anonymous
  • anonymous
how would i begin to prove the equation 2/1+cosx - tan^2 x/2=1
Trigonometry
schrodinger
  • schrodinger
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xapproachesinfinity
  • xapproachesinfinity
so it is \[\frac{1}{1+\cot x}-\tan^2x=1\]
anonymous
  • anonymous
okay i understand
xapproachesinfinity
  • xapproachesinfinity
the first thing im thinking about is multiplying by 1-cotx top and bottom for the 1/1+cotx

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xapproachesinfinity
  • xapproachesinfinity
i haven't started yet i was just asking is that the right thing? because the way you wrote it can be misleading
xapproachesinfinity
  • xapproachesinfinity
oh hold on i missed 2 on top
anonymous
  • anonymous
well it's like \[(2/1+cosx) - \tan^2 x/2=1\]
xapproachesinfinity
  • xapproachesinfinity
\[\frac{2}{1+\cot x}-\frac{\tan^2x}{2}=1\]
xapproachesinfinity
  • xapproachesinfinity
may bad! cos not cot
anonymous
  • anonymous
right right i got it
xapproachesinfinity
  • xapproachesinfinity
i must be blind or something lol
xapproachesinfinity
  • xapproachesinfinity
\[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=1\] looking at the left side \[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=\frac{2(1-\cos x)}{(1+\cos x)(1-\cos x)}-\frac{\tan^2x}{2}\]
xapproachesinfinity
  • xapproachesinfinity
let's go from there
xapproachesinfinity
  • xapproachesinfinity
\[\frac{2(1-\cos x)}{1-\cos^2x}-\frac{\tan^2x}{2}\] \[\frac{2-2\cos x}{\sin^2x}-\frac{\tan^2x}{2}\] \[\frac{2-2\cos x}{\sin^2x}-\frac{\sin^2x}{2\cos^2x}\]
xapproachesinfinity
  • xapproachesinfinity
can you go further your self
xapproachesinfinity
  • xapproachesinfinity
actually that's not an identity
anonymous
  • anonymous
i didnt think so lol
xapproachesinfinity
  • xapproachesinfinity
oh no i take it back it is! when i plugged in 0 i forgot something lol
xapproachesinfinity
  • xapproachesinfinity
hmm for pi/4 it is not good!
xapproachesinfinity
  • xapproachesinfinity
are you sure you typed the right thing!
anonymous
  • anonymous
i will show you exactly what's in the textbook but how do you get the horizontal division line in the equation like you did in all your formulas?
xapproachesinfinity
  • xapproachesinfinity
use drawing tool if you have a mouse you will do find drawing the problem
xapproachesinfinity
  • xapproachesinfinity
snap shot your problem don't you have a phone?
xapproachesinfinity
  • xapproachesinfinity
what you have there is not an identity i verified it with couple values and it fails miserably if it is an identity it should not feel even one
xapproachesinfinity
  • xapproachesinfinity
fail*
anonymous
  • anonymous
\[\frac{ 2 }{ 1 + cosx } - \tan^2\frac{ x }{ 2 } = 1\]
xapproachesinfinity
  • xapproachesinfinity
ah wow why you didn't correct me above i was dividing tan^2 over 2 2 was supposed to be within tan
xapproachesinfinity
  • xapproachesinfinity
darn you made me work really hard lol
anonymous
  • anonymous
im sorry i wouldve if i known it was wrong but i was confused on this entire problem hence why i am here asking for help
xapproachesinfinity
  • xapproachesinfinity
here what you can do \[\frac{2}{1+\cos x}=1+\tan^2(x/2)=\sec^2(x/2)\] if you prove the left is sec^2(x/2) then you are done!
xapproachesinfinity
  • xapproachesinfinity
that is the same as proving \[\frac{1+\cos x}{2}=\cos^2(x/2) \]
xapproachesinfinity
  • xapproachesinfinity
isn't that some familiar identity to yoou?
xapproachesinfinity
  • xapproachesinfinity
the last line a i wrote should be obvious to you
anonymous
  • anonymous
i get it i think the sec and tan identity was what i was thinking i would get to to prove the identity i just wasnt sure how to get there
xapproachesinfinity
  • xapproachesinfinity
1+cosx/2=cos^2(x/2) is called half angle identity which can be proven easily but you don't need to draw it all the way as long as that is an identity we know of so the initial problem is an identity as well
anonymous
  • anonymous
right right i remember that now
xapproachesinfinity
  • xapproachesinfinity
let's prove that half angle identity for the sake of argument we know that cosx=2cos^2(x/2)-1 so 1+cosx=2cos^2(x/2) cos^2(x/2)=(1+cosx)/2 hence proved [cos(2a)=2cos^2a-1 is another useful identity]
xapproachesinfinity
  • xapproachesinfinity
they are all linked!
xapproachesinfinity
  • xapproachesinfinity
oh well gotta leave!
anonymous
  • anonymous
how do i prove it's sec though?
anonymous
  • anonymous
@zepdrix I was wondering if you can demonstrate this identity 2/1+cosx - tan^2 x/2=1
anonymous
  • anonymous
I'm pretty new to this myself, so I'm trying to understand what @xapproachesinfinity did but I think I'm kinda lost...
zepdrix
  • zepdrix
Mmm can we do it in a new thread? :D I don't feel like reading through all of these comments XD lol
anonymous
  • anonymous
mm I pretty much a noob in trig identity LOL but ok xD
xapproachesinfinity
  • xapproachesinfinity
@Learner1298 we already proved it? why you need to prove sec lol read through what i did
Loser66
  • Loser66
\[\dfrac{2}{1+cos (x) }-tan^2(x/2)\] \[= \dfrac{2}{1+cos(x)} -\dfrac{sin^2(x)}{(1+cos (x))^2}\] because \(tan (x/2) = \dfrac{sin (x) }{1+cos x}\)
Loser66
  • Loser66
hence, it turns to \(\dfrac{2(1+cos(x)-sin^2(x)}{(1+cos(x))^2}\) Now, numerator only:
Loser66
  • Loser66
\(2(1+ cos(x) ) -sin^2(x)\\=2+2cosx-(1-cos^2(x) =2+2cos(x)-1+cos^2(x) \\=1+2cos(x) +cos^2(x) = (1+cos(x))^2 \)
Loser66
  • Loser66
Put it back to the fraction above, we can see that numerator = denorminator , hence the fraction =1 = R H S
Loser66
  • Loser66
@xapproach.... hey kid, the second reader didn't understand your stuff. Let old man helps you out. hahahaha.....................
xapproachesinfinity
  • xapproachesinfinity
you did a different version lol this trig stuff there are often many diff ways of doing stuff

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