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I need to graph $$y=2 \cos\left(x - \frac{\pi}{3}\right)$$ and I am having trouble figuring out what points will be on the graph. The book tells me to split it into four parts, which doesn't make sense since they split it into five parts anyways.

I know that the points will be between $\pi/3$ and $(\pi/3 + 2\pi)$ so that gives me $\pi/3 < x < 7\pi/3$

The only problem is I have no idea how to split that up into equal parts to make graphing easier.

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Indeed, it is useful to split into four parts, as follows: $$ \bigg[\frac{{2\pi }}{6},\frac{{5\pi }}{6}\bigg],\bigg[\frac{{5\pi }}{6},\frac{{8\pi }}{6}\bigg],\bigg[\frac{{8\pi }}{6},\frac{{11\pi }}{6}\bigg],\bigg[\frac{{11\pi }}{6},\frac{{14\pi }}{6}\bigg]. $$

EDIT: In response to the OP's request, here's a way to reach these numbers.

The endpoints are $\pi/3$ and $7\pi / 3$. The middle point is then given by $$ \bigg(\frac{\pi }{3} + \frac{{7\pi }}{3}\bigg)\bigg /2 = \frac{{8\pi }}{6}. $$ Next, the middle point between $\pi/3$ and $8\pi/6$ is given by $$ \bigg(\frac{\pi }{3} + \frac{{8\pi }}{6}\bigg)\bigg/2 = \bigg(\frac{{2\pi }}{6} + \frac{{8\pi }}{6}\bigg)\bigg/2 = \frac{{5\pi }}{6}, $$ and the middle point between $8\pi/6$ and $7\pi/3$ is given by $$ \bigg(\frac{8\pi }{6} + \frac{{7\pi }}{3}\bigg)\bigg/2 = \bigg(\frac{{8\pi }}{6} + \frac{{14\pi }}{6}\bigg)\bigg/2 = \frac{{11\pi }}{6}. $$

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    My book made it very confusing and used the pointed pi/3 5pi/6 4pi/3 11pi/6 and 7pi/3 How do you get those numbers though? I can't imagine a logical way to reach those numbers without mass trial and error.2011-06-13
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    Adam: $\pi/3 = 2\pi/6$, $7\pi/3 = 14\pi/6$..., the book just reduced the fractions (Shai used 6 in the denominator to help establish the four parts, you can simply reduce the fractions in those "parts" that can be reduced, and, it'll match the book.)2011-06-13
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    I'll add more details soon...2011-06-13
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    Details added...2011-06-13
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    Didn't mean to take away from the details you planned to add, Shai. My comments to Adam were getting long, and I just wanted to help him get at how you got what you got...2011-06-13
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    @amWhy: No problem, of course.2011-06-13