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Trig Identities could sum1 pleas help

  1. Sep 14, 2006 #1
    Hi, in this question i am nt sure the best way to tackle it!!
    it follows

    proove the following

    2sinxcosx=sqrt(3)-ssqrt(3)sin^2x for 0<=x<=360

    i tried using the doble angle formulae on the right, putting all on one side therfore =0 (anticipating a quadratic equation)
    having

    sin2x-sqrt(3)+2sqrt(3)sin^2(x)

    i can see that a quadratic equation is smhow possible, but dont know how to get it there....any help or tips would be greatly apprecitaed!!!

    thanks!
     
  2. jcsd
  3. Sep 14, 2006 #2
    actually the double angle formulae on the left.....
     
  4. Sep 14, 2006 #3

    Pyrrhus

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    is the identity?

    [tex] 2 \sin x \cos x = \sqrt{3} - \sqrt{3} \sin^{2} x [/tex]

    because the above equation is not an identity.
     
  5. Sep 14, 2006 #4
    sorry...not an identity...was readin the wrong stuff...it just wants me to solve for x
     
  6. Sep 14, 2006 #5

    Integral

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    I don't think you need to go that route.

    factor [itex] \sqrt {3} [/itex] from the RHS. Do you see anything that looks familiar?
     
  7. Sep 14, 2006 #6
    u mean how the (1-2sin^2x) becomes (1-2(1-cos^2x)???
    hang on i will see how that works
     
  8. Sep 14, 2006 #7
    which then looks like double angle formulae for cos
     
  9. Sep 14, 2006 #8
    gerat thanks very muc...get it down to tan2x=sqrt(3) thanks for ur help
     
  10. Sep 14, 2006 #9

    Integral

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    Did you get all of the solutions?

    My last question was in reference to the ORIGINAL equation. You do not need to use a double angle relationship to solve this.
     
  11. Sep 14, 2006 #10
    ohh kk...dunno um i got all the soltutions.... so thanks, also this is a real common identity whih i am trying to proove

    sin2x=2tanx/(tan^2x+1)

    i am trying yo simplify the RHS,but evertyhing i do makes it more complicated...i must be missing somthing simple...hat should i start with??
     
  12. Sep 15, 2006 #11

    Pyrrhus

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    one huge hint:

    [tex] \tan ^{2} x + 1 = \sec ^{2} x [/tex]
     
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