The prior section explored the (basic) graphs
of $\,y = \sin x\,$ and $\,y = \cos x\,$.
However, these trigonometric functions frequently
appear in ‘transformed’ versions.
You might see, for example,
$\,y = 1.5\sin \frac{1}{2}(x  \pi)\,$ or $\,y = \frac{1}{3}\cos (4x + 5)\,$.
The observations and techniques discussed in this section will allow you to reliably, quickly, and efficiently graph any function of the following forms, without memorizing a bunch of different formulas:
the ‘generalized functions’ addressed in this section: 
$y = a\sin k(x\pm b)$  $y = a\cos k(x\pm b)$ 
$y = a\sin (kx\pm B)$  $y = a\cos (kx\pm B)$ 
Notice, in particular, that the arguments $\,k(x\pm b)\,$ (first row) and $\,kx\pm B\,$ (second row) are different.
For example, $\,2(x5)\,$ is different from $\,2x  5\,$.
These input types are commonly seen in applications.
In mathematics, the word ‘argument’ is often used to refer to the input
of a function.
For example, the argument of the sine function in the expression ‘$\,\sin k(x\pm b)\,$’ is $\,k(x\pm b)\,$.
As a second example, the argument of the cosine function in the expression ‘$\,\cos (kx\pm B)\,$’ is $\,kx\pm B\,$.
Any argument with only two types of terms—an $\,x\,$term and a constant term—
can be handled with the fourstep graphing process discussed below.
With this fourstep technique, you don't need to identify which type of argument you're working with;
you only need to identify the coefficient of the $\,x\,$term.
For example, the fourstep technique can be used to graph (say) $\displaystyle\,y = 9\sin\frac{\pi + 5x}{8}\,$.
Why?
The argument has only an $x$ term and a constant term.
Here, $\,k = \frac 58\,$.
Why?
Rewrite the argument (as needed) to see that the coefficient of $\,x\,$ is $\,\color{red}{\frac 58}\,$:
$$\cssId{s27}{\frac{\pi +\color{red}{5}x}{\color{red}{8}} = \frac{\pi}8 + \color{red}{\frac{5}{8}}x}\,$$
The following two key observations will lead to an efficient approach.
Click the show/hide buttons for additional information.
GENERAL PROCEDURE  EXAMPLE: Graph $\,y = 7\cos (35x)\,$  
STEP 1: CHECK THAT $\,k\,$ IS POSITIVE The coefficient of the $\,x\,$term in the argument is denoted by $\,k\,$. Make sure that $\,k > 0\,$. If not, rewrite (as above), using the odd/even properties of sine/cosine. 
Initially, the coefficient of the $\,x\,$ term is $\,5\,$. Rewrite: $$\, \cssId{s62}{7\cos (35x)}\ \ \cssId{s63}{\underset{\text{law}}{\overset{\text{distributive}}{=}}}\ \ \cssId{s64}{7\cos\bigl((5x3)\bigr)}\ \ \cssId{s65}{\underset{\text{is even}}{\overset{\text{cosine}}{=}}}\ \ \cssId{s66}{7\cos (5x3)}$$ Instead, graph: $y = 7\cos (5x3)$ (So, $\,k = 5\,$.) 

STEP 2: STARTING POINT Each sine/cosine curve starts a ‘natural’ cycle when its argument is zero. Compute the value of $\,x\,$ that makes the argument equal to zero. This is your starting point; call it $\,S\,$ and mark it on a number line. 
Find $\,S\,$:
$$
\begin{gather}
\cssId{s74}{5x3 = 0}\cr
\cssId{s75}{5x = 3}\cr
\cssId{s76}{x = \frac{3}{5}}
\end{gather}
$$
So, $\,S = \frac{3}{5}\,$. 

STEP 3: ENDING POINT Compute the new period, $\displaystyle\,\frac{2\pi}{k}\,$. The cycle that starts at $\,S\,$ ends at $\displaystyle\,E := S +\frac{2\pi}{k}\,$. In other words, one complete cycle occurs on the interval $\,[S,\underbrace{S +\frac{2\pi}{k}}_{E}]\,$. Mark the ending point, $\,E\,$, on the number line. Divide the interval from STARTING POINT to ENDING POINT into four equal parts, to make it easier to draw in the basic cycle. 
$\,k = 5\,$ $\cssId{s85}{\displaystyle\text{period} = \frac{2\pi}{k} = \frac{2\pi}{5}}$ $\cssId{s86}{\displaystyle E := S + \frac{2\pi}{k} = \frac{3}{5} + \frac{2\pi}{5}}$ 

STEP 4: DRAW IN THE BASIC CYCLE; ADJUST (AS NEEDED) FOR AMPLITUDE/FLIP
As discussed in the next section, amplitude is the distance from the $x$axis to the highest (or lowest) part of the graph. The amplitude of $\,y = \sin x\,$ and $\,y = \cos x\,$ is $\,1\,$. Vertical scaling by a factor of $\,a\,$ causes the amplitude to change to $\,a\,$. Vertical scaling by a factor of $\,1\,$ causes reflection about the $x$axis (i.e., a ‘flip’).

Draw in the basic cosine cycle, flipped. The amplitude is $\,7\,$. To emphasize: In this graph of (one cycle of) $y = 7\cos (5x3)$:

The three earlier examples are repeated, using a compact version of the fourstep graphing process.
A new example is also given.
See how quick and easy it is!
In these examples, cycles are shown in a standard viewing window
($\,x\,$ between $\,20\,$ and $\,20\,$; $\,y\,$ between $\,10\,$ and $\,10\,$),
to make comparisons between graphs easier.
The cycle being graphed with the fourstep process is shown in dark green.
Additional cycles are shown to complete the viewing window.
Decimal approximations to the tenths place are included.
Remember to get exact answers first, then approximate!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
