# 5.3: Trigonometric Graphs

The previous section teased the graph of $\sin t$. Let's discuss the graphs of the trigonmetric functions, and apply transformations to them.

A function $f$ is periodic if there exists a positive number $p$ such that $f(t + p) = f(t)$ for every $t$ in the domain of $f$.
The period of a function is the least such positive number $p$, if it exists.
Draw a graph of a periodic function. What is the period?
The sine and cosine functions have period $2\pi$: $\sin(t + 2\pi) = \sin t \qquad\qquad \cos(t + 2\pi) = \cos t$
Before looking at the graph, why is $\sin$'s period $2\pi$?
Draw one period of $f(x) = \cos(x)$ by hand.

## Transformations of sine and cosine

Graph these three functions by hand: $f(x) = 2 + \cos x \qquad\qquad g(x) = -\cos x \qquad\qquad h(x) = 3\cos x$

#### Amplitude + period

The amplitude of a function tells you how much a function varies in it's range.

Given $k > 0$, the sine and cosine curves $y = a\sin kx \qquad\qquad y = a \cos kx$ have amplitude $\lvert a \rvert$ and period $\frac{2\pi}{k}$.
In the language of function transformations, what type of transformation is the amplitude $\lvert a \rvert$ and the number $k$?
Why is the period of $f(x) = \sin k x$ given by $\frac{2\pi}{k}$?
Because $k$ is horizontal stretch/shrink. Recall: $f(kx)$ stretches/shrinks horizontally $f(x)$ by $\frac{1}{k}$.
Find the amplitude, period, and sketch the graph by hand: $f(x) = 4\cos 3x \qquad\qquad g(x) = -2\sin\frac{1}{2}x$

#### Horizontal shift

Let's now throw in horizontal shifts. Recall: We must have the coefficient of $x$ to be $1$ in order to have a proper horizontal shift.

For $k > 0$, the following sine and cosine curves $y = a \sin k (x - b) \qquad y = a\cos k (x - b)$ have amplitude $\lvert a \rvert$, period $\frac{2\pi}{k}$, and horizontal shift $b$.
You can graph one complete period on the interval $\left[b , b + \frac{2\pi}{k}\right]$.
Find the amplitude, period, and horizontal shift of the following two functions. Graph one complete period. $y = 3 \sin 2\left(x - \dfrac{\pi}{4}\right) \qquad \qquad y = \dfrac{3}{4}\cos\left(2x + \dfrac{2\pi}{3}\right)$