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Trig ratios, double angles

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that sin(45+x).sin(45-x) = [tex]\frac{1}{2}[/tex]cos2x


    2. Relevant equations
    double angle formulae
    reduction formulae
    special angles
    identities


    3. The attempt at a solution
    (sin45.cosx+cos45.sinx)(sin45.cosx-cos45.sinx)
    =(sin45.cosx-cos45.sinx)2
    =([tex]\frac{1}{\sqrt{2}}[/tex]cosx-[tex]\frac{1}{\sqrt{2}}[/tex]sinx)2

    =[tex]\frac{1}{2}[/tex]cos2x-sinx.cosx+[tex]\frac{1}{2}[/tex]sinx2

    =[tex]\frac{1}{2}[/tex]-sinx.cosx
     
  2. jcsd
  3. Mar 21, 2009 #2

    danago

    User Avatar
    Gold Member

    How did you get from:

    (sin45.cosx+cos45.sinx)(sin45.cosx-cos45.sinx)

    To:

    (sin45.cosx-cos45.sinx)^2

    ?
     
  4. Mar 21, 2009 #3
    I really don't know. I must of been smoking something(hehehe):rofl:
    But thanks any way for pointing out my error I got it now.

    it should be

    (sin45.cosx+cos45.sinx)(sin45.cosx-cos45.sinx)
    =sin245.cos2x-cos245.sin2x
    =[tex]\frac{1}{2}[/tex]cos2x-[tex]\frac{1}{2}[/tex]sin2x
    =[tex]\frac{1}{2}[/tex]cos2x
     
    Last edited: Mar 21, 2009
  5. Mar 21, 2009 #4
    Start from here:

    http://http://i623.photobucket.com/albums/tt316/Saxifrage_Russell/PhysicsForumcomMarch21st.png" [Broken]

    PhysicsForumcomMarch21st.png
     
    Last edited by a moderator: May 4, 2017
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