Summation of trigonometric functions So consider a summation of ai cos (x + phi_i) where i ranges from 1 to N. Could we describe this summation as a single cosine function? Or the sum of two cosine or sine functions? How would we do this and what would be the amplitude of the cosine function in that case?
 A: Each $\cos(x+\phi_i)$ breaks into a multiple of $\cos x$ and $\sin x$, so the sum can be re-written as:
$$\begin{align}
\sum_{i=1}^N a_i \cos(x+\phi_i) &= \sum_{i=1}^N a_i\left(\cos x \cos\phi_i - \sin x\sin \phi_i \right) \\
&= \cos x \sum_{i=1}^N a_i \cos\phi_i - \sin x \sum_{i=1}^N a_i \sin \phi_i \\
&=: p \cos x - q \sin x
\end{align}$$
So, the answer to the question "Could we describe this summation as [...] the sum of a sine and cosine function [of $x$]?" is: YES!
Now, it would be awfully convenient if you could think of $p$ and $q$ as (a multiple of) the cosine and sine of an angle, because then you could use the angle addition formula to collapse the terms. That is, if ...
$$p = r \cos \theta \qquad q = r \sin \theta \qquad \qquad (1)$$
... then ...
$$p \cos x - q \sin x = r \cos x \cos \theta - r \sin x \sin\theta = r \left( \cos x \cos\theta - \sin x \sin\theta \right) = r \cos(x+\theta)$$
In that case, the answer to the question "Could we describe this summation as a single cosine function?" is also: YES! ... provided we can find the appropriate $r$ and $\theta$ to satisfy $(1)$. But that's easy when you think about how those values have to work.

Since $\sin^2 + \cos^2 = 1$, equation $(1)$ requires that
$$p^2 + q^2 = r^2 \cos^2\theta + r^2 \sin^2\theta = r^2 (1) = r^2$$
Therefore, if $r$ is to be anything, it has to be 
$$r = \pm \sqrt{p^2 + q^2}$$
Since we just need some $r$, we can go ahead and take the "$\pm$" to be "$+$".
This answers the question "[W]hat would be the amplitude of the cosine function [...]?"

Moreover, since $\sin/\cos = \tan$, equation $(1)$ also requires that
$$\frac{q}{p} = \frac{r \sin\theta}{r \cos\theta} = \tan\theta$$
so that, if $\theta$ is going to be anything, it has to be
$$\theta = \operatorname{atan}\frac{q}{p} + k \pi \quad\text{for some integer}\quad k$$ 
And since we just need some $\theta$, we can go ahead and take $k$ to be $0$.
Although you didn't ask for it, this gives you the phase shift of the cosine function.

The relation 
$$p \cos x - q \sin x = r \cos \left(x + \theta\right)$$
can be illustrated nicely with a diagram:

