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Trignometric Identities

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that
    [tex]\frac{sin3x}{sinx}[/tex]-[tex]\frac{cos3x}{cosx}[/tex] = 2


    2. Relevant equations



    3. The attempt at a solution
    LHS:[tex]\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}[/tex]

    =[tex]\frac{sin2x.cosx + cos2x.sinx}{sin}-\frac{cos2xcosx - sin2x.sinx}{cosx}[/tex]
     
  2. jcsd
  3. Sep 7, 2009 #2
    thats as far i got...im really new with trig ..caught a wake up call at school so i started working with it...
     
  4. Sep 7, 2009 #3
    By multiplying [itex]\frac{sin3x}{sinx}[/itex] with [itex]\frac{cos(x)}{cos(x)}[/itex] and [itex]\frac{cos(3x)}{cos(x)}[/itex] with [itex]\frac{sin(x)}{sin(x)}[/itex] you got:

    [tex]\frac{cos(x)sin(3x)-sin(x)cos(3x)}{sin(x)cos(x)}[/tex]

    What can you spot now? :smile:
     
  5. Sep 7, 2009 #4
    Did you try getting a common denominator at this point?
     
  6. Sep 8, 2009 #5
    can we expand sin(3x) and cos(3x) using double angle formula?
     
  7. Sep 8, 2009 #6
    It can also be done onwards from here. Do you know how to expand cos(2x) and sin(2x)?
     
  8. Sep 8, 2009 #7
    No.

    You can write the nominator:

    [tex]cos(x)sin(3x)-sin(x)cos(3x)[/tex]

    as

    [tex]sin(3x-x)=sin(2x)[/tex]

    using the sum difference formula.

    Also you can write the denominator

    [tex]cos(x)sin(x)[/tex]

    as

    [tex]\frac{sin(2x)}{2}[/tex]

    :smile:
     
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