# Further Maths - Trig Equations with Complex Numbers

> Source: https://ollybritton.com/notes/a-level/further-maths/topics/trig-equations-with-complex-numbers/ · Updated: 2021-03-03 · Tags: school, further-maths, complex-numbers

## Flashcards
##### What is $\left(z + \frac{1}{z}\right)$??
$$
2\cos\theta
$$

##### What is $\left(z - \frac{1}{z}\right)$??
$$
2i\sin\theta
$$

##### What is $\left(z - \frac{1}{z}\right)^4$??
$$
16\sin^4\theta
$$

##### What is $\left(z + \frac{1}{z}\right)^n$??
$$
2^n\cos^n\theta
$$

##### What is $\left(z^5 + \frac{1}{z^5}\right)$??
$$
2\cos 5\theta
$$

##### What is $\left(z^n - \frac{1}{z^n}\right)$??
$$
2i\sin n\theta
$$

##### If you're trying to find out an identity for $\sin^4\theta$, what should you do??
$$
\left(z - \frac{1}{z}\right)^4
$$

##### If you're trying to find out an identity for $\sin 4\theta$, what should you do??
$$
(\cos\theta + i \sin\theta)^4
$$

##### After a lot of work expanding $(\cos \theta + i\sin\theta)$ in order to work out an identity for $\sin 4\theta$, you get $3\cos^2\theta\sin\theta - \sin^3\theta$. What's the next step??
Rewriting as

$$
3(1 - \cos^2 \theta)\sin\theta - \sin^3\theta
$$

##### How should you work out the trig identity for something like $32\cos^2\theta\sin^4\theta$??
Create both the polynomials in $z$, multiply them together and simplify.

### 2021-10-12
##### What two things would you set equal to one another to turn $\cos 6\theta$ into $32\cos^6(\theta) + ... - 1$ (multiples to powers)??
$$
(\cos\6\theta + i\sin6\theta) = (\cos \theta + i \sin \theta)^6
$$

###### Trying to use $$(\cos\6\theta + i\sin6\theta) = (\cos \theta + i \sin \theta)^6$$ to come up with an identity for $\cos 6\theta$ means there is a bunch of messy imaginary parts that you don't want. How can you fix this??
Equate the real components and ignore the imaginary stuff.

##### What two things would you set equal to one another to turn $\cos^3 \theta$ into $\frac{1}{4}\cos3\theta + \frac{3}{4}\cos\theta$??
$$
(2\cos\theta)^3 = \left(z + \frac{1}{z}\right)^3
$$

And then dividing through by $8$ at the end.

##### What two things would you set equal to one another to turn $\sin^4 \theta$ into $\frac{1}{8}\cos 4\theta - ... + \frac{3}{8}$??
$$
(2i \sin\theta)^4 = \left(z - \frac{1}{z}\right)^4

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