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anonymous
 one year ago
how would i begin to prove the equation 2/1+cosx  tan^2 x/2=1
@zepdrix
anonymous
 one year ago
how would i begin to prove the equation 2/1+cosx  tan^2 x/2=1 @zepdrix

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm \frac{2}{1+\cos x}\tan^2\left(\frac{x}{2}\right)=1\]So ... we have some cosine business, ya? :) I guess we want the Tangent HalfAngle Identity that involves sines and cosines. I cant remember what they look like, I'mma have to look em up

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm \color{orangered}{\tan\left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}}\]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
 one year ago
Best ResponseYou've already chosen the best response.2No 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So 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 ^

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So 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
 one year ago
Best ResponseYou've already chosen the best response.0alright so can you make a common denominator ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2so looks like we need to give that first fraction a (1cos x) on top and bottom.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \left(\frac{1\cos x}{1\cos x}\right)\frac{2}{1+\cos x}\frac{(1\cos x)^2}{\sin^2 x}=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2 2cosx }{ (1cosx)^2 }\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops wait lol \[\frac{ 2 2cosx }{ 1 \cos^2x }\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And then our Pythagorean Identity lets us go one step further, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm \frac{22\cos x}{\sin^2x}\frac{(1\cos x)^2}{\sin^2 x}=1\]So we have,\[\Large\rm \frac{22\cos x(1\cos x)^2}{\sin^2x}=1\]Ok with that last step? They have the same denominator, so I combined them into a single fraction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0mmm 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
 one year ago
Best ResponseYou've already chosen the best response.2Ooo ya we did something silly there,\[\Large\rm (1\cos x)^2\ne 1^2\cos^2x\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm (1\cos x)^2=(1\cos x)(1\cos x)=?\]Should end up with more than just two terms when you do :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oooh lol I was taking it as \[(1 cosx) (1 +cosx)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1  2\cos x + \cos^2x\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So,\[\Large\rm \frac{22\cos x(1\cos x)^2}{\sin^2x}=1\]Becomes,\[\Large\rm \frac{22\cos x(12\cos x+\cos^2x)}{\sin^2x}=1\]Ok good.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Distribute the negative, cancel some stuff out, combine liketerms, if you're able, what are you left with on top? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2my bad :3 stealing too much of the fun lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2  2cosx  1 + 2cosx  \cos^2x }{ \sin^2x }\] \[\frac{ 1 \cos^2x }{ sin^2x }\] = 1

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Pythagorean Identity on top one more time? Oooo yay you did it \c:/ wooo good job!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you made this into a cheesecake LOL thank you! xD

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2np, ooo cheese cake sounds delic +_+

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{2}{1+cosx}=1+\tan^2\frac{x}{2}=\sec^2\frac{x}{2}=\frac{1}{\cos^2\frac{x}{2}}\]\[2\cos^2\frac{x}{2}=1+cosx\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yah that's pretty much the approach infinity was using, probably a lot easier to do it that way I suppose :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Nishant_Garg oh cool! I didn't think of that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:) When I see 1 and tan^2(x) it just automatically connects for me

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0@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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