De Moivre's theorem and trigonometric identitiesEdexcel International A Level Further Maths: Revision notes

Section 1

De Moivre's theorem

De Moivre's theorem states that for any integer $n$ : $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta.$ In exponential form this is $\left(e^{i\theta}\right)^n=e^{in\theta}$ . In modulus-argument form, $\left[r(\cos\theta+i\sin\theta)\right]^n=r^n(\cos n\theta+i\sin n\theta)$ : raise the modulus to the power $n$ and multiply the argument by $n$ . For example, with $\theta=\frac\pi9$ , $z^6=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\frac12+\frac{\sqrt3}2i$ .

Key termsDe Moivre's theorem

Common mistake

Applying the theorem to $(a+ib)^n$ directly. Write the number as $r(\cos\theta+i\sin\theta)$ first.

Section 2

Proof for any integer $n$

Positive $n$ , by induction. True for $n=1$ . Assume it is true for $n=k$ . Then $(\cos\theta+i\sin\theta)^{k+1}=(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)$ , which expands to $(\cos k\theta\cos\theta-\sin k\theta\sin\theta)+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)=\cos(k+1)\theta+i\sin(k+1)\theta$ . So true for $n=k+1$ , and by induction for all $n\ge1$ . Negative $n=-m$ . $(\cos\theta+i\sin\theta)^{-m}=\frac{1}{\cos m\theta+i\sin m\theta}=\cos m\theta-i\sin m\theta=\cos(-m\theta)+i\sin(-m\theta)$ , using the conjugate and $\cos^2+\sin^2=1$ . For $n=0$ both sides equal $1$ .

Key termsproof by induction

Exam tip

State the base case, the assumption, the inductive step and a conclusion in your proof.

Section 3

Multiple angles: $\cos n\theta$ and $\sin n\theta$ in powers of $\cos\theta$ , $\sin\theta$

Expand $(c+is)^n$ with the binomial theorem, where $c=\cos\theta$ , $s=\sin\theta$ , and equate real and imaginary parts with $\cos n\theta+i\sin n\theta$ . Remember $i^2=-1$ , $i^3=-i$ , $i^4=1$ . For $n=3$ : $(c+is)^3=\left(c^3-3cs^2\right)+i\left(3c^2s-s^3\right)$ , so $\cos3\theta=c^3-3cs^2$ and $\sin3\theta=3c^2s-s^3$ . Using $s^2=1-c^2$ , $\cos3\theta=4c^3-3c$ ; using $c^2=1-s^2$ , $\sin3\theta=3s-4s^3$ . For $n=5$ : $\cos5\theta=c^5-10c^3s^2+5cs^4=16c^5-20c^3+5c$ .

Key termsbinomial theorem

Common mistake

Dropping the powers of $i$ : the terms with an odd power of $i$ form the imaginary part.

Section 4

Powers of $\cos\theta$ and $\sin\theta$ in multiple angles

Let $z=\cos\theta+i\sin\theta$ . Then $z^n+z^{-n}=2\cos n\theta$ and $z^n-z^{-n}=2i\sin n\theta$ . Raise $z+\frac1z=2\cos\theta$ or $z-\frac1z=2i\sin\theta$ to a power, expand, then pair terms $z^k\pm z^{-k}$ . Example: $\left(z-\frac1z\right)^4=z^4-4z^2+6-4z^{-2}+z^{-4}$ , so $16\sin^4\theta=2\cos4\theta-8\cos2\theta+6$ , giving $\sin^4\theta=\frac18(\cos4\theta-4\cos2\theta+3)$ . This turns a power of sine or cosine into a sum of multiple angles that is easy to integrate: $\int_0^{\pi/2}\sin^4\theta\,d\theta=\frac{3\pi}{16}$ .

Key termsmultiple angles

Common mistake

Using $(2i\sin\theta)^4=2\sin^4\theta$ . It equals $16\sin^4\theta$ because $i^4=1$ ; for odd powers keep the $i$ .

Section 5

Choosing the method

$\cos n\theta$ or $\sin n\theta$ in powers: expand $(c+is)^n$ and take real or imaginary parts.
Powers $\cos^n\theta$ or $\sin^n\theta$ in multiple angles: expand $\left(z\pm\frac1z\right)^n$ .
Check an identity with a value such as $\theta=\frac\pi2$ , where $\sin^4\theta=1$ and $\frac18(1+4+3)=1$ .

Exam tip

Use $\cos^2\theta+\sin^2\theta=1$ at the end to write your answer in the single variable the question asks for.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on De Moivre's theorem and trigonometric identities

A complex number is $z=\cos\theta+i\sin\theta$ , where $\theta$ is real.
Given that $\theta=\frac\pi9$ , find $z^6$ in the form $a+ib$ , giving exact values of $a$ and $b$ .2 marks
Let $c=\cos\theta$ and $s=\sin\theta$ , and consider $(c+is)^3$ expanded using the binomial theorem.
Hence express $\cos3\theta$ in terms of $\cos\theta$ only.2 marks
De Moivre's theorem states that $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$ for every integer $n$ . A student proves the case $n\ge1$ by induction.
Show that if the result is true for $n=k$ , where $k\ge1$ , then it is true for $n=k+1$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).