Results 1 to 4 of 4

Math Help - Derivative of trigonometric functions

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    2

    Derivative of trigonometric functions

    1) lim as theata approaches 0 --> 1-cos theata/theata^2

    2) lim as theaata approaches 0 --> tan theata/2 theata


    Please explain how each is done please. THank yoU! I am supposed to determine the limits for each.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    first of all, it's "theta"

    it's a lot better than some people though

    i don't mind mispelling as much as some people though, but now you know, so you can spell it right next time

    Quote Originally Posted by superweirdo View Post
    1) lim as theata approaches 0 --> 1-cos theata/theata^2
    one way that everyone will hate is to apply L'Hopital's rule:

    thus, \lim_{\theta \to 0} \frac {1 - \cos \theta}{\theta ^2} = \lim_{\theta \to 0} \frac {\sin \theta}{2 \theta}

    i think you can take it from there. Hint: think "special limit"


    if you want to do it without L'Hopital's, power series is another method. too much work here though, i think

    (do you know what L'Hopital's rule is? do you know how to do it using power series?)


    2) lim as theaata approaches 0 --> tan theata/2 theata
    \lim_{\theta \to 0} \frac {\tan \theta}{2 \theta} is also a special limt (without the 2 in the denominator that is, but it doesn't matter, we can factor constants out).

    if you want to do it the hard way, you can note that: \lim_{\theta \to 0} \frac {\tan \theta}{2 \theta} = \lim_{\theta \to 0} \frac {\sin \theta}{2 \theta \cos \theta}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,711
    Thanks
    630
    Hello, superweirdo!

    We need the theorem: . \lim_{\theta\to0}\frac{\sin\theta}{\theta}\;=\;1


    1)\;\;\lim_{\theta\to0}\frac{1-\cos\theta}{\theta^2}
    Multiply by \frac{1+\cos\theta}{1+\cos\theta}: . \frac{1-\cos\theta}{\theta^2}\cdot\frac{1+\cos\theta}{1+\c  os\theta} \;=\;\frac{1-\cos^2\!\theta}{\theta^2(1+\cos\theta)} \;=\;\frac{\sin^2\!\theta}{\theta^2(1+\cos\theta)}

    then: . \lim_{\theta\to0}\left[\left(\frac{\sin\theta}{\theta}\right)^2\cdot\frac  {1}{1+\cos\theta}\right] \;=\;1^2\cdot\frac{1}{1+1} \;=\;\frac{1}{2}



    2)\;\;\lim_{\theta\to0}\frac{\tan\theta}{2\theta}

    We have: . \frac{\frac{\sin\theta}{\cos\theta}}{2\theta}\;=\;  \frac{\sin\theta}{2\theta\cos\theta}<br />

    Then: . \lim_{\theta\to0}\left[\frac{1}{2}\cdot\frac{\sin\theta}{\theta}\cdot\fra  c{1}{\cos\theta}\right] \;=\;\frac{1}{2}\cdot1\cdot\frac{1}{1}\;=\;\frac{1  }{2}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2008
    Posts
    2
    Quote Originally Posted by Soroban View Post
    Hello, superweirdo!

    We need the theorem: . \lim_{\theta\to0}\frac{\sin\theta}{\theta}\;=\;1


    Multiply by \frac{1+\cos\theta}{1+\cos\theta}: . \frac{1-\cos\theta}{\theta^2}\cdot\frac{1+\cos\theta}{1+\c  os\theta} \;=\;\frac{1-\cos^2\!\theta}{\theta^2(1+\cos\theta)} \;=\;\frac{\sin^2\!\theta}{\theta^2(1+\cos\theta)}

    then: . \lim_{\theta\to0}\left[\left(\frac{\sin\theta}{\theta}\right)^2\cdot\frac  {1}{1+\cos\theta}\right] \;=\;1^2\cdot\frac{1}{1+1} \;=\;\frac{1}{2}




    We have: . \frac{\frac{\sin\theta}{\cos\theta}}{2\theta}\;=\;  \frac{\sin\theta}{2\theta\cos\theta}<br />

    Then: . \lim_{\theta\to0}\left[\frac{1}{2}\cdot\frac{\sin\theta}{\theta}\cdot\fra  c{1}{\cos\theta}\right] \;=\;\frac{1}{2}\cdot1\cdot\frac{1}{1}\;=\;\frac{1  }{2}

    \

    THANK YOU SO MUCH! YOU'RE SO SMART!! You definitely deserve the "best helper" badge =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] derivative of rational/trigonometric functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 25th 2011, 07:11 PM
  2. Replies: 1
    Last Post: November 7th 2010, 02:37 PM
  3. Replies: 6
    Last Post: August 29th 2010, 05:23 PM
  4. Replies: 8
    Last Post: September 13th 2009, 03:34 AM
  5. Replies: 4
    Last Post: November 3rd 2008, 06:09 PM

Search Tags


/mathhelpforum @mathhelpforum