Further Maths - Trig Equations with Complex Numbers

Created: March 03, 2021

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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