Question

How do you graph `y=1-sin2x` over the interval `0<=x<=360`?

Answer 1

This is what the graph will look like.
graph{y=1-sin(2x) [-0.734, 4.852, -0.402, 2.39]}

Explanation:

You should think of this as three separate transformations of the function `f(x) = sinx`.

Firstly we will write the function you are looking to sketch as `y=-sin2x+1`.

Working through from `sinx` the first transformation in an enlargement in the x-direction scale factor `1/2` to give `sin2x`. Effectively this increases the frequency by 2.

Now we deal with the negative. Since this is outside the function it affects the y-axis, in this case a reflection about the x-axis. We now have `-sin2x`.

Finally we must make an addition outside the function, this again affects the y-axis and in this case is a translation vertically by 1.