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by **Aaahhhh Fuuu.....** on Sun Apr 17, 2011 1:39 pm

This equation cos2theta = 1, the closest equation using cos2theta double angle formula is the cos^2x - sin^2x version. But it is not exact. Can anyone show me the steps on how to complete the equation? The answer of 0 = 2y^2 is the nearest to y = 0 that I can solve without completing using dividing by two then square rooting.

**Sponsor**

by **maggiemagnet** on Sun Apr 17, 2011 10:00 pm

Aaahhhh Fuuu..... wrote:This equation cos2theta = 1, the closest equation using cos2theta double angle formula is the cos^2x - sin^2x version. But it is not exact. Can anyone show me the steps on how to complete the equation? The answer of 0 = 2y^2 is the nearest to y = 0 that I can solve without completing using dividing by two then square rooting.

Why not solve the equation as it is? You have cos(2@) = 1. For what values of 2@ is the cosine equal to 1? Then divide by 2 to find the values for @.

:clap:

by **Aaahhhh Fuuu.....** on Sun Apr 17, 2011 10:33 pm

"Throw yo' hands up!?"
Attach yourself to this aspect of the equation if I did not mention it: The problem is asking to convert from polar to rectangular. Looking over my question I see I did not mention this part.

by **maggiemagnet** on Mon Apr 18, 2011 1:29 pm

Aaahhhh Fuuu..... wrote:"Throw yo' hands up!?"
Attach yourself to this aspect of the equation if I did not mention it: The problem is asking to convert from polar to rectangular. Looking over my question I see I did not mention this part.

I didn't know. Sorry. I didn't mean to offend you. :oops:

by **Aaahhhh Fuuu.....** on Tue Apr 19, 2011 2:33 am

The only thing I need some instruction for this problem of cos2theta = 1 using polar to rectangular form. The result shown in the book is y = 0 but without any methods just the answer is useless. Thank you for your sympathies :thumb:

by **Martingale** on Tue Apr 19, 2011 2:48 am

Aaahhhh Fuuu..... wrote:The only thing I need some instruction for this problem of cos2theta = 1 using polar to rectangular form. The result shown in the book is y = 0 but without any methods just the answer is useless. Thank you for your sympathies

hmmmm...

well....we 'know' right....that

$x^2+y^2=r^2$

and

$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$

we also have $x=r\cos(\theta)$ and $y=r\sin(\theta)$ so $\left(x^2+y^2=r^2\right)$

thus

$\frac{x}{r}=\cos(\theta)$ and $\frac{y}{r}=\sin(\theta)$

sooooooo.....

$\cos(2\theta)=1$

$\cos^2(\theta)-\sin^2(\theta)=1$

... start replacing the above with what we know and the solution will follow...

by **Aaahhhh Fuuu.....** on Tue Apr 19, 2011 3:56 pm

Thank you Martingale.

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