#### allegansveritatem

##### Full Member

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- Thread starter allegansveritatem
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Could you add correct parentheses?

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You know better than this after 892 posts. You wrote:

1 - cos(Θ)/1 - cos(Θ)

and then

Please show us what you have tried and

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Please share your work/thoughts about this problem

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If you mean (1- cos(theta))/(1- cos(theta))^2 then it is (1- cos(theta))/((1- cos(theta)(1- cos(theta))= 1/(1- cos(theta)) because the "1- cos(theta)" in the numerator cancels one of the two in the denominator.

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Do you know about the difference of two squares?How can I come from 1-cos(theta)/1-cos(theta)^2 to 1/1-cos(theta). I know these two are equivalent because they both generate the same graph...but try as I may I can't seem to massage number 1 into number 2.

\(1-\cos^2(\theta)=(1-\cos(\theta))(1+\cos(\theta))\)

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no, it is 1-cos(theta)^2--just cos(theta) is squared but as PLA says. 1-cos(theta)^2 is the difference between two squares and that solves my dilemma.

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I see it! ThanksDo you know about the difference of two squares?

\(1-\cos^2(\theta)=(1-\cos(\theta))(1+\cos(\theta))\)

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sorry. You are right. I usually do my work on a sheet of paper and then photograph that and upload the photo. It is less cumbersome than typing it out. I did a lot of work on this to get it into the state it is in. What I was given was: r= csc(theta)(csc(theta)cot(theta) and I was asked to put this into the form r=de/1+/-cos or sin(theta). But the work was done so all over the place on the page that I didn't think it would be helpful to photo it and upload--my usual procedure.You know better than this after 892 posts. You wrote:

1 - cos(Θ)/1 - cos(Θ)^{2}

and then

Please show us what you have tried andexactlywhere you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

Please share your work/thoughts about this problem

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