How to graph sinusoidal functions I understand how to graph sinusoidal functions, but how do you decide to choose an input? For $\cos(x)$, people choose $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$, etc. but for $\cos(4x)$, choosing those same inputs would give the outputs: $1, 1, 1, 1$, etc. to get the outputs $0$ and $-1$, you need other inputs. How do I decide/find those inputs?
 A: A possible hint may be: The numbers as $$x=k\pi/8, k\in\mathbb{Z}$$ would get you out of problems.
A: Consider the graph of the cosine function $f: \mathbb{R} \to [-1, 1]$ defined by $f(x) = \cos x$.

Observe that at $0$, the function obtains its maximum value of $1$.  Its value falls to $0$ at $x = \pi/2$, continues to decrease to its minimum value of $-1$ at $x = \pi$, increases to $0$ at $x = 3\pi/2$, and continues to increase to its maximum value of $1$ at $x = 2\pi$.  The cycle then repeats itself.  In fact, the graph repeats itself every $2\pi$ radians.  
If there exists a positive number $p$ such that $f(x) = f(x + p)$ for each $x$ in its domain, then we say that the function $f$ is periodic. If there exists a smallest such $p$, the function is said to have period $p$.
The cosine function has period $2\pi$.
The function $g: \mathbb{R} \to [-1, 1]$ defined by $g(x) = \cos(4x)$ has frequency $4$, meaning that it complete four full cycles in one period of the cosine function 
$f: \mathbb{R} \to [-1, 1]$ defined by $f(x) = \cos x$.  Consequently, it has period 
$$T = \frac{2\pi}{4} = \frac{\pi}{2}$$
Notice that $g(4 \cdot 0) = g(0) = 1$.  Thus, the function $g$ also assumes it maximum value of $1$ at $x = 0$.  The value of the function $g(x) = \cos(4x)$ then decreases to $0$, which it reaches when
$$4x = \frac{\pi}{2} \implies x = \frac{\pi}{8}$$
It continues to decrease to its minimum value of $-1$, which it reaches when
$$4x = \pi \implies x = \frac{\pi}{4}$$
The function then increases to $0$, which it reaches when 
$$4x = \frac{3\pi}{2} \implies x = \frac{3\pi}{8}$$
The function continues to increase to its maximum value of $1$, which it reaches when
$$4x = 2\pi \implies x = \frac{\pi}{2}$$
Since $g$ has period $\pi/2$, the graph then repeats itself, as shown below.

Notice that we have divided a period of the graph into four subintervals by determining the values at which the cosine function has its maximum and minimum values and $x$-intercepts, then used periodicity to draw the graph.  
Alternatively, we could solve for the maxima, minima, and $x$-intercepts of $g(x) = \cos(4x)$, then draw a smooth curve through the points we obtain. 
maxima: 
\begin{align*}
\cos(4x) & = 1\\
\cos(4x) & = \cos(2n\pi), n \in \mathbb{Z}\\
4x & = 2n\pi, n \in \mathbb{Z}\\
x & = \frac{n\pi}{2}, n \in \mathbb{Z}
\end{align*}
minima: 
\begin{align*}
\cos(4x) & = -1\\
\cos(4x) & = \cos(\pi + 2n\pi), n \in \mathbb{Z}\\
4x & = \pi + 2n\pi, n \in \mathbb{Z}\\
x & = \frac{\pi}{4} + \frac{n\pi}{2}, n \in \mathbb{Z}
\end{align*}
$x$-intercepts:
\begin{align*}
\cos(4x) & = 0\\
\cos(4x) & = \cos\left(\frac{\pi}{2} + n\pi\right), n \in \mathbb{Z}\\
4x & = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}\\
x & = \frac{\pi}{8} + \frac{n\pi}{4}, n \in \mathbb{Z}
\end{align*}
A: You see that the inputs you gave as examples for $\cos x$ showed what might be considered as "interesting" values such as the maximum and minimum points as well as the roots (where the graph crosses the $x$ axis). Well when you draw any graph try to choose those points. So for $\cos (4x)$ do the same.
To find where the graph crosses the $x-$axis set $y=0$ and solve for $x$.
To find the maxima/minima/inflexions differentiate and set equal to zero and solve for $x$. 
