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Math Help - Trigonometric Identity

  1. #1
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    Trigonometric Identity

    Hi, I need help on this problem:

    Use the identity  2\sin{\frac{x}{2}}\cos{kx} = <br />
\sin{(2k+1)}\frac{x}{2} - \sin{(2k-1)}\frac{x}{2}  and the telescoping property of finite sums to prove that if  x \neq 2m\pi (m an integer), we have  \sum_{k=1}^{n}  \cos{kx}  = \frac{\sin{\frac{nx}{2}}\cos{\frac{1}{2}(n+1)x}}{\  sin{\frac{1}{2}x}}

    Here is what I have so far:

    Taking  k= 1,2,...,n
     2\sin{\frac{x}{2}}\cos{kx}= 2\sin{\frac{x}{2}}\sum_{k=1}^{n}\cos{kx} =

    [\sin{\frac{3x}{2}}-\sin{\frac{x}{2}} ] +  [\sin{\frac{5x}{2}}-\sin{\frac{3x}{2}}]+ ... + [\sin{\frac{(2n-1)x}{2}}-\sin{\frac{(2n-3)x}{2}}] + [\sin{\frac{(2n+1)x}{2}} -\sin{\frac{(2n-1)x}{2}}]

     = \sin{\frac{(2n+1)x}{2}} - \sin{\frac{x}{2}} =  \sin{(n+ \frac{1}{2})x } - \sin{\frac{x}{2}}

    This reduces to  \sin{nx}\cos{\frac{x}{2}} + \cos{nx}\sin{\frac{x}{2}} -\sin{\frac{x}{2}}

    Now I'm stuck.
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  2. #2
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    Hi !
    Quote Originally Posted by PersaGell View Post
    Hi, I need help on this problem:

    Use the identity  2\sin{\frac{x}{2}}\cos{kx} = <br />
\sin{(2k+1)}\frac{x}{2} - \sin{(2k-1)}\frac{x}{2}  and the telescoping property of finite sums to prove that if  x \neq 2m\pi (m an integer), we have  \sum_{k=1}^{n}  \cos{kx}  = \frac{\sin{\frac{nx}{2}}\cos{\frac{1}{2}(n+1)x}}{\  sin{\frac{1}{2}x}}

    Here is what I have so far:

    Taking  k= 1,2,...,n
    {\color{red}\sum_{k=1}^n} 2\sin{\frac{x}{2}}\cos{kx}= 2\sin{\frac{x}{2}}\sum_{k=1}^{n}\cos{kx} =

    [\sin{\frac{3x}{2}}-\sin{\frac{x}{2}} ] +  [\sin{\frac{5x}{2}}-\sin{\frac{3x}{2}}]+ ... + [\sin{\frac{(2n-1)x}{2}}-\sin{\frac{(2n-3)x}{2}}] + [\sin{\frac{(2n+1)x}{2}} -\sin{\frac{(2n-1)x}{2}}]

     = \sin{\frac{(2n+1)x}{2}} - \sin{\frac{x}{2}}
    (There's the red part that was missing )

    Once you're here, try to think "how can I transform a difference of sines into a product of a sine and a cosine ?" and then use this formula (which is quite the same as the identity you're given at the very beginning) :

    \sin(a)-\sin(b)=2 \cos \left(\frac{a+b}{2} \right) \cdot \sin \left(\frac{a-b}{2}\right)

    And you're done

    (this formula is available, as well as many others, here : Trigonometry)
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  3. #3
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    It is even easier.
    Use complex numbers and geometric series.
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  4. #4
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    Your last line is wrong, which is
    <br />
\sin{nx}\cos{\frac{x}{2}} + \cos{nx}\sin{\frac{x}{2}} -\sin{\frac{x}{2}}<br />

    See here,
    <br />
=\sin{(n+ \frac{1}{2})x } - \sin{\frac{x}{2}}<br />
    <br />
=2\cos{(\frac{n+1}{2})x }\sin{\frac{nx}{2}}<br />
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