# How do you graph y=sin(3x)?

Mar 21, 2018

Per. T = $\frac{2 \pi}{3}$
Amp. = $1$ #### Explanation:

The best thing about sinusoidal functions is that you don't have to plug in random values or make a table. There's only three key parts:

Here's the parent function for a sinusoidal graph:

color(blue)(f(x)=asin(wx) color(red)((- phi) + k) Ignore the part in red

First, you need to find the period, which is always $\frac{2 \pi}{w}$ for $\sin \left(x\right) , \cos \left(x\right) , \csc \left(x\right) , \mathmr{and} \sec \left(x\right)$ functions. That $w$ in the formula is always the term next to the $x$. So, let's find our period:

$\frac{2 \pi}{w} = \frac{2 \pi}{3}$. $\textcolor{b l u e}{\text{Per. T} = \frac{2 \pi}{3}}$

Next, we have the amplitude, which is $a$, and generally in front of the trigonometric term, and what the y-coordinates will be every other point. The amplitude can be regarded as the max and min of the graph, as seen above.

So, now we have our amplitude. $\textcolor{b l u e}{\text{Amp.} = 1}$

When you make a sinusoidal graph, the period will be four x-coordinates to the right and left.

Start with the fourth point, as seen above, which is your period, $\textcolor{b l u e}{\frac{2 \pi}{3}}$

Then go to the second point, which is half the period: $\textcolor{b l u e}{\frac{\frac{2 \pi}{3}}{2} = \frac{\pi}{3}}$

Then go to the first point, which is one fourth the period (or half the second point: $\textcolor{b l u e}{\frac{\frac{\pi}{3}}{2} = \frac{\pi}{6}}$

Now we have our five key points in terms of $\textcolor{b l u e}{\frac{\pi}{6}} :$

$\textcolor{b l u e}{\left(0 , 0\right) \left(\frac{\pi}{6} , 1\right) \left(\frac{\pi}{3} , 0\right) \left(\frac{\pi}{2} , - 1\right) \left(\frac{2 \pi}{3} , 0\right)}$

This is the same as:

$\textcolor{b l u e}{\left(0 , 0\right) \left(\frac{\pi}{6} , 1\right) \left(\frac{2 \pi}{6} , 0\right) \left(\frac{3 \pi}{6} , - 1\right) \left(\frac{4 \pi}{6} , 0\right)}$

Just notice that the top values are simplified to what the graph shows.

Another important thing to remember is that $S \in \left(x\right)$ graphs start at the origin and progress upward, unless the amplitude is negative, then they would progress downward. $C o s \left(x\right)$ graphs start at $\left(0 , \text{Amplitude}\right)$ and move downward, unless the amplitude is negative, then it would start at $\left(0 , \text{-Amplitude}\right)$ and move upward.