Question

How do you graph sine and cosine functions when it is translated?

Answer 1

I think you'll find a useful answer here: http://socratic.org/trigonometry/graphing-trigonometric-functions/translating-sine-and-cosine-functions

Vertical translation

Graphing `y=sinx+k` Which is the same as `y=k+sinx`:

In this case we start with a number (or angle) `x`. We find the sine of `x`, which will be a number between `-1` and `1`. The after that, we get `y` by adding `k`. (Remember that `k` could be negative.)

This gives us a final `y` value betwee `-1+k` and `1+k`.

This will translate the graph up if `k` is positive (`k>0`)
or down if `k` is negative (`k<0`)

Examples:

`y=sinx`
graph{y=sinx [-5.578, 5.52, -1.46, 4.09]}

`y=sinx+2 = 2+sinx`
graph{y=sinx+2 [-5.578, 5.52, -1.46, 4.09]}

`y=sinx-4=-4+sinx`
graph{y=sinx-4 [-5.58, 5.52, -5.17, 0.38]}

The reasoning is the same for `y=cosx+k=k+cosx`, but the starting graph looks different, so the final graph is also different:

`y=cosx`
graph{y=cosx [-5.578, 5.52, -1.46, 4.09]}

`y=cosx+2 = 2+cosx`
graph{y=cosx+2 [-5.578, 5.52, -1.46, 4.09]}

Answer 2

Horizontal Translation

One way to think about horizontal translations of a function is to think about the value of `x` that will cause us to find `f(0)`.

We know the graph of `y=f(x)=sin(x)`.

To graph `y=sin(x-4)`, we can think of it as graphing `y=f(x-4)`.
Now, what value of `x` will make me find `f(0)=sin(0)`? Clearly, it is `x=4`.
So "`4` is the new `0`". Everything moves `4` to the right.

graph{y=sin(x-4) [-0.498, 7.295, -2.302, 1.596]}

To graph `y=sin(x+ pi/3)`, we ask ourselves, "What value of `x` will cause us to find `sin(0)`?
That will be `x=- pi/3`
So, "`- pi/3` is the new `0`". Everything moves `- pi/3` to the right. Wait a minute, surly it's more clear to say: Everything moves `pi/3` to the left.

(For the graph below, remember that `pi/3` is a little more than `1`)

graph{y=sin(x+pi/3) [-3.02, 1.845, -1.192, 1.241]}

To start the graph of `y=sin(bx- c)` , we'll need to change the period and also translate.

Cosine

The reasoning for graphing `y=cos(x-h)` is the same. The difference is that
`sin(0)=0`, so the point corresponding to the 'new `0`' goes on the `x`-axis
`cos0=1`, so the point corresponding to the 'new `0`' goes at `y=1`

`y = cos (x+ pi/3)`
graph{y=cos(x+pi/3) [-3.02, 1.845, -1.192, 1.241]}