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

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources


Write a one sentence answer...



Explain in detail...


I want someone to double check my answer

Describe your changes (optional) 200

Alex P. Share
Apr 27, 2017


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


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.

Was this helpful? Let the contributor know!