Results 1 to 3 of 3

Thread: Tan - equation

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3

    Tan - equation

    Prove :
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    May 2009
    Posts
    612
    Thanks
    309
    $\displaystyle \tan^2\left(\frac{\pi}{12}\right) + \tan^2\left(\frac{5\pi}{12}\right)$
    $\displaystyle \begin{aligned}
    &= \tan^2\left(\frac{\pi}{12}\right) + \cot^2\left(\frac{\pi}{12}\right) \\
    &= \left[\csc \left(\frac{\pi}{6}\right) - \cot \left(\frac{\pi}{6}\right)\right]^2 + \left[\csc \left(\frac{\pi}{6}\right) + \cot \left(\frac{\pi}{6}\right)\right]^2
    \end{aligned}$
    $\displaystyle \begin{aligned}
    &= \csc^2 \left(\frac{\pi}{6}\right) - 2\csc \left(\frac{\pi}{6}\right)\cot \left(\frac{\pi}{6}\right) + \cot^2 \left(\frac{\pi}{6}\right) \\
    &\;\;+ \csc^2 \left(\frac{\pi}{6}\right) + 2\csc \left(\frac{\pi}{6}\right)\cot \left(\frac{\pi}{6}\right) + \cot^2 \left(\frac{\pi}{6}\right) \\
    &= 2\csc^2 \left(\frac{\pi}{6}\right) + 2\cot^2 \left(\frac{\pi}{6}\right) \\
    \end{aligned}$
    $\displaystyle \begin{aligned}
    &= 2\csc^2 \left(\frac{\pi}{6}\right) + 2\left(\csc^2 \left(\frac{\pi}{6}\right) - 1\right) \\
    &= 4\csc^2 \left(\frac{\pi}{6}\right) - 2 \\
    &= 4(2^2) - 2 \\
    &= 14
    \end{aligned}$

    Note: this was not meant to be an efficient way.


    01
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, dhiab!

    My method is not very efficient either . . .

    . . $\displaystyle \begin{array}{ccc}\sin^2\!\theta &=& \dfrac{1-\cos2\theta}{2} \\ \\[-4mm] \cos^2\!\theta &=& \dfrac{1+\cos2\theta}{2} \end{array}$



    $\displaystyle \text{Prove: }\;\tan^2\left(\frac{\pi}{12}\right) + \tan^2\left(\frac{5\pi}{12}\right) \:=\:14$

    We have: .$\displaystyle \frac{\sin^2(\frac{\pi}{12})}{\cos^2(\frac{\pi}{12 })} + \frac{\sin^2(\frac{5\pi}{12})}{\cos^2(\frac{5\pi}{ 12})} $ . $\displaystyle =\;\; \frac{\dfrac{1-\cos\frac{\pi}{6}}{2}} {\dfrac{1 + \cos\frac{\pi}{6}}{2}} + \frac{\dfrac{1-\cos\frac{5\pi}{6}}{2}} {\dfrac{1 + \cos\frac{5\pi}{6}}{2}} $

    . . . . . .$\displaystyle = \;\frac{1-\cos\frac{\pi}{6}}{1 + \cos\frac{\pi}{6}} + \frac{1-\cos\frac{5\pi}{6}}{1 + \cos\frac{5\pi}{6}} \;=\;\frac{1-\frac{\sqrt{3}}{2}} {1 + \frac{\sqrt{3}}{2}} + \frac{1+\frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}} $

    . . . . . .$\displaystyle =\; \frac{\left(1-\frac{\sqrt{3}}{2}\right)^2 + \left(1 + \frac{\sqrt{3}}{2}\right)^2} {1 - \left(\frac{\sqrt{3}}{2}\right)^2} \;=\; \frac{\left(1 - \sqrt{3} + \frac{3}{4}\right) + \left(1 + \sqrt{3} + \frac{3}{4}\right)}{1 - \frac{3}{4}}$

    . . . . . .$\displaystyle =\;\frac{\frac{14}{4}}{\frac{1}{4}} \;=\;\boxed{14}$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Apr 11th 2011, 01:17 AM
  2. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Aug 28th 2009, 11:39 AM
  3. