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Trig Identities and Double Angle Formulas

  1. Nov 5, 2007 #1
    [SOLVED] Trig Identities and Double Angle Formulas

    1. The problem statement, all variables and given/known data
    If tan(a) = 1/5 and tan(b) = 1/239, find tan(4a - b)


    2. Relevant equations
    tan2x = 2tanx/1-tan^2x

    tan(x-y) = tanx - tany/ 1 +tanx tany


    3. The attempt at a solution

    tan (4a - b) = (tan4a - tanb)/(1 + tan4a tanb)
    = ((2tan2a/1-tan^2 2a) - tanb)/ (1-(2tan^2 2a/1-tan^2 2a)tanb)

    = ((2tan2(1/5)/1-tan^2 2(1/5)) - tan(1/239)) / (1 - (2tan^2 2(1/5))tan(1/239))
    = ((2tan(2/5))/1-tan^2 (2/5)) - tan(1/239)) / (1 - 2tan^2 (2/5)tan(1/239))

    After this I get stuck. Please someone help. I need this really quick.
     
  2. jcsd
  3. Nov 5, 2007 #2

    rock.freak667

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    Homework Helper

    tan(a) = 1/5 means that where ever you see tan(a) you can replace it with 1/5

    tan2(a)=2tan(a)/{1-tan^2(a)}
    you can find a value for tan2a...but if I were you, I would not bother to expand out tan4a. I would just find the value for tan4a and substitute it
     
  4. Nov 5, 2007 #3
    ok, Here is my new try:

    tan4a - tanb / 1 + tan4a tanb

    tan(4/5) - tan(1/239) / 1 + tan4(1/5) tan(1/239)
    tan(951/1195) / 1 + tan(951/1195)

    After this, what would I do? Please help. Thanks.
     
  5. Nov 5, 2007 #4

    rock.freak667

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    well you still aren't getting it

    [tex]tan4a=\frac{2tan2a}{1-tan^22a}[/tex] ....call this (*)

    now separately you know that [tex]tan2a=\frac{2tana}{1-tan^2a}[/tex]

    so then substituting [itex]tana=\frac{1}{5}[/itex]

    [tex] tan2a= \frac{\frac{2}{5}}{1-\frac{1}{25}}[/tex]
    [itex]tan2a=\frac{5}{12}[/itex]


    now that you have a value for tan2a can you sub this value into * and find the value for tan4a ?
     
  6. Nov 6, 2007 #5
    Yes, I finally got it. Thanks to rock.freak667.

    tan (4a - b) = tan4a - tanb / 1 + tan4a tanb

    tan4a = 2tan2a / 1-2tan^2 a

    tan2a = 2tana / 1- tan^2a
    = 2(1/5) / 1 - (1/25)
    = (2/5) / (24/25)
    = 5/12

    tan4a = 2tan2a / 1-2tan^2 a
    = 2(5/12) / (1 - 25/144)
    = (5/6) / (119/144)
    = 120/119

    tan (4a - b) = tan4a - tanb / 1 + tan4a tanb
    = (120/119) - (1/239) / 1 + (120/119)(1/239)
    = (28561/28441) / 1 + 120/28441
    = (28561/28441) / (28561/28441)
    = 1

    Thanks very much to rock.freak667. I really needed help on this.
     
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