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.

The

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.

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

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:
\[f(x) = 4\cos 3x \qquad\qquad g(x) = -2\sin\frac{1}{2}x\]

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 $\left[b , b + \frac{2\pi}{k}\right]$.

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.
\[y = 3 \sin 2\left(x - \dfrac{\pi}{4}\right) \qquad \qquad y = \dfrac{3}{4}\cos\left(2x + \dfrac{2\pi}{3}\right)\]