Further Maths - Trig Equations with Complex Numbers

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?

sin-four-theta-substitute-pythagoras

Rewriting as

\[3(1 - \cos^2 \theta)\sin\theta - \sin^3\theta\]

How should you work out the trig identity for something like $32\cos^2\theta\sin^4\theta$?

mixed-power-identity-multiply-polynomials

Create both the polynomials in $z$, multiply them together and simplify.

Using de Moivre for multiple-angle identities

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-six-theta-de-moivre-setup

\[(\cos 6\theta + i\sin6\theta) = (\cos \theta + i \sin \theta)^6\]

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?

cos-six-theta-equate-real-parts

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 $\frac{1}{4}\cos3\theta + \frac{3}{4}\cos\theta$?

cos-cubed-identity-setup

\[(2\cos\theta)^3 = \left(z + \frac{1}{z}\right)^3\]

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 $\frac{1}{8}\cos 4\theta - ... + \frac{3}{8}$?

sin-fourth-identity-setup

\[(2i \sin\theta)^4 = \left(z - \frac{1}{z}\right)^4\]

Flashcards

The $z \pm 1/z$ identities

Powers of sin and cos to multiple angles

Using de Moivre for multiple-angle identities

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