle four times, then read off the height

The fourth power of $\cos(\theta)+i\sin(\theta)$ sits at angle $4\theta$. Watch the five terms of the expansion land on that same point, and see which of them make $\sin(4\theta)$.

Start with cos(θ) + i sin(θ): a point on the unit circle at angle θDe Moivre: the 4th power turns the angle θ into 4θExpand (cos(θ) + i sin(θ))⁴: five terms, laid end to end, land on the same pointEquate imaginary parts: the up and down steps make sin(4θ)Divide by sin(θ), then swap sin²(θ) for 1 − cos²(θ)ReImOθzz²z³4θcos⁴(θ)4i cos³(θ)sin(θ)−6cos²(θ) sin²(θ)−4i cos(θ) sin³(θ)z⁴sin(4θ)sin(θ)z = cos(θ) + i sin(θ)(cos(θ) + i sin(θ))⁴= cos(4θ) + i sin(4θ)cos⁴(θ)+ 4i cos³(θ) sin(θ)− 6cos²(θ) sin²(θ)− 4i cos(θ) sin³(θ)+ sin⁴(θ)sin(4θ) = 4cos³(θ) sin(θ)− 4cos(θ) sin³(θ)sin(4θ) ÷ sin(θ), sin(θ) ≠ 0= 4cos³(θ) − 4cos(θ) sin²(θ)= 4cos³(θ) − 4cos(θ)(1 − cos²(θ))= 8cos³(θ) − 4cos(θ)

Drawn to scale for one angle $\theta$. Terms without $i$ run across and terms with $i$ run up and down, so the height of the end of the chain is the imaginary part.

Now for the mathematics

Pick an angle and follow the terms to sin(4θ)

Move $\theta$. The five terms of $\left(\cos(\theta)+i\sin(\theta)\right)^4$ are laid end to end from $O$: terms without $i$ run across and terms with $i$ run up and down. Then try a common error and see where it goes wrong.

Re Im O 1 z z4
Terms without $i$ (real) Terms with $i$ (imaginary) $z^4=\cos(4\theta)+i\sin(4\theta)$ End of the chain
Try a common error
Both sides of the identity

Here $\sin(\theta)=0$. Division is allowed because you are given $\sin(\theta)\neq0$.

The chain lands exactly on $z^4$, so its height, $4\cos^{3}(\theta)\sin(\theta)$ ${}-4\cos(\theta)\sin^{3}(\theta)$, is $\sin(4\theta)$. Divide by $\sin(\theta)$ and use $\sin^{2}(\theta)=1-\cos^{2}(\theta)$, and both sides agree: $8\cos^{3}(\theta)-4\cos(\theta)$.

The chain no longer lands on $z^4$, so its height is not $\sin(4\theta)$.

At this angle the chain still happens to land on $z^4$. Move $\theta$ to see it miss.

With $i^3=+i$ the fourth term changes sign. The imaginary part becomes $4\cos^{3}(\theta)\sin(\theta)$ ${}+4\cos(\theta)\sin^{3}(\theta)$, and after dividing by $\sin(\theta)$ and substituting you get $4\cos^{3}(\theta)+4\cos(\theta)\left(1-\cos^{2}(\theta)\right)$ ${}=4\cos(\theta)$, not $8\cos^{3}(\theta)-4\cos(\theta)$. In fact $i^3=-i$.

The expansion is right, so the chain still lands on $z^4$. The slip comes later: expanding $-4\cos(\theta)\left(1-\cos^{2}(\theta)\right)$ as $-4\cos(\theta)-4\cos^{3}(\theta)$ gives $-4\cos(\theta)$. Watch the sign: $-4\cos(\theta)\left(1-\cos^{2}(\theta)\right)$ ${}=-4\cos(\theta)+4\cos^{3}(\theta)$.

The expansion is right, so the chain still lands on $z^4$. The slip is writing $\sin^{2}(\theta)=\cos^{2}(\theta)-1$, which gives $4\cos^{3}(\theta)-4\cos(\theta)\left(\cos^{2}(\theta)-1\right)$ ${}=4\cos(\theta)$. From $\sin^2(\theta)+\cos^2(\theta)=1$ you get $\sin^{2}(\theta)=1-\cos^{2}(\theta)$.

Using $3$ instead of $4$ comes from row 3 of Pascal’s triangle. The imaginary part becomes $3\cos^{3}(\theta)\sin(\theta)$ ${}-3\cos(\theta)\sin^{3}(\theta)$, and you reach $6\cos^{3}(\theta)-3\cos(\theta)$. Power $4$ needs row 4: $1,4,6,4,1$.

Marking guide · 6 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · De Moivre’s theorem1 mark

Apply De Moivre’s theorem with $n=4$.

$$\begin{aligned}&\left(\cos(\theta)+i\sin(\theta)\right)^4\\&\quad=\cos(4\theta)+i\sin(4\theta)\end{aligned}$$

Marking

A response that does not use De Moivre’s theorem (e.g. double-angle identities applied twice) does not gain the first, second or third marks.

