How to graph y = floor{cos|2x|} on [-pi, pi], y = sin(pi x) on [-2, 2]

[caspian]

Hi i'm having a little problem with these graphs.

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1. Sketch the graph of \(\displaystyle y\, =\, \lfloor \cos|2x| \rfloor\) on \(\displaystyle [-\pi,\, \pi]\)

2. Sketch the graph of \(\displaystyle y\, =\, \sin(\pi x)\) on \(\displaystyle [-2,\, 2]\)

 

[ksdhart2]

Hi i'm having a little problem with these graphs.

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1. Sketch the graph of \(\displaystyle y\, =\, \lfloor \cos|2x| \rfloor\) on \(\displaystyle [-\pi,\, \pi]\)

2. Sketch the graph of \(\displaystyle y\, =\, \sin(\pi x)\) on \(\displaystyle [-2,\, 2]\)

It'd probably be easiest to take it step-by-step and introduce one change to the graph at a time. For the first one, you (should) know what the graph of cos(x) looks like. So, what does the graph of cos(2x) look like? If you're unsure, perhaps making a chart of some points would be helpful...

• x = -pi; cos(2x) cos(2 * -pi) = ???
• x = -3pi/4; cos(2x) = cos(2 * -3pi/4) = ???
• x = -pi/2; cos(2x) = cos(2 * -pi/2) = ???
• x = -pi/4; cos(2x) = cos(2 * -pi/4) = ???
• x = 0; cos(2x) = cos(2 * 0) = ???
• x = pi/4; cos(2x) = cos(2 * pi/4) = ???
• x = pi/2; cos(2x) = cos(2 * pi/2) = ???
• x = 3pi/4; cos(2x) = cos(2 * 3pi/4) = ???
• x = pi; cos(2x) cos(2 * pi) = ???

Next, add in the absolute value sign. How does that change the graph? What does the graph of cos(|2x|) look like? Finally, add in the "floor" function (the strange not-quite brackets). This functions means to take an input and round down to the nearest integer. So floor(4.5) = 4, floor(pi) = 3, floor(sqrt(2)) = 1, etc. How does this change the graph? Then for the second function, use what you found with regard to cos(2x) to figure out how sin(pi * x) looks when compared to the graph of sin(x).

Go ahead and give these problems another go. If you get stuck again, that's alright, but when you reply back, please include any and all efforts you've made on this problem. Please show us the attempt(s) you've made so we can critique where you might have gone wrong. Thank you.