Results 1 to 6 of 6

Thread: proof by induction, trigonometric identity help:

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    1

    proof by induction, trigonometric identity help:

    first up, sorry if this is the wrong forum to post but i was unsure where to post it under.

    i am trying to prove that:

    $\displaystyle
    1 + 2* sum( cos(ku), k=1 to n) = sin( (2n+1)u/2) / sin(u/2):
    $
    (ment to read sum( cos(ku), k=1 to n))

    my induction assumption is that:
    $\displaystyle
    1 + 2( cos(u) +.....+ cos(ku) ) = sin( ku + u/2)/sin(u/2)
    $

    and so i am trying to reduce:
    $\displaystyle
    sin(ku + u/2)/sin(u/2) + 2cos( ku + u)
    $
    to become
    $\displaystyle
    sin( ku + 3u/2) / sin(u/2)
    $

    this is where i am stuck. any help would be appreciated greatly
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    A solution without using induction!


    $\displaystyle cos(u) + cos(2u) + cos(3u) + ... + cos(nu)
    $

    Let$\displaystyle T = \Sigma_{k = 1}^ {n}[ cos(ku) ]$


    Multiplying by $\displaystyle 2 sin(\frac{u}{ 2})$:


    $\displaystyle 2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n}[2 sin(\frac{u}{ 2})cos(ku) ]
    $

    Now, using the identity $\displaystyle 2 sin(\frac{u}{ 2}) cos(ku) = sin( [ \frac{(2k + 1) }{ 2} ]u ) - sin( [ \frac{(2k - 1)}{ 2} ]u )$:


    $\displaystyle 2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n}[ sin( [ \frac{(2k + 1)}{ 2 }]u ) - sin( [\frac{(2k - 1)} { 2} ]u ) ]$


    Splitting the R.H.S into two separate summations, we will get:

    $\displaystyle 2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n} [ sin( [\frac{ (2k + 1)} { 2} ]u ) ] - \Sigma_{k = 1}^ {n} sin( [ \frac{(2k - 1)} { 2} ]u ) ]$

    Or:

    $\displaystyle 2T sin(\frac{u}{ 2})) = \Sigma _{k = 2}^ {n } [ sin( [ \frac{(2k - 1)} { 2} ]u ) ] - \Sigma_{k = 1}^ {n} [ sin( [\frac{ (2k - 1)} { 2} ]u ) ]
    $

    The above is telescopic series:

    $\displaystyle 2T sin(\frac{u}{ 2})
    $
    $\displaystyle = sin( [ \frac{(2n + 1)} { 2} ]u ) - sin(\frac{u}{ 2})
    $
    $\displaystyle = sin [\frac{ (2n + 1)} { 2} ]u ) - sin(\frac{u}{ 2})
    $
    $\displaystyle = 2 sin(\frac{nu }{ 2}) cos( \frac{[ n + 1 ]u} { 2} )
    $

    Hence:
    $\displaystyle T =\frac{ sin(\frac{nu}{ 2}) cos( \frac{[ n + 1 ]u} { 2} ) }{sin(\frac{u}{ 2})}$
    Last edited by Also sprach Zarathustra; Aug 8th 2010 at 01:53 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    First of all, let's write this in a form that is readable...

    You're asked to prove that

    $\displaystyle 1 + 2\sum_{k = 1}^{n}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 1)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$.


    Base step $\displaystyle n = 1$:

    $\displaystyle LHS = 1 + 2\cos{u}$.

    $\displaystyle RHS = \frac{\sin{\left(\frac{3u}{2}\right)}}{\sin{\left( \frac{u}{2}\right)}}$

    $\displaystyle = \frac{\sqrt{\frac{1 - \cos{3u}}{2}}}{\sqrt{\frac{1 - \cos{u}}{2}}}$

    $\displaystyle = \sqrt{\frac{\frac{1 - \cos{3u}}{2}}{\frac{1 - \cos{u}}{2}}}$

    $\displaystyle = \sqrt{\frac{1 - \cos{3u}}{1 - \cos{u}}}$

    $\displaystyle = \sqrt{\frac{1 - (4\cos^3{u} - 3\cos{u})}{1 - \cos{u}}}$

    $\displaystyle = \sqrt{\frac{1 + 3\cos{u} - 4\cos^3{u}}{1 - \cos{u}}}$

    $\displaystyle = \sqrt{\frac{(1 - \cos{u})(2\cos{u} + 1)^2}{1 - \cos{u}}}$

    $\displaystyle = \sqrt{(2\cos{u} + 1)^2}$

    $\displaystyle = 2\cos{u} + 1$

    $\displaystyle = LHS$.


    Inductive step - Assume the statement is true for $\displaystyle n$.

    So we assume $\displaystyle 1 + 2\sum_{k = 1}^{n}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 1)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$.

    Now we need to show that the statement is true for $\displaystyle n + 1$, i.e. that
    $\displaystyle 1 + 2\sum_{k = 1}^{n+1}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 3)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$.


