Graphing trigonometric functions

**Wolvenmoon** · September 15th 2009, 10:32 AM

Okay, so my book spends a whole lot of paper and ink talking about these, but it never goes and explicitly breaks down what's going on.

What would help my understanding immensely is if someone took the following functions and broke down what each variable did on the graph.

$Y= c + a sin 2\pi/b (x - d)$
$Y= C + a tan \pi/b (x-d)$

and the equivelents for the secant and cosecant functions. (cos and cot behave the same way as sin and tangent if I'm getting anything at all from the book) I'm having a lot of trouble because my book and the online video lectures split the sections up differently and emphasize different parts.

**Grandad** · September 15th 2009, 10:54 PM

Hello Wolvenmoon

Originally Posted by Wolvenmoon

Okay, so my book spends a whole lot of paper and ink talking about these, but it never goes and explicitly breaks down what's going on.

What would help my understanding immensely is if someone took the following functions and broke down what each variable did on the graph.

$Y= c + a sin 2\pi/b (x - d)$
$Y= C + a tan \pi/b (x-d)$

and the equivelents for the secant and cosecant functions. (cos and cot behave the same way as sin and tangent if I'm getting anything at all from the book) I'm having a lot of trouble because my book and the online video lectures split the sections up differently and emphasize different parts.

I'll show you how the first one works, and leave you to sort out the second for yourself - it's basically similar.

In the attachments, I've built up the function step by step.

The first shows $y=\sin(2\pi x)$ . This is a single cycle of a sine wave for values of $x$ from $0$ to $1$ . Note that $y$ has values between $\pm1$ .

The second shows $y=\sin\Big(\frac{2\pi}{4}\, x\Big)$ ; in other words $y=\sin\Big(\frac{2\pi}{b}\, x\Big)$ , with $b =4$ . You'll see that the difference is that $x$ now needs to take values from 0 to 4 to give one complete cycle. So $b$ gives the wavelength of the sine wave.

The third graph shows $y=\sin\Big(\frac{2\pi}{4} (x-.05)\Big)$ . So I've given $d$ the value $0.5$ in your formula $y=c+a\sin\Big(\frac{2\pi}{b} (x-d)\Big)$ . If you look closely, you'll see that the graph has been shifted $0.5$ units to the right. So $d$ gives the phase-shift.

Graph #4 is $y=3\sin\Big(\frac{2\pi}{4} (x-0.5)\Big)$ . So I've put $a=3$ . You'll notice that $y$ now takes values between $\pm3$ . So $a$ is the amplitude of the sine wave.

Finally, I've shown you the graph of $y=1+3\sin\Big(\frac{2\pi}{4} (x-.05)\Big)$ . So I've put $c=1$ . You'll see that $y$ now takes values between $-2$ and $+4$ , the graph having been shifted upwards by $1$ unit. $c$ , then, is the vertical shift.

Grandad