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VeroZarate Group Title

Verify that cot x sec4x = cot x + 2 tan x + tan3x

  • one year ago
  • one year ago

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  1. VeroZarate Group Title
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    so i know cotx is cos/sin

    • one year ago
  2. VeroZarate Group Title
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    idk if we can change the tan though to sin/cos?

    • one year ago
  3. binarymimic Group Title
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    this one is hard but try this

    • one year ago
  4. binarymimic Group Title
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    first divide everything by cot x

    • one year ago
  5. binarymimic Group Title
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    \[\sec^{4} x = 1 + 2\frac{ \tan x }{ \cot x } + \frac{ \tan^{3} x }{ \cot x }\]

    • one year ago
  6. binarymimic Group Title
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    and we know that \[\frac{ \tan x }{ \cot x } = \frac{ \frac{ \sin x }{ \cos x } }{ \frac{ \cos x }{ \sin x } } = \frac{\sin^{2} x }{ \cos^{2} x } = \tan^{2} x\]

    • one year ago
  7. VeroZarate Group Title
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    would it apply to the second one but it would be (sin^3x/cos^3x)/cosx/sinx ?

    • one year ago
  8. binarymimic Group Title
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    that would simplify it to: \[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]

    • one year ago
  9. binarymimic Group Title
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    yeah the second one is: \[\frac{ \tan^{3} x }{ \cot x } = \frac{ \tan x }{ \cot x } * \tan^{2} x = \tan^{2} x * \tan^{2} x = \tan^{4} x\]

    • one year ago
  10. VeroZarate Group Title
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    where did you get the second tan^2x

    • one year ago
  11. binarymimic Group Title
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    I divided tan^3 x into tan^x * tan^2 x

    • one year ago
  12. binarymimic Group Title
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    |dw:1357511455432:dw|

    • one year ago
  13. VeroZarate Group Title
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    oh ok

    • one year ago
  14. binarymimic Group Title
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    So this is why we have the following equation so far: \[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]

    • one year ago
  15. binarymimic Group Title
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    next step is to subtract both sides by tan^4 x

    • one year ago
  16. binarymimic Group Title
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    \[\sec^{4} x - \tan^{4} x = 1 + 2 \tan^{2} x\]

    • one year ago
  17. VeroZarate Group Title
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    and divide both side by2?

    • one year ago
  18. binarymimic Group Title
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    no there is a trick here now. notice the left side is the difference of two squares

    • one year ago
  19. binarymimic Group Title
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    \[\sec^{4} x - \tan^{4} x = (\sec^{2} x)^{2} - (\tan^{2} x)^{2}\]

    • one year ago
  20. binarymimic Group Title
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    remember that a^2 - b^2 = (a - b)(a + b) so we can now write the left side as: \[\sec^{4} x - \tan^{4} x = (\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x)\]

    • one year ago
  21. VeroZarate Group Title
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    right you foil it?

    • one year ago
  22. binarymimic Group Title
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    we kind of did the reverse of FOIL, we factored it

    • one year ago
  23. binarymimic Group Title
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    So now we have: \[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\]

    • one year ago
  24. binarymimic Group Title
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    but do you remember the identity sec^2 x = tan^2 x + 1

    • one year ago
  25. VeroZarate Group Title
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    yes but there is the 2?

    • one year ago
  26. binarymimic Group Title
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    we'll get to that :) so if sec^2 x = tan^2 x + 1 the if we subtract tan^2 from both sides we get sec^2 x - tan^2 x = 1

    • one year ago
  27. binarymimic Group Title
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    Then we see how

    • one year ago
  28. binarymimic Group Title
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    \[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\] becomes \[(1)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^2{x}\]

    • one year ago
  29. VeroZarate Group Title
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    isnt it sec^4

    • one year ago
  30. binarymimic Group Title
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    No we factored \[\sec^{4} x - \tan^{4} x \] into \[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x)\] because it is a difference of sqaures

    • one year ago
  31. VeroZarate Group Title
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    ok so when you subtract tan^2x?

    • one year ago
  32. binarymimic Group Title
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    There is a trig identity: \[\sec^{2} x = 1 + \tan^{2} x\] i am saying, if we rewrite it as: \[\sec^{2} x - \tan^{2} x = 1\] then the left factor on our left hand side reduces to just 1: \[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = (1)(\sec^{2} x + \tan^{2} x) \]

    • one year ago
  33. binarymimic Group Title
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    so we are now left with \[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\]

    • one year ago
  34. VeroZarate Group Title
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    ok you had lost me. now i get it sorry

    • one year ago
  35. binarymimic Group Title
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    its a complicated problem compared to the last one o_o

    • one year ago
  36. VeroZarate Group Title
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    yes it is. i still have to do 5 more assignments. lol

    • one year ago
  37. binarymimic Group Title
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    so now we have \[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\] can you see the next step ?

    • one year ago
  38. VeroZarate Group Title
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    substitute the sec^2x with 1+ tan^2x?

    • one year ago
  39. binarymimic Group Title
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    we could, or we could save a step and just subtract tan^2 x from both sides

    • one year ago
  40. binarymimic Group Title
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    if we subtract tan^2 x from both sides we get \[\sec^{2} x = 1 + \tan^{2} x\]

    • one year ago
  41. binarymimic Group Title
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    which is an identity

    • one year ago
  42. VeroZarate Group Title
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    ok

    • one year ago
  43. VeroZarate Group Title
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    but how does this equal to cotxsec^4x?

    • one year ago
  44. binarymimic Group Title
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    the whole idea is to reduce the original equation into a trig identity

    • one year ago
  45. binarymimic Group Title
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    if you follow the steps then that's precisely what we did

    • one year ago
  46. VeroZarate Group Title
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    to verify that it equals on both sides?

    • one year ago
  47. binarymimic Group Title
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    if you manipulate an equation using algebra then whatever you end up with is equal to what you started with.

    • one year ago
  48. VeroZarate Group Title
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    oh ok thanks so much!

    • one year ago
  49. binarymimic Group Title
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    np

    • one year ago
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