Using Euler’s and De Moivre’s formulas with multiple angles and powers of cos or sin

eulerf

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 = cos θ + i sin θ and how to work with it).


equations-euler-3-titanium-ring-formulas

Example: cos3θ and cos 3θ

The fundamental equation here is Euler’s amazing formula

E: e = cos θ + i sin θ


To work out cos3θ, 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 + e−iθ, or cos θ = ½(e + e−iθ) (C)

E − E* ⇒ 2i sin θ = e − e−iθ, or sin θ = 12i (e − e−iθ) (S)

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

cos3θ = (½)3 (e + e−iθ)3

Expand the right hand side using Pascal’s triangle

cos3θ = (½)3 (e3iθ + 3e2iθ.e−iθ + 3e.e−2iθ + e−3iθ)

Now use index laws to simplify

cos3θ = (½)3 (e3iθ + 3e + 3e−iθ + e−3iθ)

Now group terms so we can use C again

cos3θ = (½)3 (e3iθ + e−3iθ + 3[e + e−iθ])

Using C again

cos3θ = (½)3 (2cos3θ + 6cosθ)

  = ¼ (cos3θ + 3 cosθ)

The textbook substitutes z for e in this working, I suppose to make it simpler. You can do that if you like, but you have to switch from e 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 = e3iθ = (e)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 = cos3θ + 3i cos2θ.sinθ − 3 cosθ.sin2θ − i sin3θ

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

cos 3θ = cos3θ − 3 cosθ.sin2θ

You can then use cos2θ + sin2θ = 1 to get the right−hand side all in terms of cos θ

cos 3θ = cos3θ − 3 cosθ.(1 − cos2θ) = 4 cos3θ − 3 cosθ