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

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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2
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]}

#### Explanation:

You should think of this as three separate transformations of the function $f \left(x\right) = \sin x$.

Firstly we will write the function you are looking to sketch as $y = - \sin 2 x + 1$.

Working through from $\sin x$ the first transformation in an enlargement in the x-direction scale factor $\frac{1}{2}$ to give $\sin 2 x$. 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 $- \sin 2 x$.

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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