Complex exp. Fourier series, finding $x(t)$ when $X(j\omega)$ is given as magnitude and phase plot
The video could've stopped at the answer:

$x(t)=4+ 2e^{j(3\omega t + \pi /2)} + 4e^{j(4\omega t - \pi /2)} + 2e^{-j(3\omega t + \pi /2)} + 4e^{-j(4\omega t - \pi /2)}$

And this answer can be read directly from the plots. The idea is that the $n^{th}$ term in the sum is equal to $|C_n|e^{j(n\omega t+\angle C_n)}$. You get this from the Fourier series equation which is $x(t)=\sum_n C_n e^{jn\omega t}$, where $C_n$ may be a complex coefficient meaning it has a magnitude and a phase. Re-write $C_n$ as $C_n = |C_n|e^{j\angle C_n}$, plug into the Fourier series equation and you get what I wrote above. 

The extra steps that the video does is just to convert the complex exponential expression into cos/sin functions using Euler's formula. One result of Euler's formula is that $2cos(\theta)=e^{j\theta}+e^{-j\theta}$, and that is why in the end it turns out to be cosine and not sine.