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anonymous

  • one year ago

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

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  1. xapproachesinfinity
    • one year ago
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    so it is \[\frac{1}{1+\cot x}-\tan^2x=1\]

  2. anonymous
    • one year ago
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    okay i understand

  3. xapproachesinfinity
    • one year ago
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    the first thing im thinking about is multiplying by 1-cotx top and bottom for the 1/1+cotx

  4. xapproachesinfinity
    • one year ago
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    i haven't started yet i was just asking is that the right thing? because the way you wrote it can be misleading

  5. xapproachesinfinity
    • one year ago
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    oh hold on i missed 2 on top

  6. anonymous
    • one year ago
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    well it's like \[(2/1+cosx) - \tan^2 x/2=1\]

  7. xapproachesinfinity
    • one year ago
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    \[\frac{2}{1+\cot x}-\frac{\tan^2x}{2}=1\]

  8. xapproachesinfinity
    • one year ago
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    may bad! cos not cot

  9. anonymous
    • one year ago
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    right right i got it

  10. xapproachesinfinity
    • one year ago
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    i must be blind or something lol

  11. xapproachesinfinity
    • one year ago
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    \[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=1\] looking at the left side \[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=\frac{2(1-\cos x)}{(1+\cos x)(1-\cos x)}-\frac{\tan^2x}{2}\]

  12. xapproachesinfinity
    • one year ago
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    let's go from there

  13. xapproachesinfinity
    • one year ago
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    \[\frac{2(1-\cos x)}{1-\cos^2x}-\frac{\tan^2x}{2}\] \[\frac{2-2\cos x}{\sin^2x}-\frac{\tan^2x}{2}\] \[\frac{2-2\cos x}{\sin^2x}-\frac{\sin^2x}{2\cos^2x}\]

  14. xapproachesinfinity
    • one year ago
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    can you go further your self

  15. xapproachesinfinity
    • one year ago
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    actually that's not an identity

  16. anonymous
    • one year ago
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    i didnt think so lol

  17. xapproachesinfinity
    • one year ago
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    oh no i take it back it is! when i plugged in 0 i forgot something lol

  18. xapproachesinfinity
    • one year ago
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    hmm for pi/4 it is not good!

  19. xapproachesinfinity
    • one year ago
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    are you sure you typed the right thing!

  20. anonymous
    • one year ago
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    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?

  21. xapproachesinfinity
    • one year ago
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    use drawing tool if you have a mouse you will do find drawing the problem

  22. xapproachesinfinity
    • one year ago
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    snap shot your problem don't you have a phone?

  23. xapproachesinfinity
    • one year ago
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    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

  24. xapproachesinfinity
    • one year ago
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    fail*

  25. anonymous
    • one year ago
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    \[\frac{ 2 }{ 1 + cosx } - \tan^2\frac{ x }{ 2 } = 1\]

  26. xapproachesinfinity
    • one year ago
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    ah wow why you didn't correct me above i was dividing tan^2 over 2 2 was supposed to be within tan

  27. xapproachesinfinity
    • one year ago
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    darn you made me work really hard lol

  28. anonymous
    • one year ago
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    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

  29. xapproachesinfinity
    • one year ago
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    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!

  30. xapproachesinfinity
    • one year ago
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    that is the same as proving \[\frac{1+\cos x}{2}=\cos^2(x/2) \]

  31. xapproachesinfinity
    • one year ago
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    isn't that some familiar identity to yoou?

  32. xapproachesinfinity
    • one year ago
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    the last line a i wrote should be obvious to you

  33. anonymous
    • one year ago
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    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

  34. xapproachesinfinity
    • one year ago
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    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

  35. anonymous
    • one year ago
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    right right i remember that now

  36. xapproachesinfinity
    • one year ago
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    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]

  37. xapproachesinfinity
    • one year ago
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    they are all linked!

  38. xapproachesinfinity
    • one year ago
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    oh well gotta leave!

  39. anonymous
    • one year ago
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    how do i prove it's sec though?

  40. anonymous
    • one year ago
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    @zepdrix I was wondering if you can demonstrate this identity 2/1+cosx - tan^2 x/2=1

  41. anonymous
    • one year ago
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    I'm pretty new to this myself, so I'm trying to understand what @xapproachesinfinity did but I think I'm kinda lost...

  42. zepdrix
    • one year ago
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    Mmm can we do it in a new thread? :D I don't feel like reading through all of these comments XD lol

  43. anonymous
    • one year ago
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    mm I pretty much a noob in trig identity LOL but ok xD

  44. xapproachesinfinity
    • one year ago
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    @Learner1298 we already proved it? why you need to prove sec lol read through what i did

  45. Loser66
    • one year ago
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    \[\dfrac{2}{1+cos (x) }-tan^2(x/2)\] \[= \dfrac{2}{1+cos(x)} -\dfrac{sin^2(x)}{(1+cos (x))^2}\] because \(tan (x/2) = \dfrac{sin (x) }{1+cos x}\)

  46. Loser66
    • one year ago
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    hence, it turns to \(\dfrac{2(1+cos(x)-sin^2(x)}{(1+cos(x))^2}\) Now, numerator only:

  47. Loser66
    • one year ago
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    \(2(1+ cos(x) ) -sin^2(x)\\=2+2cosx-(1-cos^2(x) =2+2cos(x)-1+cos^2(x) \\=1+2cos(x) +cos^2(x) = (1+cos(x))^2 \)

  48. Loser66
    • one year ago
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    Put it back to the fraction above, we can see that numerator = denorminator , hence the fraction =1 = R H S

  49. Loser66
    • one year ago
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    @xapproach.... hey kid, the second reader didn't understand your stuff. Let old man helps you out. hahahaha.....................

  50. xapproachesinfinity
    • one year ago
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    you did a different version lol this trig stuff there are often many diff ways of doing stuff

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