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Solution to this trigonometric equation

  1. Apr 29, 2016 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    ##tanx=\frac{(1+tan1)(1+tan2)-2}{(1-tan1)(1-tan2)-2}## find x
    2. Relevant equations
    3. The attempt at a solution

    I tried multiplying through the paranthesis and arrived at ##tanx=\frac{(tan1tan2-1)+(tan2+tan1)}{(tan1tan2-1)-(tan2+tan1)}## and i dont know if this is any simpler than what was originally set. Is there a trigonometric identity at play here?
     
  2. jcsd
  3. Apr 29, 2016 #2

    SammyS

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    Is that really tan(1) and tan(2), as in 1 and 2 radians?
     
  4. Apr 29, 2016 #3

    SammyS

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    Whatever it is, divide the numerator & denominator by ##\ tan1\,tan2 -1 \ ##.
     
  5. Apr 30, 2016 #4

    diredragon

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    It's in degrees. So i divided both num and den by ##tan1tan2 - 1## and i get
    ##tanx=\frac{(tan1)^2(tan2)^2-2tan1tan2+1+(tan2)^2tan1-tan2+(tan1)^2tan2-tan1}{(tan1)^(tan2)^2-2tan1tan2+1-(tan2)^2tan1+tan2-(tan1)^2-tan1}##
    What should i do now?
     
  6. Apr 30, 2016 #5

    NascentOxygen

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    From post #1 and using SammyS's suggestion you should get it into the form: (1 + A)/(1 - A)

    Can you try it again.
     
  7. Apr 30, 2016 #6

    diredragon

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    Oh, well then it is
    ##tanx=\frac{1+\frac{tan1+tan2}{tan1tan2-1}}{1-\frac{tan1+tan2}{tan1tan2-1}}##
    Is there a way to simplify this?
     
  8. Apr 30, 2016 #7

    NascentOxygen

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    Ask google to search on tan(a+b) (or some trig expression like that) and see whether you can substitute something for that expression you have for the term I designated by A.
     
  9. Apr 30, 2016 #8

    diredragon

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    ##tanx=\frac{1-tan3}{1+tan3}## but thats as far as the simolification goes right? Now the use of calculator is required. It says that the answer is ##x=42## in degrees.
     
  10. Apr 30, 2016 #9

    diredragon

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    Yeah i solved it. I used ##tan45-tan3## and the identity to arrive at ##tanx=tan42##
     
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