Results 1 to 4 of 4

Math Help - Trigonometric Identity

  1. #1
    Newbie
    Joined
    May 2007
    Posts
    4

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    It is even easier.
    Use complex numbers and geometric series.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Aug 2008
    Posts
    530
    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 />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometric Identity
    Posted in the Trigonometry Forum
    Replies: 0
    Last Post: August 20th 2010, 09:23 PM
  2. Trigonometric identity help
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 1st 2010, 02:35 PM
  3. trigonometric identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 05:35 PM
  4. trigonometric identity
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: May 19th 2008, 08:18 PM
  5. trigonometric identity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 5th 2008, 03:42 AM

Search Tags


/mathhelpforum @mathhelpforum