Hi, how can I divide 2pi into four equal parts? :)

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- Jun 25th 2005, 11:35 PMcoopsterdudeDividing 2pi
Hi, how can I divide 2pi into four equal parts? :)

- Jun 25th 2005, 11:41 PMMath HelpPlease clarify
I assume you do not simply mean $\displaystyle \frac{2\pi}{4}$

Do you mean cutting an angle into four equal parts?

2 pi radians implies a circle. You could bisect the circle and then bisect each portion once more. - Jun 26th 2005, 05:01 PMcoopsterdude
I am trying to follow the logic of an example in my book. The example is Using Key Points to Sketch a Sine Curve.

Sketch the graph of y = 2 sib x on interval [-pi, 4pi].

it goes on to say: Note that y = 2 sub x = 2(sin x) indicates that the y-values for the key points will have twice the magnitude of the graph of y = sin x. Divide the preiod 2pi into four equal parts to get a set of key points. The points are: (0,0) , (pi/2,2) , (pi,0) , (3pi/-2) , and (2pi,0)

How did the book come up with this set of points. Do I have to use a graphing calulator to find these points? Can I do this without a calculator?

Thanks for any help with this one. - Jun 26th 2005, 11:19 PMticbol
Let us forget about the

y = 2 sib x

and

y = 2 sub x

They just add to the confusion.

(What are those, anyway? )

You mean

y = 2 sin x

for the key points?

Well, that sure means the y-values for the key points are twice those corresponding y-values on the y = sin x.

Umm, wait a minute,.....you mean

y sub 2 = 2 sin x?

Or,

y_2 = 2 sin x?

Because y = sin x?

And so, y_2 = 2*y = 2 sin x.

Figures.

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And then you are asking how the book got the points

(0,0), (pi/2,2), (pi,0), (3pi/2, -2) and (2pi,0)

for y sub 2?

-----------------

In graphing y = sin x, usually the points at x = 0, pi/2, pi, 3pi/2 and 2pi are plotted on the x,y axes.

(Like the 2pi was divided into 4 equal parts.

2pi/4 = pi/2.

So, 1st part is from 0 to pi/2

2nd part is from pi/2 to pi

3rd part is from pi to 3pi/2

4th part is from 3pi/2 to 2pi.)

Then,

when x=0, y = sin(0) = 0. ...............so point (0,0)

when x=pi/2, y = sin(pi/2) = 1. .......so point (pi/2,1)

when x=pi, y = sin(pi) = 0. ..............so point (pi,0)

when x=3pi/2, y = sin(3pi/2) = -1. ....so point (3pi/2,-1)

when x=2pi, y = sin(2pi) = 0. ...........so point (2pi,0)

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Now, since y sub 2 = 2*y, then,

at x=0, y sub 2 = 2*0 = 0 .............so point (0,0)

at x=pi/2, y sub 2 = 2*1 = 2 ..........so point (pi/2,2)

at x=pi, y sub 2 = 2*0 = 0 .............so point (pi,0)

at x=3pi/2, y sub 2 = 2*-1 = -2 ......so point (3pi/2,-2)

at x=2pi, y sub 2 = 2*0 = 0 ............so point (2pi,0)

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That is how. - Jun 27th 2005, 08:13 PMcoopsterdude
Thank you. That helps me so much. I appreciate all the information.