Further Maths/Lessons/Complex Numbers (Year 2)/Trigonometric Identities
Core Pure · Complex Numbers · Year 2

Trigonometric Identities

Using De Moivre's theorem to derive multiple-angle identities, and the $z+\frac1z$ trick for expressing powers of $\cos\theta$ and $\sin\theta$ as sums of cosines and sines.

11 min read 2 worked examples
Recap — De Moivre's theorem

Remember from the previous lesson: for any integer $n$, $\big(r(\cos\theta+i\sin\theta)\big)^n=r^n(\cos n\theta+i\sin n\theta)$. This lesson puts that theorem to work deriving trig identities.

Deriving trigonometric identities

Method — expanding $(\cos\theta+i\sin\theta)^n$ two ways

De Moivre's theorem gives one expression for $(\cos\theta+i\sin\theta)^n$. Expanding the same power using the binomial theorem gives a second expression, in terms of powers and products of $\cos\theta$ and $\sin\theta$. Equating real parts (or imaginary parts) of these two expressions produces an identity for $\cos n\theta$ (or $\sin n\theta$) purely in terms of powers of $\cos\theta$ and $\sin\theta$.

Example 1 · Deriving a multiple-angle identity
Show that $\cos4\theta=8\cos^4\theta-8\cos^2\theta+1$.

By De Moivre's theorem, $\cos4\theta+i\sin4\theta=(\cos\theta+i\sin\theta)^4$.

Expanding the right-hand side by the binomial theorem, writing $c=\cos\theta$ and $s=\sin\theta$ for brevity:

$(c+is)^4=c^4+4c^3(is)+6c^2(is)^2+4c(is)^3+(is)^4$

$=c^4+4ic^3s-6c^2s^2-4ics^3+s^4$

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

Equating real parts: $\cos4\theta=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta$.

Using $\sin^2\theta=1-\cos^2\theta$ throughout to eliminate $\sin\theta$: with $c=\cos\theta$,

$\cos4\theta=c^4-6c^2(1-c^2)+(1-c^2)^2=c^4-6c^2+6c^4+1-2c^2+c^4=8c^4-8c^2+1$

as required.

The $z+\frac1z$ and $z^n+\frac1{z^n}$ identities

Theorem 1.2 — reciprocal identities

Let $z=\cos\theta+i\sin\theta$. Since $|z|=1$, De Moivre's theorem gives $z^n=\cos n\theta+i\sin n\theta$ and $z^{-n}=\cos n\theta-i\sin n\theta$ (using $\cos(-n\theta)=\cos n\theta$, $\sin(-n\theta)=-\sin n\theta$). Adding and subtracting these:

$$z^n+\frac1{z^n}=2\cos n\theta$$

$$z^n-\frac1{z^n}=2i\sin n\theta$$

The special case $n=1$ gives the frequently-used pair $z+\dfrac1z=2\cos\theta$ and $z-\dfrac1z=2i\sin\theta$.

Why this is useful

These identities let you express powers like $\cos^n\theta$ or $\sin^n\theta$ as a sum of cosines/sines of multiple angles, by writing $(2\cos\theta)^n=\left(z+\dfrac1z\right)^n$, expanding with the binomial theorem, and grouping matching pairs $z^k+\dfrac1{z^k}$ back into $2\cos k\theta$. This is the reverse process to Example 1, and is the key technique for integrating powers of trig functions and for summing certain trigonometric series (next lesson).

Example 2 · Expressing $\cos^6\theta$ as a sum of cosines
Show that $\cos^6\theta=\dfrac1{32}\cos6\theta+\dfrac6{32}\cos4\theta+\dfrac{15}{32}\cos2\theta+\dfrac{10}{32}$.

Let $z=\cos\theta+i\sin\theta$, so $z+\dfrac1z=2\cos\theta$, giving $\left(z+\dfrac1z\right)^6=64\cos^6\theta$.

Expand by the binomial theorem: $\left(z+\dfrac1z\right)^6=z^6+6z^4+15z^2+20+\dfrac{15}{z^2}+\dfrac6{z^4}+\dfrac1{z^6}$ (using $\binom60=1,\binom61=6,\binom62=15,\binom63=20$ and symmetry).

