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Math Help - lagrange's trignometric identity

  1. #1
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    lagrange's trignometric identity

    use the identity  1+ z+ {z^2} + {z^3}+ {z^4}+...+ {z^n} = \frac{1-{z^ (n+1)} }{1-z} , z \neq 0 to prove lagrange's trignometric identity :
    1+ cos\ominus +  cos2\ominus+...+ cosn\ominus = \frac {1}{2} + \frac {sin (n+\frac {\ominus}{2})}{2sin(\frac{\ominus}{2})}  , 0<   \ominus <     2 \pi
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  2. #2
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    Hello,

    We know that the cosine is the real part of e^{i\theta}

    So by some changes, \cos n\theta=\Re((e^{i\theta})^n)

    We then have :
    F=1+\cos\theta+\cos 2\theta+\dots+\cos n\theta=\Re\left(1+e^{i\theta}+(e^{i\theta})^2+\do  ts+(e^{i\theta})^n\right)

    Which is, by the identity you're given, equal to :

    F=\Re\left(\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\right)=\Re\left(\frac{1-\cos((n+1)\theta)-i\sin((n+1)\theta)}{1-\cos\theta-i\sin\theta}\right)

    Now, since you want to take the real part of this fraction, transform the denominator into a real number : multiply the numerator and the denominator by the complex conjugate of 1-\cos\theta-i\sin\theta, which is 1-\cos\theta+i\sin\theta

    By using the identity (a+ib)(a-ib)=a^2+b^2, we get in the denominator : (1-\cos\theta)^2+\sin^2\theta=1+\cos^2\theta-2\cos\theta+\sin^2\theta=2-2\cos\theta=2(1-\cos\theta)


    And in the numerator, using basic algebraic operations and rearranging, we get :
    1-\cos\theta-\cos((n+1)\theta)+{\color{red}\cos((n+1)\theta)\co  s\theta+\sin((n+1)\theta)\sin\theta} +i \left[\sin\theta-\sin((n+1)\theta)-\sin\theta\cos((n+1)\theta)+\cos\theta\sin((n+1)\t  heta)\right]

    Since we only want the real part, we may only keep the first line.

    And then the red part is exactly \cos((n+1)\theta-\theta)=\cos n\theta


    So finally, we have :

    F=\frac{1-\cos\theta-\cos((n+1)\theta)+\cos(n\theta)}{2(1-\cos\theta)}=\frac 12+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2(1-\cos\theta)}


    By the half angle formula, we have 1-\cos\theta=2\sin^2 \tfrac \theta 2

    And by the identity \cos a-\cos b=-2\sin\tfrac{a+b}{2}\sin\tfrac{a-b}{2}=2\sin\tfrac{a+b}{2}\sin\tfrac{b-a}{2} (because the sine function is odd)

    So \cos(n\theta)-\cos((n+1)\theta)=2\sin\left(\tfrac{2n+1}{2} \cdot\theta\right)\sin\tfrac\theta 2=2\sin\left(n\theta+\tfrac \theta 2\right)\sin\tfrac\theta 2


    And we can simplify... :

    F=\frac 12+\frac{\sin\left(n\theta+\tfrac \theta 2\right)}{2\sin\tfrac \theta 2}




    Note : Some steps have been left to your understanding...
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  3. #3
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     \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}} = \frac{ 1-e^{i(n+1)\theta}}{(e^{i \frac{\theta}{2}}) ( e^{- i \frac{\theta}{2}}  - e^{i \frac{\theta}{2}})}

     = \frac{ e^{-i \frac{\theta}{2}} - e^{i(n+\frac{1}{2})\theta}}{-2i \sin{\frac{\theta}{2}}}

    The series is equal to the real part of it , eg

      \frac{1}{2} + \frac{ \sin{(n+\frac{1}{2})\theta}}{2\sin{\frac{\theta}{2  }}}
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  4. #4
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    thanks for moo"i understand your proof" and simplependulum
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