    $\displaystyle LHS = 1 + 2\sum_{k = 1}^{n+1}\cos{(ku)}$.


    $\displaystyle RHS = \frac{\sin{\left[\frac{(2n + 3)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$

    $\displaystyle = \frac{\sin{\left[\frac{(2n+1)u}{2} + \frac{2u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$

    $\displaystyle = \frac{\sin{\left[\frac{(2n+1)u}{2} + u\right]}}{\sin{\left(\frac{u}{2}\right)}}$

    $\displaystyle = \frac{\sin{\left[\frac{(2n+1)u}{2}\right]}\cos{(u)}+\cos{\left[\frac{(2n+1)u}{2}\right]}\sin{(u)}}{\sin{\left(\frac{u}{2}\right)}}$


    Now I'm a little stuck too, will keep thinking about it...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    Also sprach, you are looking for $\displaystyle 1+2T$, not just $\displaystyle T$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    BTW, this sum is known as the Dirichlet Kernel Clicky!

    EDIT: Ahh, double post lol. I should have just edited!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by stalker123 View Post
    first up, sorry if this is the wrong forum to post but i was unsure where to post it under.

    i am trying to prove that:

    $\displaystyle
    1 + 2* sum( cos(ku), k=1 to n) = sin( (2n+1)u/2) / sin(u/2):
    $
    (ment to read sum( cos(ku), k=1 to n))

    my induction assumption is that:
    $\displaystyle
    1 + 2( cos(u) +.....+ cos(ku) ) = sin( ku + u/2)/sin(u/2)
    $

    and so i am trying to reduce:
    $\displaystyle
    sin(ku + u/2)/sin(u/2) + 2cos( ku + u)
    $
    to become
    $\displaystyle
    sin( ku + 3u/2) / sin(u/2)
    $

    this is where i am stuck. any help would be appreciated greatly
    P(n)

    $\displaystyle 1+2\displaystyle\huge\sum_{k=1}^ncos(ku)=\frac{sin \left[(2n+1)\frac{u}{2}\right]}{sin\left(\frac{u}{2}\right)}$

    P(n+1)

    $\displaystyle 1+2\displaystyle\huge\sum_{k=1}^{n+1}cos(ku)=\frac {sin\left[(2[n+1]+1)\frac{u}{2}\right]}{sin\left(\frac{u}{2}\right)}\ ?$

    Proof

    $\displaystyle 1+2\displaystyle\huge\sum_{k=1}^{n+1}cos(ku)=1+2\s um_{k=1}^ncosku+2cos[(n+1)u]$

    If P(n) is valid, this will equal

    $\displaystyle \displaystyle\huge\frac{sin\left[(2n+1)\frac{u}{2}\right]}{sin\frac{u}{2}}+2cos[(n+1)u]$

    Hence, the question is

    $\displaystyle \displaystyle\huge\frac{sin\left([2(n+1)+1]\frac{u}{2}\right)}{sin\frac{u}{2}}=\frac{sin\left ([2n+1]\frac{u}{2}\right)+\left(sin\frac{u}{2}\right)2cos[(n+1)u]}{sin\frac{u}{2}}\ ?$

    $\displaystyle sin\left([2n+3]\frac{u}{2}\right)=sin\left([2n+1]\frac{u}{2}\right)+2cos\left([n+1]u\right)sin\frac{u}{2}\ ?$

    Expanding the rightmost term

    $\displaystyle 2cos\left([n+1]u\right)sin\frac{u}{2}=sin\left([n+1]u+\frac{u}{2}\right)-sin\left([n+1]u-\frac{u}{2}\right)$

    $\displaystyle =sin\left([2n+3]\frac{u}{2}\right)-sin\left([2n+1]\frac{u}{2}\right)$

    The inter-term relationship is proven,
    hence the trail of gunpowder has been placed all the way to that bank.
    "What bank?"
    That bank at infinity!

    All that's required is to prove the equality is true for n=1..

    $\displaystyle 1+2cosu=\displaystyle\huge\frac{sin\frac{3u}{2}}{s in\frac{u}{2}}=\frac{sin\left(u+\frac{u}{2}\right) }{sin\frac{u}{2}}$

    and since

    $\displaystyle sinu=2sin\frac{u}{2}cos\frac{u}{2}$

    and using

    $\displaystyle sin(A+B)=sinAcosB+cosAsinB$

    the result follows

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] To divide by zero in a trigonometric identity proof
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Jul 10th 2011, 03:46 AM
  2. Trigonometric identity - HELP please
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jun 17th 2010, 09:12 PM
  3. Proof by induction of an identity.
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Dec 7th 2009, 06:50 AM
  4. Proof by induction stuck on trigonometric sum :help:
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Dec 15th 2008, 11:45 AM
  5. trigonometric identity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 5th 2008, 03:42 AM

Search Tags


/mathhelpforum @mathhelpforum