Group into reciprocal pairs: $=\left(z^6+\dfrac1{z^6}\right)+6\left(z^4+\dfrac1{z^4}\right)+15\left(z^2+\dfrac1{z^2}\right)+20$.

Using $z^n+\dfrac1{z^n}=2\cos n\theta$: $=2\cos6\theta+12\cos4\theta+30\cos2\theta+20$.

So $64\cos^6\theta=2\cos6\theta+12\cos4\theta+30\cos2\theta+20$, and dividing by $64$ gives $\cos^6\theta=\dfrac1{32}\cos6\theta+\dfrac6{32}\cos4\theta+\dfrac{15}{32}\cos2\theta+\dfrac{10}{32}$, as required.

Watch out — sign changes with $\sin^n\theta$

When $n$ is even, $\left(z-\dfrac1z\right)^n=(2i\sin\theta)^n$ picks up a factor of $i^n$, which is real when $n$ is even but $\pm1$ or $\pm i$ depending on $n\bmod4$ — always simplify $i^n$ carefully before comparing real parts, or you'll introduce a sign error. When $n$ is odd, the whole expression is purely imaginary, and you should compare imaginary parts (dividing through by $i$) instead of real parts.

Exam-style questions

Question 1 6 marks

Use De Moivre's theorem to show that $\sin3\theta=3\sin\theta-4\sin^3\theta$.

$\cos3\theta+i\sin3\theta=(\cos\theta+i\sin\theta)^3$. (M1)

Expanding with $c=\cos\theta,\,s=\sin\theta$: $(c+is)^3=c^3+3ic^2s-3cs^2-is^3=(c^3-3cs^2)+i(3c^2s-s^3)$. (M1, A1)

Equating imaginary parts: $\sin3\theta=3c^2s-s^3=3(1-s^2)s-s^3=3s-3s^3-s^3=3\sin\theta-4\sin^3\theta$. (M1, A1, A1)

Question 2 6 marks

Express $\sin^5\theta$ in the form $a\sin5\theta+b\sin3\theta+c\sin\theta$, where $a$, $b$ and $c$ are constants to be found.

Let $z=\cos\theta+i\sin\theta$. Then $z-\dfrac1z=2i\sin\theta$, so $\left(z-\dfrac1z\right)^5=32i^5\sin^5\theta=32i\sin^5\theta$ (using $i^5=i$). (M1, A1)

Binomial expansion: $z^5-5z^3+10z-\dfrac{10}z+\dfrac5{z^3}-\dfrac1{z^5}$, grouped as $\left(z^5-\dfrac1{z^5}\right)-5\left(z^3-\dfrac1{z^3}\right)+10\left(z-\dfrac1z\right)$. (M1, A1)

$=2i\sin5\theta-5(2i\sin3\theta)+10(2i\sin\theta)=2i\sin5\theta-10i\sin3\theta+20i\sin\theta$. (M1)

So $32i\sin^5\theta=2i\sin5\theta-10i\sin3\theta+20i\sin\theta$; dividing by $32i$: $\sin^5\theta=\dfrac1{16}\sin5\theta-\dfrac5{16}\sin3\theta+\dfrac{10}{16}\sin\theta$, so $a=\dfrac1{16},\,b=-\dfrac5{16},\,c=\dfrac58$. (A1)

Question 3 5 marks

Given $\cos4\theta=8\cos^4\theta-8\cos^2\theta+1$ (Example 1), find all solutions of $8\cos^4\theta-8\cos^2\theta+1=0$ in the interval $0\leq\theta<\pi$, giving your answers as exact multiples of $\pi$.

The equation is equivalent to $\cos4\theta=0$. (M1)

For $0\leq\theta<\pi$, $4\theta$ ranges over $[0,4\pi)$. Solving $\cos4\theta=0$: $4\theta=\dfrac\pi2,\dfrac{3\pi}2,\dfrac{5\pi}2,\dfrac{7\pi}2$ (the values in $[0,4\pi)$ where cosine is zero). (M1, A1)

So $\theta=\dfrac\pi8,\dfrac{3\pi}8,\dfrac{5\pi}8,\dfrac{7\pi}8$. (A1, A1)

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De Moivre's Theorem