Question

sketch the graph of the function over the interval [0,2π], by comparing it to the graph of y=sinx

a.) y=sin2x

Answer 1

looks just like sine but period is \[\pi\] instead of
\[2\pi\]

Answer 2

why is that?

Answer 3

because as x goes from 0 to pi, 2x goes from 0 to 2pi

Answer 4

okay gotcha

Answer 5

so what will the graph of sine look like?

Answer 6

period of
\[\sin(bx)\] is
\[\frac{2\pi}{b}\]

Answer 7

sorry, why?

Answer 8

http://www.wolframalpha.com/input/?i=y%3Dsin%282x%29

Answer 9

I don't think I understand periods

Answer 10

same reason. sine is periodic with period 2pi
as x goes from 0 to 2pi/b
bx goes from 0 to 2pi

Answer 11

period of sine is 2pi means
\[\sin(x)=\sin(x+2\pi)\] you have just gone around the circle again

Answer 12

okay

Answer 13

that is why graph goes up and down and up and down and up and down like it does

Answer 14

i'm just so confused as to how I graph it knowing that it is periodic with 2π

Answer 15

here is the graph of
\[y=\sin(x)\]
http://www.wolframalpha.com/input/?i=y%3Dsin%28x%29

Answer 16

oh okay I got it because it crosses the x-axis everytime sine is 0 on the circle

Answer 17

goes through (0,0) then up to 1, then down to 0, then down to -1, then back up to 0 at 2pi, then it repeats itself

Answer 18

exactly. 0 at 0 then up to 1 at pi/2, then back to 0 at pi, then down to -1 at 3pi/2 then up to 0 at 2pi

Answer 19

oh okay I got it because it crosses the x-axis everytime sine is 0 on the circle