Equation to graph a sine wave that acts like a point on a unit circle

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  • #1
btb4198
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I need an equation to graph a sine wave that act like a unit circle but only positive numbers.
so I need it to be 0 at 0, A at 90 , 0 at 180, A at 270, 0 at 360, and A at 450 and so on and so on...

Now I know sin(0) is 0 in degrees and sin(90) 1
and I know if you Square a number is will always get you a positive number
so Y = A* sin(x)^2 should work, but x has to be in degrees, so you have to convert from radians to degrees.

with is 180/π

so now I have
Y = A * sin(x*(180/π) ^2
but this is not working on my computer's graphing Calculator:
1633288714715.png


I made A = 5
I tried doing adding - 90
Y = A * sin(x*(180/π -90) ^2

but that did not work
I never trying adding 90, - 360 and 360 nothing is working.
why is this ? what am I missing ?
ago the graph should be 0 on 0, 180 , and 360 and should be A( in this case 5) only on 90, 270 and 450 and so on and so on
 
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Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
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Do you just want something like ##|sin(x)|##?

Also, your calculator probably assumes x is in radians to begin with, so when you try to convert it to degrees, you're shortening the period a lot more than you intended.
 
  • #3
btb4198
569
10
Do you just want something like ##|sin(x)|##?

Also, your calculator probably assumes x is in radians to begin with, so when you try to convert it to degrees, you're shortening the period a lot more than you intended.
no that give me this:
1633290723767.png
 
  • #4
btb4198
569
10
I need it to go down on 90 and not 180
 
  • #5
Office_Shredder
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Try multiplying by ##\pi/180## instead of ##180/\pi## inside of the sine function to get the right conversion to degrees.
 
  • #6
btb4198
569
10
Try multiplying by ##\pi/180## instead of ##180/\pi## inside of the sine function to get the right conversion to degrees.
Thanks !
that work.
really weird google said this:
1633296394316.png


is it wrong then?
how you were right!
my graph is working on
 
  • #7
pasmith
Homework Helper
2022 Award
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I need an equation to graph a sine wave that act like a unit circle but only positive numbers.
so I need it to be 0 at 0, A at 90 , 0 at 180, A at 270, 0 at 360, and A at 450 and so on and so on...

Now I know sin(0) is 0 in degrees and sin(90) 1
and I know if you Square a number is will always get you a positive number
so Y = A* sin(x)^2 should work, but x has to be in degrees, so you have to convert from radians to degrees.

with is 180/π

so now I have
Y = A * sin(x*(180/π) ^2
but this is not working on my computer's graphing Calculator:
View attachment 290112

I made A = 5
I tried doing adding - 90
Y = A * sin(x*(180/π -90) ^2

but that did not work
I never trying adding 90, - 360 and 360 nothing is working.
why is this ? what am I missing ?
ago the graph should be 0 on 0, 180 , and 360 and should be A( in this case 5) only on 90, 270 and 450 and so on and so on

From the basic trig identities [tex]
\begin{align*}
\cos^2 x+ \sin^2 x &= 1 \\
\cos^2 x - \sin^2 x &= \cos(2x)
\end{align*}
[/tex] it follows that [itex]\sin^2 x = (1 - \cos (2x))/2[/itex]. That's not what you want.

It sounds like what you want is [tex]y(x) = \frac{A}{2}\left(1 + \sin\left( \frac{\pi x}{180}\right)\right).[/tex]
 
  • #8
btb4198
569
10
From the basic trig identities [tex]
\begin{align*}
\cos^2 x+ \sin^2 x &= 1 \\
\cos^2 x - \sin^2 x &= \cos(2x)
\end{align*}
[/tex] it follows that [itex]\sin^2 x = (1 - \cos (2x))/2[/itex]. That's not what you want.

It sounds like what you want is [tex]y(x) = \frac{A}{2}\left(1 + \sin\left( \frac{\pi x}{180}\right)\right).[/tex]
No
Office_Shredder was right...
I got the graph I wanted
 

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