Results 1 to 3 of 3

Math Help - Proof by induction stuck on trigonometric sum :help:

  1. #1
    Member ssadi's Avatar
    Joined
    Oct 2008
    Posts
    104

    Proof by induction stuck on trigonometric sum :help:

    How do I convert:

    tan(k+1)\theta cot\theta-k-1+tan(k+1)\theta tan(k+2)\theta

    =tan(k+1)\theta[cot\theta+tan(k+2)\theta]-k-1(As far as I could get)

    Into:

    tan(k+2)\theta cot\theta-k-1

    Code:
    Original sum:
      Use the method of induction to prove that:
       
      
    
    
    
    
    \displaystyle\sum_{r=1}^n tanr\theta tan(r+1)\theta=tan(n+1)\theta cot\theta-n-1 
    Help will be appreciated big time
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by ssadi View Post
    How do I convert:

    tan(k+1)\theta cot\theta-k-1+tan(k+1)\theta tan(k+2)\theta

    =tan(k+1)\theta[cot\theta+tan(k+2)\theta]-k-1(As far as I could get)

    Into:

    tan(k+2)\theta cot\theta-{\color{red}(k+1)}-1

    Code:
    Original sum:
      Use the method of induction to prove that:
       
      
    
    
    
    
    \displaystyle\sum_{r=1}^n tanr\theta tan(r+1)\theta=tan(n+1)\theta cot\theta-n-1 
    Notice that for the inductive step you need that (k+1) in place of k.

    Use the formula \tan(\phi-\psi) = \frac{\tan\phi-\tan\psi}{1+\tan\phi\tan\psi}, taking \phi=(k+1)\theta, \psi=\theta. That tells you that \tan(k+1)\theta = \frac{\tan(k+2)\theta-\tan\theta}{1+\tan\theta\tan(k+2)\theta}.

    Next, \tan(k+1)\theta\bigl(\cot\theta+\tan(k+2)\theta\bi  gr) = \cot\theta\tan(k+1)\theta\bigl(1+\tan\theta\tan(k+  2)\theta\bigr). Substitute the above fractional expression for \tan(k+1)\theta into that, and it will become \tan(k+2)\theta\cot\theta-1, which is what you want.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member ssadi's Avatar
    Joined
    Oct 2008
    Posts
    104
    Quote Originally Posted by Opalg View Post
    Notice that for the inductive step you need that (k+1) in place of k.

    Use the formula \tan(\phi-\psi) = \frac{\tan\phi-\tan\psi}{1+\tan\phi\tan\psi}, taking \phi=(k+1)\theta, \psi=\theta. That tells you that \tan(k+1)\theta = \frac{\tan(k+2)\theta-\tan\theta}{1+\tan\theta\tan(k+2)\theta}.

    Next, \tan(k+1)\theta\bigl(\cot\theta+\tan(k+2)\theta\bi  gr) = \cot\theta\tan(k+1)\theta\bigl(1+\tan\theta\tan(k+  2)\theta\bigr). Substitute the above fractional expression for \tan(k+1)\theta into that, and it will become \tan(k+2)\theta\cot\theta-1, which is what you want.
    This trigonmetric stuff is tough, i may need some time to get the hang of it properly.
    Many thanks for your effort, even if it is childplay for you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stuck on proof by induction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 17th 2010, 08:21 AM
  2. Proof by induction stuck at very end!!
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 25th 2009, 03:38 AM
  3. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 09:33 PM
  4. Proof by induction stuck on a sum:help:
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 15th 2008, 10:27 AM
  5. Induction proof...I'm stuck!
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 18th 2008, 10:24 AM

Search Tags


/mathhelpforum @mathhelpforum