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Trigometric Identies Rearrangement

  1. Jun 3, 2008 #1
    1. The problem statement, all variables and given/known data

    I have the term:

    2tan2x + (2x-1)(2sec²2x) = 0

    I need to rearrange it into the form

    4x + sin4x - 2 = 0

    I tried:
    [tex]\frac{2sin2x}{cos2x} + \frac{2(2x-1)}{cos²2x} = 0[/tex]

    i multiply each side by cos²2x

    [tex]2.Sin2x.Cos2x + 4x - 2= 0[/tex]

    but now where?

  2. jcsd
  3. Jun 3, 2008 #2


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    Look at the sine double angle identity.
  4. Jun 3, 2008 #3
    i thought about that but you cant use it... can you?

    sin2x = 2sinxcosx

    you could do

    2(2sinxcosx)cos2x + 4x - 2

    or can you do somthing with the 2sin2xCos2x ??? it's 2x not x so you cant equate it to sin2x

    Not sure :/

  5. Jun 3, 2008 #4


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    Let u = 2x

    2sin(u)cos(u) = sin(2u)
  6. Jun 3, 2008 #5
    but to equal sin4x doesnt it need to be 4sinxcosx not 2sin2xcos2x?

  7. Jun 3, 2008 #6


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    Not at all.
    Obviously 2sin(u)cos(u) = sin(2u) is true since the variable inside the formula doesn't matter. Now, replace u with 2x.
    2sin(2x)cos(2x) = sin(2*(2x)) = sin(4x).
    There's no need for another 2 out in front.
  8. Jun 3, 2008 #7
    and that's ture for all double angle formulas
    and similar for the half angle formulas?

    Cheerz :)
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