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Proving Trig Identities

  1. Jan 17, 2008 #1
    [SOLVED] Proving Trig Identities

    1. The problem statement, all variables and given/known data

    Prove that

    tanx - sinx---=---- tanxsinx
    ________------- _________
    tanxsinx -------- tanx + sinx



    2. Relevant equations

    Is this impossible, so far it has been for me, what about you?

    3. The attempt at a solution

    I litterlly have 4 pages of written of attempts, I just can't figure it out
     
  2. jcsd
  3. Jan 17, 2008 #2

    rock.freak667

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    Actually it isn't that hard.

    I used the LHS and just simply converted all the tanx to sinx/cosx and simplified it. Did you do that?
     
  4. Jan 17, 2008 #3
    What does LHS stand for?, and I have converted the tanx's to sinx/cosx, but there are so many ways to go from there, I just can't seem to get one, Can you possibly tell me the final simplified version of either side that you got?
     
  5. Jan 17, 2008 #4

    rock.freak667

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    LHS is Left Hand Side.

    [tex]\frac{tanx-sinx}{tanxsinx}= \frac{\frac{sinx}{cosx}-sinx}{\frac{sin^2x}{cosx}}[/tex]

    then multiply the numerator and denominator by cosx then what do you get?

    And always remember that [itex]sin^2 x+cos^2 x=1[/itex]
     
  6. Jan 17, 2008 #5
    Am I aloud to multiply the sinx/cosx before finding common denominators with -sinx?

    [tex]\frac{\{sinx-sinxcosx}{cosx}[/tex]

    Is this what you mean?
     
    Last edited: Jan 17, 2008
  7. Jan 17, 2008 #6

    rock.freak667

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    Well if you multiply the Numerator and denominator by cos x then it is the same as multiplying by 1. So that it remains the same.

    But you were supposed to get


    [tex]\frac{\{sinx-sinxcosx}{sin^2 x}[/tex]


    then recall that [itex]sin^2 x=1-cos^2 x[/itex]
     
  8. Jan 17, 2008 #7

    So I'm suppose to reduce the left side to

    [tex]\frac{\{sinx-sinxcosx}{sin^2 x}[/tex]

    and then do that to the right?
     
  9. Jan 17, 2008 #8

    rock.freak667

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    No. Work with one side only. After you reduce it to what is above. Just substitute sin^2x and you will get the answer in another 4 lines
     
  10. Jan 17, 2008 #9
    I'm sorry, I don't follow any of this, Normally I understand this stuff, but this "puzzle" is just confusing me

    [tex]\frac{\frac{sinx}{cosx}-\frac{sinxcosx}{cosx}}{\frac{sinx^{2}}{cosx}}[/tex]
     
    Last edited: Jan 17, 2008
  11. Jan 17, 2008 #10

    rock.freak667

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    Using the LHS
    [tex]\frac{tanx-sinx}{tanxsinx}[/tex]

    sub [itex]tanx=\frac{sinx}{cosx}[/itex]


    [tex] \frac{\frac{sinx}{cosx}-sinx}{\frac{sin^2x}{cosx}}[/tex]

    multiply both the numerator and denominator by cosx


    [tex] \frac{cosx (\frac{sinx}{cosx}-sinx)}{(\frac{sin^2x}{cosx})cosx}[/tex]


    [tex] \frac{sinx-sinxcosx}{sin^2x}}[/tex]

    sub [itex]sin^2x=1-cos^2x[/itex]

    Following better now?
     
  12. Jan 17, 2008 #11
    Then I want to get

    [tex]\frac{sinx(1-cosx)}{(1+cosx)(1-cosx)}[/tex]

    correct?
     
  13. Jan 17, 2008 #12

    rock.freak667

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    yes, what do you have left now when the 1-cosx cancels out?
     
  14. Jan 17, 2008 #13
    you get

    [tex]\frac{sinx}{1+cosx}[/tex] Then I reduce the right side to the same thing, right?
     
  15. Jan 17, 2008 #14

    rock.freak667

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    Nope. you need to somehow make the Left side the same as the right side. Which is what you are doing

    What you want is

    [tex]\frac{sinx}{1+cosx}[/tex]

    to somehow become

    [tex]\frac{tanxsinx}{tanx+sinx}[/tex]


    Now if you take what you have and multiply the numerator and denominator by tanx

    the numerator will become tanxsinx right? Isn't that the numerator that you want?
    Deal with the denominator now.
     
  16. Jan 17, 2008 #15
    (1+cosx)(tanx)---------------tanx + cosxtanx---------------tanx + cosxsinx/cosx
    cancel out the cosx and you get tanx + sinx
     
  17. Jan 17, 2008 #16

    rock.freak667

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    Yes, the denominator is now tanx+sinx and the numerator is tanxsinx

    so you now have what they wanted...
     
  18. Jan 17, 2008 #17
    Thank you so much for your help, You were the 3rd person I asked, and the only one who actually knew how to solve this, you are the man, thanks again.
     
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