Results 1 to 3 of 3

Thread: Trigonometric Calculus

  1. #1
    Zel is offline
    Newbie Zel's Avatar
    Feb 2010

    Question Trigonometric Calculus

    I am having a bit of trouble with these, so if anyone could help me, I would greatly appreciate it.
    Thanks in advance

    1. Find from first principles the derivative of the function f(x) = tan(2x)

    a) Find the area between the graph of y = tan x and the x axis for -1≤ x ≤ 1, giving your answer to two decimal places
    b) Using calculus, show that ∫ tan x dx = -ln| cos x| + c, where c is an arbitrary constant
    c) Find a number k, such that ∫ tan x dx = 2, for 0≤ x≤ k, giving your answer to two decimal places

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Aug 2008
    2. b) $\displaystyle \int{\tan{x}\,dx} = \int{\frac{\sin{x}}{\cos{x}}\,dx}$

    $\displaystyle = -\int{\left(\frac{1}{\cos{x}}\right)(-\sin{x})\,dx}$.

    Now let $\displaystyle u = \cos{x}$ so that $\displaystyle \frac{du}{dx} = -\sin{x}$ and the integral becomes

    $\displaystyle -\int{\left(\frac{1}{u}\right)\,\frac{du}{dx}\,dx}$

    $\displaystyle = -\int{\frac{1}{u}\,du}$

    $\displaystyle = -\ln{|u|} + C$

    $\displaystyle = -\ln{|\cos{x}|} + C$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Apr 2005
    For 1, I presume you mean using $\displaystyle \lim_{h\to 0}\frac{tan(x+h)- tan(x)}{h}$.

    $\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$ so that is $\displaystyle \lim_{h\to 0}\frac{\frac{sin(x+h)}{cos(x+h)}- \frac{sin(x)}{cos(x)}}{h}$$\displaystyle =\lim_{h\to 0}\frac{sin(x+h)cos(x)- sin(x)cos(x+h)}{h cos(x)cos(x+h)}$$\displaystyle = \lim_{h\to 0}\frac{sin(x+h)cos(x)- sin(x)cos(x)+ sin(x)cos(x)- cos(x)sin(x+ h)}{h cos(x)cos(x+h)}$$\displaystyle = \left(cos(x)\lim(h\to 0}\frac{sin(x+h)- sin(x)}{h}- cos(x+h)\lim_{h\to 0}\frac{sin(x+h)- sin(x)}{h}\right)\left(\lim_{h\to 0}\frac{1}{cos(x)cos(x+h)}$

    Use limits you know about sine and cosine.

    Prove It showed how to do 2b. That should make 2a easy. For 2c, set the formula for $\displaystyle \int_0^k tan(x)dx$, that you get from 2b, equal to 2 and solve for k.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometric Calculus
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Nov 28th 2010, 08:37 AM
  2. Replies: 6
    Last Post: Aug 29th 2010, 05:23 PM
  3. Replies: 2
    Last Post: May 12th 2009, 08:36 AM
  4. Replies: 1
    Last Post: Jun 23rd 2008, 09:17 AM
  5. Pre-Calculus: Trigonometric forumlas and more
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Apr 5th 2007, 04:03 AM

Search Tags

/mathhelpforum @mathhelpforum