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

1 Answer
Alex P.
Apr 27, 2017

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.

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