Sinusoidal Functions

Start with $f(\theta )=\sin \theta$. Stretch its graph, squeeze it, slide it up or down or sideways, and whatever you get is a sinusoidal function. Those moves are the additive and multiplicative transformations you met in Unit 1, applied to sine. In topic 3.6 you will see each one written into the form $f(\theta )=a\sin (b(\theta +c))+d$.

Cosine is already in the family. Slide the sine graph $\frac{\pi }{2}$ units to the left and it lands exactly on the cosine graph:

Test one input. At $\theta =0$ the right side is $\sin \frac{\pi }{2}=1$, and $\cos 0=1$. So cosine is sine with a phase shift, and both are sinusoidal.

Where the two graphs differ is symmetry. The sine graph has rotational symmetry about the origin: $\sin (-\theta )=-\sin \theta$, so sine is an odd function. The cosine graph has reflective symmetry over the $y$-axis: $\cos (-\theta )=\cos \theta$, so cosine is an even function.

A sinusoidal graph swings between a highest output, its maximum, and a lowest output, its minimum. Two numbers come straight from those.

The amplitude is half the difference between the maximum and the minimum. The midline is the horizontal line through their average:

In the diagram the graph tops out at 8 and bottoms out at 2. The amplitude is $\frac{8-2}{2}=3$ and the midline is $y=\frac{8+2}{2}=5$. Check it: $5+3=8$ and $5-3=2$. The graph climbs 3 over the midline and drops 3 under it.

So amplitude is the distance from the midline to a maximum. It is not the distance from a minimum to a maximum; that is twice the amplitude. Amplitude is a distance, so it is never negative.

For $\sin \theta$ and $\cos \theta$ the maximum is 1 and the minimum is $-1$. That makes the amplitude 1 and the midline $y=0$, the horizontal axis.

A sinusoidal graph repeats itself. The period is the horizontal length of one full cycle. Read it from a maximum to the next maximum, or from a minimum to the next minimum. In the diagram the maxima sit at $t=3$ and $t=15$, so the period is $15-3=12$.

A maximum to the next minimum is only half a cycle. In the diagram that runs from $t=3$ to $t=9$, which is 6, half of 12.

For $\sin \theta$ and $\cos \theta$ the period is $2\pi$. One full turn around the unit circle brings the point back to where it started, so the outputs start over.

The frequency is the reciprocal of the period:

It tells you what fraction of a cycle the graph completes per unit of input. The diagram's function has period 12, so its frequency is $\frac{1}{12}$. Sine and cosine have frequency $\frac{1}{2\pi }$. A shorter period means a higher frequency, and the period times the frequency is always 1.

Follow $\sin \theta$ through one period, from $0$ to $2\pi$. It starts on the midline, rises to its maximum at $\frac{\pi }{2}$, falls back through the midline at $\pi$ to its minimum at $\frac{3\pi }{2}$, and climbs back to the midline at $2\pi$. So it is increasing on $(0,\frac{\pi }{2})$ and $(\frac{3\pi }{2},2\pi )$, and decreasing on $(\frac{\pi }{2},\frac{3\pi }{2})$.

Watch how it bends, too. Around a maximum the graph is concave down, curving like an upside-down bowl. Around a minimum it is concave up. As the input increases, every sinusoidal graph switches back and forth between concave down and concave up.

The switch happens on the midline. The graph is concave down wherever it is higher than its midline and concave up wherever it is lower, so each midline crossing is a point of inflection. For $\sin \theta$ those are at $0$, $\pi$ and $2\pi$. In the diagram they are at $t=0$, $6$ and $12$, each halfway between a maximum and the neighboring minimum.

Increasing and concavity are separate questions. Between $\frac{\pi }{2}$ and $\pi$, sine is decreasing and concave down; between $\pi$ and $\frac{3\pi }{2}$ it is still decreasing, but concave up.