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Christians4 Group Title

Prove:

  • 2 years ago
  • 2 years ago

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  1. Christians4 Group Title
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    \[Tan^2x - \cos^2x = 1/\cos^2 - 1 - \cos^2x\]

    • 2 years ago
  2. lgbasallote Group Title
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    can you rewrite that right side? is that \[\frac 1{cos^2 x - 1} - \cos ^2 x\]

    • 2 years ago
  3. Christians4 Group Title
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    |dw:1352207602501:dw|

    • 2 years ago
  4. lgbasallote Group Title
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    you know this identity right? \[\sin^2 x = 1 - \cos^2 x\]

    • 2 years ago
  5. Christians4 Group Title
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    yes

    • 2 years ago
  6. lgbasallote Group Title
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    wait..nevermind what i said....

    • 2 years ago
  7. Christians4 Group Title
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    alright

    • 2 years ago
  8. lgbasallote Group Title
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    \[\frac 1{\cos^2x} - 1 - \cos^2 x\] convert to common denominators first \[\implies \frac 1{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} - \frac{\cos^4 x}{\cos^2 x}\] agree?

    • 2 years ago
  9. Christians4 Group Title
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    yes :) I was on the right track whew :P

    • 2 years ago
  10. lgbasallote Group Title
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    now combine the numerators \[\frac{1 - \cos^2 x - \cos ^ 4 x}{\cos^2 x}\] is this what you did?

    • 2 years ago
  11. Christians4 Group Title
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    yes but Instead if making it cos^4x I cancelled out the top bracket Cos^2x with the bottom one...which is probably where I went wrong?

    • 2 years ago
  12. lgbasallote Group Title
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    no. that works too \[\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x\] yes?

    • 2 years ago
  13. Christians4 Group Title
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    hmm, why didnt you just totally cancel out the denominator?

    • 2 years ago
  14. Christians4 Group Title
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    I was left with 1-cos^2x-cos^2x :S

    • 2 years ago
  15. lgbasallote Group Title
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    ah. that's different

    • 2 years ago
  16. lgbasallote Group Title
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    you can't do that

    • 2 years ago
  17. HorribleAtMath Group Title
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    |dw:1352208030577:dw|

    • 2 years ago
  18. Christians4 Group Title
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    hmm whats the reason I couldnt?

    • 2 years ago
  19. lgbasallote Group Title
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    observe what i did earlier \[\frac 1{\cos^2 x} - 1 - \cos^2 x \implies \frac 1{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} - \frac{\cos^4 x}{\cos^2 x}\] each term has its own denominator

    • 2 years ago
  20. HorribleAtMath Group Title
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    |dw:1352208100756:dw|

    • 2 years ago
  21. HorribleAtMath Group Title
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    ap calc etcher tell me that

    • 2 years ago
  22. Christians4 Group Title
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    OH I understand

    • 2 years ago
  23. lgbasallote Group Title
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    if you cancel out the cos^2 x below cos^4 x, you can only do it there, and not in the other terms. make sense?

    • 2 years ago
  24. Christians4 Group Title
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    yes! I love you :P

    • 2 years ago
  25. lgbasallote Group Title
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    so anyway, the RHS simplifies to \[\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x\] i assume you know how to proceed form tehre?

    • 2 years ago
  26. Christians4 Group Title
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    should i move to the left side?

    • 2 years ago
  27. Christians4 Group Title
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    wo think i got it? L.S) will equel 1-cos^2/cos^2x - Cos^2x

    • 2 years ago
  28. Christians4 Group Title
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    Ls = Rs Sweet! Thanks for your help! I also found a few more I am having trouble with...this next question has a tan^3x which I haven't used in a while i forget what i can do with it

    • 2 years ago
  29. lgbasallote Group Title
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    tan^3 x = tan^2 x * tan x

    • 2 years ago
  30. lgbasallote Group Title
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    that can work

    • 2 years ago
  31. lgbasallote Group Title
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    or maybe tan^3 x = sin^3 x / cos^3 x

    • 2 years ago
  32. Rosh007 Group Title
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    \[\tan ^{2}x - \cos^{2}x =\frac{1}{ \cos^2(x)} -1- \cos^{2}x \] \[\tan ^{2}x - \cos^{2}x =\frac{(1-\cos^{2}x) }{ \cos^2(x)} - \cos^{2}x \] which is \\[\tan ^{2}x =\frac{(1-\cos^{2}x) }{ \cos^2(x)} \] which is \[\tan ^{2}x =\frac{(\sin^{2}x) }{ \cos^2(x)}\] which satisfis the relation

    • 2 years ago
  33. Christians4 Group Title
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    Thanks guys, lemme try this other one and i cant get it ill post a new question

    • 2 years ago
  34. Rosh007 Group Title
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    all the best :)

    • 2 years ago
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