how would i begin to prove the equation 2/1+cosx - tan^2 x/2=1
@zepdrix

- anonymous

how would i begin to prove the equation 2/1+cosx - tan^2 x/2=1
@zepdrix

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- zepdrix

\[\Large\rm \frac{2}{1+\cos x}-\tan^2\left(\frac{x}{2}\right)=1\]So ... we have some cosine business, ya? :)
I guess we want the Tangent Half-Angle Identity that involves sines and cosines.
I cant remember what they look like, I'mma have to look em up

- zepdrix

\[\Large\rm \color{orangered}{\tan\left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}}\]Do you see how this one might end up being really useful for us?
It gives us a common denominator, that might be helpful at some point.

- zepdrix

No no no we want the other one actually :)
AHHH sorry
sine on the bottom is going to be a lot more useful.
But, I'm thinking ahead, this can be a tough problem to try and work out on your own.

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## More answers

- zepdrix

So using our identity,\[\Large\rm \color{orangered}{\tan\left(\frac{x}{2}\right)=\frac{1-\cos x}{\sin x}}\]notice that in our problem, the tangent is being squared,\[\Large\rm \color{orangered}{\tan^2\left(\frac{x}{2}\right)=\left(\frac{1-\cos x}{\sin x}\right)^2}\]This is what we're going to replace our tangent with ^

- anonymous

Ooh ok

- zepdrix

So our problem,\[\Large\rm \frac{2}{1+\cos x}-\color{orangered}{\tan^2\left(\frac{x}{2}\right)}=1\]Becomes,\[\Large\rm \frac{2}{1+\cos x}-\color{orangered}{\left(\frac{1-\cos x}{\sin x}\right)^2}=1\]We'll write it like this:\[\Large\rm \frac{2}{1+\cos x}-\frac{(1-\cos x)^2}{\sin^2 x}=1\]ya? :o

- anonymous

alright
so can you make a common denominator ?

- zepdrix

Yes and we'll use this idea of "difference of squares" to achieve that.
Do you remember what that would look like?
We're trying to do something like this:\[\Large\rm (1-\cos x)(1+\cos x)=1-\cos^2x=\sin^2x\]We multiply the denominator by it's conjugate to get these squares.

- anonymous

Ohh ok. yes!

- zepdrix

so looks like we need to give that first fraction a (1-cos x) on top and bottom.

- zepdrix

\[\large\rm \left(\frac{1-\cos x}{1-\cos x}\right)\frac{2}{1+\cos x}-\frac{(1-\cos x)^2}{\sin^2 x}=1\]

- anonymous

\[\frac{ 2- 2cosx }{ (1-cosx)^2 }\]
?

- zepdrix

Top looks ok.

- anonymous

Oops wait lol
\[\frac{ 2- 2cosx }{ 1- \cos^2x }\]

- zepdrix

Ah there we go :)

- anonymous

I feel dumb lol

- zepdrix

And then our Pythagorean Identity lets us go one step further, ya?

- anonymous

yup !

- zepdrix

\[\Large\rm \frac{2-2\cos x}{\sin^2x}-\frac{(1-\cos x)^2}{\sin^2 x}=1\]So we have,\[\Large\rm \frac{2-2\cos x-(1-\cos x)^2}{\sin^2x}=1\]Ok with that last step?
They have the same denominator, so I combined them into a single fraction.

- anonymous

mmm I see! just one thing so will it be:
\[\frac{ 2 - 2cosx +1 -\cos^2x }{ \sin^2x }\]
I hope I didn't make another silly mistake =_=
lol

- zepdrix

Ooo ya we did something silly there,\[\Large\rm (1-\cos x)^2\ne 1^2-\cos^2x\]

- zepdrix

gotta FOIL that out.

- zepdrix

\[\Large\rm (1-\cos x)^2=(1-\cos x)(1-\cos x)=?\]Should end up with more than just two terms when you do :)

- anonymous

Oooh lol I was taking it as
\[(1- cosx) (1 +cosx)\]

- zepdrix

:3

- anonymous

\[1 - 2\cos x + \cos^2x\]

- anonymous

T_T

- zepdrix

So,\[\Large\rm \frac{2-2\cos x-(1-\cos x)^2}{\sin^2x}=1\]Becomes,\[\Large\rm \frac{2-2\cos x-(1-2\cos x+\cos^2x)}{\sin^2x}=1\]Ok good.

- anonymous

waaaait lemme do it

- zepdrix

Distribute the negative,
cancel some stuff out, combine like-terms, if you're able,
what are you left with on top? :)

- zepdrix

my bad :3 stealing too much of the fun lol

- anonymous

\[\frac{ 2 - 2cosx - 1 + 2cosx - \cos^2x }{ \sin^2x }\]
\[\frac{ 1- \cos^2x }{ sin^2x }\] = 1

- anonymous

Yasssss *___*

- zepdrix

Pythagorean Identity on top one more time?
Oooo yay you did it \c:/ wooo good job!

- anonymous

you made this into a cheesecake
LOL thank you! xD

- zepdrix

np,
ooo cheese cake sounds delic +_+

- anonymous

\[\frac{2}{1+cosx}=1+\tan^2\frac{x}{2}=\sec^2\frac{x}{2}=\frac{1}{\cos^2\frac{x}{2}}\]\[2\cos^2\frac{x}{2}=1+cosx\]

- anonymous

Which is the same

- zepdrix

Yah that's pretty much the approach infinity was using,
probably a lot easier to do it that way I suppose :)

- anonymous

@Nishant_Garg oh cool!
I didn't think of that

- anonymous

:) When I see 1 and tan^2(x) it just automatically connects for me

- xapproachesinfinity

@Nishant_Garg same here i like to make it sec whne i see 1+tan
i did the same approach with this question in a different post

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