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Trigonometric Series

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Reduce the following series to the most simplified form:

    1. 2 sin[tex]^{2}[/tex]([tex]\theta[/tex]/2)*(1+cos[tex]\theta[/tex])*(1+cos[tex]^{2}[/tex][tex]\theta[/tex])(1+cos[tex]^{4}[/tex][tex]\theta[/tex])............

    2. tan ([tex]\theta[/tex]/2)*(1+sec [tex]\theta[/tex])(1+2[tex]\theta[/tex])(1+sec 4[tex]\theta[/tex])...................................

    2. Relevant equations
    For problem 1:
    2 sin[tex]^{2}[/tex]([tex]\theta[/tex]/2)=1-cos[tex]\theta[/tex])

    3. The attempt at a solution

    For problem 1:
    the result I got:
    y=2[tex]^{(n+1)}[/tex]

    Result=
    1-cos[tex]^{y}[/tex][tex]\theta[/tex]




    is it correct?????
    For problem 2:
    If I try to expand
    sec 2 [tex]\theta[/tex]
    I get huge denominators which are difficult to solve. Similarly with tan ([tex]\theta[/tex]/2),I again get a denominator with no pattern following any rule.




    Please help me for the second one. and tell me if I am correct in solving the first problem.
    regards,
    Ritwik
     
    Last edited: May 4, 2008
  2. jcsd
  3. May 4, 2008 #2
    So I have finally got that latex code right. I have been trying the second problem for hours. Please help me.
    Thanks in advance.
    Regards,
    Ritwik
     
  4. May 4, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I presume you mean
    tan ([tex]\theta[/tex]/2)*(1+sec [tex]\theta[/tex])(1+sec(2[tex]\theta[/tex]))(1+sec 4[tex]\theta[/tex])

    There is no "n" in the problem so I have no idea what "y= 2n+1" means!

    Since you say "huge denominators" I presume you switched to sine and cosine. That shouldn't be necessary. Use the corresponding identities as in (1) for tangent and secant.
     
  5. May 4, 2008 #4
    'n' stands for the number of terms
    Yes I switched to sine and cosine. But I wonder which idntitis to use. I have been working over thse problms for 6 hours now. Could u please help???
    Thanks
     
  6. May 4, 2008 #5
    can this ever help 1+tan^2=sec^2
    this is the only identity i know between sec and tan. Please help me someone
     
  7. May 4, 2008 #6
    I really have been spending a wretched time to solve that
    tan (x/2)*(1+sec x)(1+sec 2x)(1+sec 4x)...................................
    How can I apply any identity here??
    tan (x/2)=2 tan (x/4) / 1- tan^2 (x/4)
    Please help!!!!!!
     
    Last edited: May 4, 2008
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