Marker: correctly applies De Moivre’s theorem

Step 2 · Expand the power1 mark

Expand with the binomial theorem, using coefficients 1, 4, 6, 4, 1 and $i^2=-1$, $i^3=-i$, $i^4=1$.

$$\begin{aligned}&\left(\cos(\theta)+i\sin(\theta)\right)^4\\&\quad=\cos^{4}(\theta)+4i\cos^{3}(\theta)\sin(\theta)\\&\qquad-6\cos^{2}(\theta)\sin^{2}(\theta)\\&\qquad-4i\cos(\theta)\sin^{3}(\theta)+\sin^{4}(\theta)\end{aligned}$$

Marker: correctly expands $\left(\cos(\theta)+i\sin(\theta)\right)^4$

Step 3 · Equate imaginary parts1 mark

Match the terms containing $i$ with $i\sin(4\theta)$ from Step 1.

$$\begin{aligned}\sin(4\theta)&=4\cos^{3}(\theta)\sin(\theta)\\&\quad-4\cos(\theta)\sin^{3}(\theta)\end{aligned}$$

Why?

$\sin(4\theta)$ is the imaginary part of the right side of Step 1, so it equals the imaginary part of the expansion: the two terms carrying $i$.

Marker: correctly determines $\sin(4\theta)$ by equating imaginary parts

Step 4 · Divide by sin(θ)1 mark

Divide both sides by $\sin(\theta)$, which is allowed because you are given $\sin(\theta)\neq0$.

$$\begin{aligned}\frac{\sin(4\theta)}{\sin(\theta)}&=4\cos^{3}(\theta)-4\cos(\theta)\sin^{2}(\theta),\\&\quad\text{since } \sin(\theta)\neq0\end{aligned}$$

Marking

A reference to $\sin(\theta)\neq0$ is not required for this mark.

Marker: correctly divides by $\sin(\theta)$

Step 5 · Pythagorean identity1 mark

Replace $\sin^{2}(\theta)$ with $1-\cos^{2}(\theta)$, then expand, watching the sign, and collect like terms.

$$\begin{aligned}\frac{\sin(4\theta)}{\sin(\theta)}&=4\cos^{3}(\theta)\\&\quad-4\cos(\theta)\left(1-\cos^{2}(\theta)\right)\\&=8\cos^{3}(\theta)-4\cos(\theta)\end{aligned}$$

Marking

This mark is for the correct substitution $\sin^2(\theta)=1-\cos^2(\theta)$. An algebraic error after this substitution does not gain the sixth mark.

Marker: correctly applies the Pythagorean identity

Step 6 · State the result1 mark

Finish with the exact statement asked for.

$$\begin{aligned}\frac{\sin(4\theta)}{\sin(\theta)}&=8\cos^{3}(\theta)-4\cos(\theta)\\&\quad\text{as required}\end{aligned}$$

Marker: shows the required result based on evidence of prior mathematical reasoning

Putting it all together

Each phrase of the question gave you something

“Given $\sin(\theta)\neq0$”
This is what lets you divide by $\sin(\theta)$.
“use De Moivre’s theorem”
Start from $\left(\cos(\theta)+i\sin(\theta)\right)^4$ $=\cos(4\theta)+i\sin(4\theta)$. The multiple angle $4\theta$ becomes a $4$th power you can expand with the binomial theorem.
“$\dfrac{\sin(4\theta)}{\sin(\theta)}$”
$\sin(4\theta)$ is the imaginary part of $\cos(4\theta)+i\sin(4\theta)$, so you equate imaginary parts: $\sin(4\theta)=4\cos^{3}(\theta)\sin(\theta)$ ${}-4\cos(\theta)\sin^{3}(\theta)$. Then you divide by $\sin(\theta)$ to get $4\cos^{3}(\theta)$ ${}-4\cos(\theta)\sin^{2}(\theta)$.
“$8\cos^{3}(\theta)-4\cos(\theta)$”
The target contains only $\cos(\theta)$, so the remaining $\sin^{2}(\theta)$ becomes $1-\cos^{2}(\theta)$. Expanding and collecting gives $8\cos^{3}(\theta)-4\cos(\theta)$.
“prove the identity”
End with the exact statement asked for, linked to the steps above: “as required”.
Common errors

If I had to guess where you might lose marks…

  • Taking $i^3=+i$. The imaginary part becomes $4\cos^{3}(\theta)\sin(\theta)$ ${}+4\cos(\theta)\sin^{3}(\theta)$. After dividing by $\sin(\theta)$ and substituting, you get $4\cos^{3}(\theta)+4\cos(\theta)\left(1-\cos^{2}(\theta)\right)$ ${}=4\cos(\theta)$, not $8\cos^{3}(\theta)-4\cos(\theta)$.
  • A sign slip expanding $-4\cos(\theta)\left(1-\cos^{2}(\theta)\right)$. Writing it as $-4\cos(\theta)-4\cos^{3}(\theta)$ gives $-4\cos(\theta)$.
  • Pythagoras backwards. Writing $\sin^{2}(\theta)=\cos^{2}(\theta)-1$ gives $4\cos^{3}(\theta)-4\cos(\theta)\left(\cos^{2}(\theta)-1\right)$ ${}=4\cos(\theta)$.
  • Wrong binomial coefficients. Using row 3 of Pascal’s triangle ($3$ instead of $4$) makes the imaginary part $3\cos^{3}(\theta)\sin(\theta)$ ${}-3\cos(\theta)\sin^{3}(\theta)$, which leads to $6\cos^{3}(\theta)-3\cos(\theta)$.
  • Proving it without De Moivre’s theorem. Double-angle formulas applied twice can give valid algebra, but the question says “use De Moivre’s theorem”, so this loses Marks 1–3.

Only mark this done when you could do it without help. Reading the solution does not count.

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.

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