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

[btb4198]

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:

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

 

[Office_Shredder]

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.

 

[btb4198]

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:

 

[btb4198]

I need it to go down on 90 and not 180

 

[Office_Shredder]

Try multiplying by $\pi/180$ instead of $180/\pi$ inside of the sine function to get the right conversion to degrees.

 

[btb4198]

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:

is it wrong then?
how you were right!
my graph is working on

 

[pasmith]

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 <br /> \begin{align*}<br /> \cos^2 x+ \sin^2 x &amp;= 1 \\<br /> \cos^2 x - \sin^2 x &amp;= \cos(2x)<br /> \end{align*}<br /> it follows that \sin^2 x = (1 - \cos (2x))/2. That's not what you want.

It sounds like what you want is y(x) = \frac{A}{2}\left(1 + \sin\left( \frac{\pi x}{180}\right)\right).

 

[btb4198]

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

It sounds like what you want is y(x) = \frac{A}{2}\left(1 + \sin\left( \frac{\pi x}{180}\right)\right).

No
Office_Shredder was right...
I got the graph I wanted

 

[btb4198]

Office_Shredder I posted a follow up question here:
https://www.physicsforums.com/threads/mathematically-modeling-a-real-system-in-c.1007740/
can you take a look at it ?
I did not know if it should go under here too or under computer science