How would you go about finding the figures to sketch a circular function?

Attachment 28429

Thanks

James

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- May 22nd 2013, 03:56 AMJamesTCircular function
How would you go about finding the figures to sketch a circular function?

Attachment 28429

Thanks

James - May 22nd 2013, 04:04 AMProve ItRe: Circular function
I'd start by remembering that $\displaystyle \displaystyle \cos{ \left( X + \pi \right) } = -\cos{(X)}$.

- May 22nd 2013, 04:07 AMMINOANMANRe: Circular function
see the figure below

the rest is easy

Attachment 28431 - May 22nd 2013, 07:15 PMjohngRe: Circular function
Minoanman,

I haven't a clue as to what curve you have drawn. Could you be more explicit by giving parametric equations or an implicit equation for the curve? In what sense does your curve represent

$\displaystyle y=1-2\text{cos}(3\theta+\pi)$? - May 22nd 2013, 09:04 PMMINOANMANRe: Circular function
f(θ)=y=1-2cos(3θ+π) .............this is the curve as supplied by James .......read the second line of his POST...

if you substitute θ ith x you will get .

Attachment 28434

if you substitute y for r ( Polar forms) then you will get..

Attachment 28435 - May 22nd 2013, 10:34 PMjohngRe: Circular function
Minoan,

But what is the curve of your first response? - May 23rd 2013, 08:29 AMMINOANMANRe: Circular function
y=1-2cos(3θ+π) as given by James....

- May 23rd 2013, 07:32 PMjohngRe: Circular function
Minoan,

I'll try to make my question perfectly clear. Presumably, you used some software to generate the image in your first response. What was the software and exactly what did you type to get the curve?

Neither Geogebra nor Graph will accept the given equation: y=1-2cos(t+pi) (or theta for t). Wolfram Alpha interprets the given equation as plotting the function y=f(t) with parameter t; that is, the first graph of your second response.