Find SIN given COS and Radius

markraz

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Hi is it possible to find the SIN of circle if I only know the radius and the COS?

Thanks in advance
 

pka

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Hi is it possible to find the SIN of circle if I only know the radius and the COS?
Do you mean can one find \(\sin(\theta))\) if we know \(\cos(\theta)=x~\&~ r=\rho~\)?
 

Dr.Peterson

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Hi is it possible to find the SIN of circle if I only know the radius and the COS?
You can find the sine of an angle if you only know the cosine; you probably know that identity. I imagine you mean something a little more complicated, but I can't guess what. (A circle doesn't have a sine.)
 

markraz

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You can find the sine of an angle if you only know the cosine; you probably know that identity. I imagine you mean something a little more complicated, but I can't guess what. (A circle doesn't have a sine.)
Thanks. I thought a circle had a sin and a cos like this picture?

1595991276235.png
 

Dr.Peterson

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Those are the sine and cosine of the angle in standard position, if it is a unit circle. We never talk about "the sine of a circle" (unless you have been taught something I've never seen).

Maybe you need to tell us how much trigonometry you have learned.

But if what you were asking about is how to find the length you labeled SIN if you know the radius and the length you labeled COS, then a guy named Pythagoras has some good ideas for you ...
 
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pka

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Thanks. I thought a circle had a sin and a cos like this picture?
@markraz, you cannot look at a single diagram and then generalize.
First we use polar coordinates to describe circles.
The set \(\{(\rho\cos(t),\rho\sin(t)): 0\le t\le 2\pi\}\) is a circle centered at \((0,0)\) with radius \(\rho\).
In that setup we have \(x=\rho\cos(t)~\&~y=\rho\sin(t)\) Now note that \(x^2+y^2=\rho^2\) which is the classic equation of a circle.
 

markraz

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(unless you have been taught something I've never seen).
Thanks for the reply , TBH I have never taken a real math class. I quit Highschool in 9th grade in 1981. I'm trying to teach myself all this stuff now
 

markraz

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@markraz, you cannot look at a single diagram and then generalize.
First we use polar coordinates to describe circles.
The set \(\{(\rho\cos(t),\rho\sin(t)): 0\le t\le 2\pi\}\) is a circle centered at \((0,0)\) with radius \(\rho\).
In that setup we have \(x=\rho\cos(t)~\&~y=\rho\sin(t)\) Now note that \(x^2+y^2=\rho^2\) which is the classic equation of a circle.
Ok thanks, so since if I know X and P can I just plug in the numbers use basic algebra to solve?
thanks
 

markraz

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so can I use the pythagorean trig identity? cos^2(x) + sin^2(y) = 1 or
cos^2(x) - 1 = sin^2(y) ??

thanks
 

Subhotosh Khan

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so can I use the pythagorean trig identity? cos^2(x) + sin^2(y) = 1 or
cos^2(x) - 1 = sin^2(y) ??
You have that wrong. The Pythagorean Trig Identity is:

cos^2(x) + sin^2(x) = 1

or

1 - cos^2(x) = sin^2(x)

Note that here arguments of the sine and cosine functions are same (x).
 

Dr.Peterson

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so can I use the pythagorean trig identity? cos^2(x) + sin^2(y) = 1 or
cos^2(x) - 1 = sin^2(y) ??

thanks
Or use the Pythagorean Theorem itself, x^2 + y^2 = r^2, which is also the equation of the circle. Here x would be your "COS", and y would be your "SIN". That's if you really want distances, not trig function values.

I recommend going through at least the first few chapters of a trigonometry textbook (or the trig part of a precalculus textbook), in order to get all these ideas in an orderly way.
 

markraz

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Or use the Pythagorean Theorem itself, x^2 + y^2 = r^2, which is also the equation of the circle. Here x would be your "COS", and y would be your "SIN". That's if you really want distances, not trig function values.

I recommend going through at least the first few chapters of a trigonometry textbook (or the trig part of a precalculus textbook), in order to get all these ideas in an orderly way.
Hi Thanks, this all makes sense now. Appreciate it
 
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