1. Sin and Cos Graph

2. Originally Posted by Mr_Green
The general form will be
$y = A sin(kx + \phi)$

Since A is the amplitude, this part is easy: A = 3.

$y = 3 sin(kx + \phi)$

Now, we want a 0 of the sine function to be positioned at $x = - \frac{\pi}{2}$. We can do this by setting $kx + \phi = 0$ at $x = - \frac{\pi}{2}$, so we have the equation:
$k \left ( - \frac{\pi}{2} \right ) + \phi = 0$

Again, we want a 0 of the sine function to be positioned at $x = \frac{3 \pi}{2}$. We can do this by setting $kx + \phi = \pi$ ( $\pi$ is the next zero for the sine function after 0) at $x = \frac{3 \pi}{2}$, so we have the equation:
$k \left ( \frac{3 \pi}{2} \right ) + \phi = 0$

Solving the first equation for $\phi$ gives:
$\phi = \frac{\pi}{2}k$

Inserting this into the second equation gives:
$\frac{3\pi}{2}k + \frac{\pi}{2}k = \pi$
or
$k = \frac{1}{2}$

This gives
$\phi = \frac{\pi}{4}$.

Thus
$y = 3 sin \left ( \frac{1}{2}x + \frac{\pi}{4} \right )$.

-Dan

3. Thanks!

4. Originally Posted by Mr_Green
THANKS!

FOR COS WOULD IT BE:

3cos(.5x+pi/4)

???
Close. You can do that in two ways. The first is to do it the same way I did below, except to use
$-\frac{\pi}{2}k + \phi = -\frac{\pi}{2}$ ( $-\frac{\pi}{2}$ is a convenient first zero for cos)

$\frac{3\pi}{2}k + \phi = \frac{\pi}{2}$ (the next zero for cos after $-\frac{\pi}{2}$)

The other way is to recall that $sin(\theta) = cos(\theta - \pi/2)$ so to switch to cosine we need to "shave" a $\pi/2$ off the argument.

Either way I got:
$y = 3 cos \left ( \frac{1}{2}k - \frac{\pi}{4} \right )$

-Dan