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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

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  1. zepdrix
    • one year ago
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    \[\Large\rm \frac{2}{1+\cos x}-\tan^2\left(\frac{x}{2}\right)=1\]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

  2. zepdrix
    • one year ago
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    \[\Large\rm \color{orangered}{\tan\left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}}\]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.

  3. zepdrix
    • one year ago
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    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.

  4. zepdrix
    • one year ago
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    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 ^

  5. anonymous
    • one year ago
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    Ooh ok

  6. zepdrix
    • one year ago
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    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

  7. anonymous
    • one year ago
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    alright so can you make a common denominator ?

  8. zepdrix
    • one year ago
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    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.

  9. anonymous
    • one year ago
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    Ohh ok. yes!

  10. zepdrix
    • one year ago
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    so looks like we need to give that first fraction a (1-cos x) on top and bottom.

  11. zepdrix
    • one year ago
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    \[\large\rm \left(\frac{1-\cos x}{1-\cos x}\right)\frac{2}{1+\cos x}-\frac{(1-\cos x)^2}{\sin^2 x}=1\]

  12. anonymous
    • one year ago
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    \[\frac{ 2- 2cosx }{ (1-cosx)^2 }\] ?

  13. zepdrix
    • one year ago
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    Top looks ok.

  14. anonymous
    • one year ago
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    Oops wait lol \[\frac{ 2- 2cosx }{ 1- \cos^2x }\]

  15. zepdrix
    • one year ago
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    Ah there we go :)

  16. anonymous
    • one year ago
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    I feel dumb lol

  17. zepdrix
    • one year ago
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    And then our Pythagorean Identity lets us go one step further, ya?

  18. anonymous
    • one year ago
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    yup !

  19. zepdrix
    • one year ago
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    \[\Large\rm \frac{2-2\cos x}{\sin^2x}-\frac{(1-\cos x)^2}{\sin^2 x}=1\]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.

  20. anonymous
    • one year ago
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    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

  21. zepdrix
    • one year ago
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    Ooo ya we did something silly there,\[\Large\rm (1-\cos x)^2\ne 1^2-\cos^2x\]

  22. zepdrix
    • one year ago
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    gotta FOIL that out.

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

  24. anonymous
    • one year ago
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    Oooh lol I was taking it as \[(1- cosx) (1 +cosx)\]

  25. zepdrix
    • one year ago
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    :3

  26. anonymous
    • one year ago
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    \[1 - 2\cos x + \cos^2x\]

  27. anonymous
    • one year ago
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    T_T

  28. zepdrix
    • one year ago
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    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.

  29. anonymous
    • one year ago
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    waaaait lemme do it

  30. zepdrix
    • one year ago
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    Distribute the negative, cancel some stuff out, combine like-terms, if you're able, what are you left with on top? :)

  31. zepdrix
    • one year ago
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    my bad :3 stealing too much of the fun lol

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

  33. anonymous
    • one year ago
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    Yasssss *___*

  34. zepdrix
    • one year ago
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    Pythagorean Identity on top one more time? Oooo yay you did it \c:/ wooo good job!

  35. anonymous
    • one year ago
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    you made this into a cheesecake LOL thank you! xD

  36. zepdrix
    • one year ago
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    np, ooo cheese cake sounds delic +_+

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

  38. anonymous
    • one year ago
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    Which is the same

  39. zepdrix
    • one year ago
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    Yah that's pretty much the approach infinity was using, probably a lot easier to do it that way I suppose :)

  40. anonymous
    • one year ago
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    @Nishant_Garg oh cool! I didn't think of that

  41. anonymous
    • one year ago
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    :) When I see 1 and tan^2(x) it just automatically connects for me

  42. xapproachesinfinity
    • one year ago
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    @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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