How to substitute graph cosine and sine period? for example the period of a normal function is: $$\text{period} = 2\pi $$
but in our graph the period is $$ \text{period} = 8\pi $$
to substitute it we make this:
$$ f(x) =\cos\left(\frac14x\right) $$
but why it is multiplied by $1/4$, and not $4x$?
 A: The period is the least number $T$ such that $f(x) = f(x+T)$ for all $x$. So for $f(x) = \cos (x)$ the period is $2\pi$ as you say. Then you want a function $g(x) = \cos (kx)$ such that $g(x+8\pi) = \cos[ k(x+8\pi)] = g(x)$. In other words you want $k*8\pi = 2\pi$.
You see that in general, if $T$ is the new period, the coefficient $k$ is given by $k = \frac{2\pi}{T}$. In other words, the longer the period, the smaller the coefficient.
A: Because if for the period of a regular cosine is $2\pi$, we need to stretch the "transition" of cos from 0 to $8\pi$. So we make the coefficient small, so $x$ needs to be larger than $2\pi$.
A: for $a>1$, $f(ax)$ makes the function $f(x)$ compress in $x$ coordinate and for $a<1$ expands. So here you want to expand the function in $x$ coordinate, you have to multiply by $a=\frac{1}{4}<1$.
A: if you view $\cos t$ as the $x$-coordinate of a bug running around the unit circle with angular velocity one, then the bug needs $2\pi$ units of time, called the period of the motion, to go around the circle once.
if you want the bug to take as much as four times to go around, then the bug must move with a quarter of the angular speed as before. that is you need $\cos\left( \frac14 t\right).$ 
