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Period of a Graph
- Thread starter quader
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HallsofIvy
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- Jan 27, 2012
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The period of cos(x) is \(\displaystyle 360\) degrees (it is hard to do this in degrees rather than radians!) so the period of cos(ax) is \(\displaystyle 360\) divided by a. But this problem has \(\displaystyle cos^2(x)\). cos(x+ 180)= -cos(x) so that \(\displaystyle cos^2(x+ 180)= cos^2(x)\).
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HallsofIvy
Elite Member
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- Jan 27, 2012
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No, the period is NOT 180 degrees and I did not say it was!
As for the maximum, a fraction, with a fixed numerator will be largest when the denominator is smallest. What is the smallest possible value of \(\displaystyle 1+ cos^2(2x)[/itex]?\)
As for the maximum, a fraction, with a fixed numerator will be largest when the denominator is smallest. What is the smallest possible value of \(\displaystyle 1+ cos^2(2x)[/itex]?\)
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D
Deleted member 4993
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.Write down the period of the graph of View attachment 4482.... please fix your function. You know x0 = 1
and also the coordinates of a maximum value of y.
So, I thought the period was 360 or 2Pi divided by the angle, but I'm not sure what to do here.