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VeroZarate

  • one year ago

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

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  1. VeroZarate
    • one year ago
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    so i know cotx is cos/sin

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

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

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

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

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

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

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

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

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

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

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

  13. VeroZarate
    • one year ago
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    oh ok

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

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

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

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

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

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

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

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

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

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

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

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

  26. binarymimic
    • one year ago
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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

  27. binarymimic
    • one year ago
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    Then we see how

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

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

  30. binarymimic
    • one year ago
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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

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

  32. binarymimic
    • one year ago
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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: \[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = (1)(\sec^{2} x + \tan^{2} x) \]

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

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

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

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

  37. binarymimic
    • one year ago
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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 ?

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

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

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

  41. binarymimic
    • one year ago
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    which is an identity

  42. VeroZarate
    • one year ago
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    ok

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

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

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

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

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

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

  49. binarymimic
    • one year ago
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    np

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