Sum of waves with equal amplitudes and frequencies but in different phases

Question

I have a problem where I need to sum four cos wave functions with equal amplitudes and frequencies, but different in phases. Presumably it should be solved using complex numbers:

$f(t) = \cos(\omega t) + \cos(\omega t + \delta) + \cos(\omega t + 2 \delta) \ + \cos(\omega t + 3 \delta)$

I know of the formula for imaginary numbers in trigonometric terms: $e^{i\theta} = \cos(\theta) + i \sin(\theta)$ and since the $\theta$ is given in the cos terms, maybe somehow it can be turned to exponential form?

Answer 1

With outside help, I found a solution which uses formulas

$ cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} $; $sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}$; $\sum_{i=0}^N r^i = \frac{1-r^{N+1}}{1-r}$ for r < 1

The original can be reexpressed as a sum of exponentials and factorized:
$\frac{1}{2}(e^{i\omega t}(1 + e^{i\delta}+ e^{2i\delta}+ e^{3i\delta})+e^{-i\omega t}(1 + e^{-i\delta}+ e^{-2i\delta}+ e^{-3i\delta}))$
Using the geometric progression formula:
$\frac{1}{2}(\frac{e^{i\omega t}(1-e^{4i\delta})}{1-e^{i\delta}} + \frac{e^{-i\omega t}(1-e^{-4i\delta})}{1-e^{-i\delta}})$
Now making the exponents equal, but with opposite signs:
$\frac{1}{2}(\frac{e^{i\omega t} e^{2i\delta}(e^{-2i\delta}-e^{2i\delta})}{e^{0.5i\delta}(e^{-0.5i\delta}-e^{0.5i\delta})} + \frac{e^{-i\omega t} e^{-2i\delta}(e^{2i\delta}-e^{-2i\delta})}{e^{-0.5i\delta}(e^{0.5i\delta}-e^{-0.5i\delta})})$
After this we can factorize again even if the sign is opposite:
$\frac{e^{2i\delta}-e^{-2i\delta}}{2(e^{0.5i\delta} - e^{-0.5i\delta})}(\frac{e^{i\omega t} e^{2i\delta}}{e^{0.5i\delta}} + \frac{e^{-i \omega t} e^{-2i\delta}}{e^{-0.5i\delta}})$
This we can already interpret with the formulas given:
$\frac{sin(2\delta)}{2sin(0.5\delta)}(e^{i\omega t + 1.5i\delta} + e^{-i\omega t - 1.5i\delta})$

Expressing the right hand side as cos and simplifying:
$\frac{sin(2\delta)}{sin(0.5\delta)}cos(\omega t + 1.5 \delta)$