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(2x - 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(2x - 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(2x - 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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