# How to bring this about?

#### allegansveritatem

##### Full Member
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. Any hints?

#### lev888

##### Senior Member
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. Any hints?

#### Subhotosh Khan

##### Super Moderator
Staff member
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. Any hints?
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 and exactly where you are stuck.

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

#### HallsofIvy

##### Elite Member
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. Any hints?
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.

#### pka

##### Elite Member
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.
Do you know about the difference of two squares?
$$1-\cos^2(\theta)=(1-\cos(\theta))(1+\cos(\theta))$$

#### HallsofIvy

##### Elite Member
But that's not what the original post has! It has 1- cos(theta)^2 which, because there were other parentheses missing, I interpreted as (1- cos(theta))^2. Then it is just a/a^2= 1/a

#### allegansveritatem

##### Full Member
But that's not what the original post has! It has 1- cos(theta)^2 which, because there were other parentheses missing, I interpreted as (1- cos(theta))^2. Then it is just a/a^2= 1/a
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.

#### allegansveritatem

##### Full Member
Do you know about the difference of two squares?
$$1-\cos^2(\theta)=(1-\cos(\theta))(1+\cos(\theta))$$
I see it! Thanks

#### allegansveritatem

##### Full Member
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 and exactly where you are stuck.

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