Let's take this one a step at a time.

We'll start by simply drawing the graph of `y = cos\ 2x`.

`a = 1`; `"period" = (2π)/2 = π.`

The graph of `y= cos(2x)` for `0 ≤ x ≤ 2pi`

Now, let's shift the curve (we are now considering the "minus π" part of the question.

`"displacement" = -c/b = -(-π)/2 = π/2`

So we have to shift every point on the curve to the right (because phase shift is a positive number) by `pi/2`. This give us *y* = cos(2*x* - *π*), the blue dotted curve.

The graph of `y= cos(2x-pi)` for `0 ≤ x ≤ 2pi`

Now we need to consider the minus out the front of the expression *y* = −cos(2*x* - *π*). The minus will just give us a mirror image in the *x*-axis, since every positive value becomes negative and every negative value becomes positive. In other words, the minus turns it upside down.

The answer for the graph of *y* = −cos(2*x* - *π*) is the green curve:

The graph of `y= -cos(2x-pi)` for `0 ≤ x ≤ 2pi`

But wait, this is what we started with! It looks the same as `y = cos\ 2x`.

This example shows an interesting thing about phase shift and periodic functions. If you shift far enough, you can easily obtain equivalent sine or cosine expressions.

You will get a better understanding of why this works in Sums and Differences of Angles, which we meet later.

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