See also these videos

(the bit about writing c and s as abbreviations for cos θ and sin θ is a good idea, I think)

(Despite what the video says, I really don’t think you need to *memorise* anything here: just know the Euler formula e^{iθ} = cos θ + i sin θ and how to work with it).

**Example: cos ^{3}θ and cos 3θ**

The fundamental equation here is Euler’s amazing formula

**E: e ^{iθ} = cos θ + i sin θ**

To work out cos

^{3}θ, for example:

*First, reverse-engineer E to get equations for cos θ and sin θ*

E ⇒ E*: e^{−iθ} = cos θ − i sin θ

*You can see this either from cos −θ = cos θ and sin −θ = −sin θ, or from the symmetry of the complex numbers: any general equation remains true when you replace i throughout by −i.*

E + E* ⇒ 2 cos θ = e^{iθ} + e^{−iθ}, or cos θ = ½(e^{iθ} + e^{−iθ}) (**C**)

E − E* ⇒ 2i sin θ = e^{iθ} − e^{−iθ}, or sin θ = ^{1}⁄_{2i} (e^{iθ} − e^{−iθ}) (**S**)

*Now use ***C*** − which you will remember without effort after enough practice*

cos^{3}θ = (½)^{3} (e^{iθ} + e^{−iθ})^{3}

*Expand the right hand side using Pascal’s triangle*

cos^{3}θ = (½)^{3} (e^{3iθ} + 3e^{2iθ}.e^{−iθ} + 3e^{iθ}.e^{−2iθ} + e^{−3iθ})

*Now use index laws to simplify*

cos^{3}θ = (½)^{3} (e^{3iθ} + 3e^{iθ} + 3e^{−iθ} + e^{−3iθ})

*Now group terms so we can use C again*

cos^{3}θ = (½)^{3} (e^{3iθ} + e^{−3iθ} + 3[e^{iθ} + e^{−iθ}])

*Using C again*

cos^{3}θ = (½)^{3} (2cos3θ + 6cosθ)

= ¼ (cos3θ + 3 cosθ)

*The textbook substitutes z for e ^{iθ} in this working, I suppose to make it simpler. You can do that if you like, but you have to switch from e^{iθ} to z and then back again, which adds a complication. With enough practice, you get used to the grouping of terms, and you can make this working shorter.*

To work out cos 3θ

You use **E** directly, without the reverse-engineering

E ⇒ cos 3θ + i sin 3θ = e^{3iθ} = (e^{iθ})^{3} = (cos θ + i sin θ)^{3} (**De Moivre**)

*(Or, rather, the equation above is De Moivre’s formula for n=3. De Moivre’s formula is the general one with n instead of 3)*

*Again, use Pascal’s triangle to multiply out the right−hand side*

cos 3θ + i sin 3θ = cos^{3}θ + 3i cos^{2}θ.sinθ − 3 cosθ.sin^{2}θ − i sin^{3}θ

*Equate real parts to find cos 3θ. You could equate imaginary parts to find sin 3θ*

cos 3θ = cos^{3}θ − 3 cosθ.sin^{2}θ

*You can then use cos ^{2}θ + sin^{2}θ = 1 to get the right−hand side all in terms of cos θ*

cos 3θ = cos^{3}θ − 3 cosθ.(1 − cos^{2}θ) = 4 cos^{3}θ − 3 cosθ