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Math Help - trig prove

  1. #1
    Member srirahulan's Avatar
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    Post trig prove

    \frac{sin(n+1)A-sin(n-1)A}{cos(n+1)A+2cos(nA)+cos(n-1)A}=tan\frac{A}{2} when i simplify this equation i can't understand the sin(n+1)A,sin(n-1)A.
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  2. #2
    Junior Member Sarasij's Avatar
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    Re: trig prove

    Quote Originally Posted by srirahulan View Post
    \frac{sin(n+1)A-sin(n-1)A}{cos(n+1)A+2cos(nA)+cos(n-1)A}=tan\frac{A}{2} when i simplify this equation i can't understand the sin(n+1)A,sin(n-1)A.
    Numerator
    -------------
    N = sin(n+1)A - sin(n-1)A
    = 2cos[{(n+1)A+(n-1)A}/2]sin[{(n+1)A-(n-1)A}/2]
    = 2cos(2nA/2)sin(2A/2)
    = 2cos(nA)sin(A)

    Denominator
    -------------
    D = cos(n+1)A+2cos(nA)+cos(n-1)A
    = 2cos(nA)+cos(n+1)A+cos(n-1)A
    = 2cos(nA)+2cos[{(n+1)A+(n-1)A}/2]cos[{(n+1)A-(n-1)A}/2]
    = 2cos(nA)+2cos(nA)cosA

    So D=2cos(nA)(1+cosA)

    So,

    N/D
    =2cos(nA)sin(A)/2cos(nA)(1+cosA)
    = sinA/(1+cosA)
    = 2sin(A/2)cos(A/2)/2cos2(A/2)
    =sin(A/2)/cos(A/2)
    =tan(A/2)
    Last edited by Sarasij; June 6th 2012 at 10:00 PM.
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