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Show that secx + rt3 cosecx = 4

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that secx + rt3 cosecx = 4 can be written in the form sinx + rt3cos x = 2sin2x

    2. Relevant equations



    3. The attempt at a solution

    Didnt really know where to start with this one. trig identities maybe?
    cheers
     
  2. jcsd
  3. Dec 19, 2008 #2

    tiny-tim

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    Hi matt_crouch! :smile:
    oh come on!

    whenever you see sec or cosec, multiply everything by cos and/or sin :smile:
     
  4. Dec 19, 2008 #3
    Re: Trig

    Im probably missing a really obvious step. But i still dont know where to start.

    If i write the equation as 1/cosx + 1/rt3sinx = 4 and i multiply by sinx and cosx to get

    1 + 1/rt3 =4cosxsinx

    and im not sure but can the RHS be written as 2sin2x?
    am i heading in the right direction cos im still stuck as to how i can get 1+ 1/rt3 can be written as sinx + rt3cosx
     
  5. Dec 19, 2008 #4

    Dick

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    Re: Trig

    That's some pretty nasty algebra on the RHS. You are multiplying by sin(x)*cos(x). How did you get the 1's and why did sqrt(3) move into the denominator?
     
  6. Dec 19, 2008 #5

    tiny-tim

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    erm … sinx + cosx/rt3 =4cosxsinx :redface:

    but how did that rt3 get on the bottom? :confused:
    Yes!!!

    You must learn these trigonometric identities and be sure of them! :smile:
     
  7. Dec 20, 2008 #6
    Re: Trig

    ya i really have.. =]

    basically i had no idea what i was doing. Confused the hell out of myself so tore the page out an started again.. =]
     
  8. Dec 20, 2008 #7
    Re: Trig

    Sorry for asking, but what is rt3?
     
  9. Dec 20, 2008 #8

    tiny-tim

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    i assumed it was √3 … but does it matter?
     
  10. Dec 21, 2008 #9
    Re: Trig

    In this case, it doesn't. I wanted to know in case if I find rt3 again somewhere. Thanks tiny-tim.
     
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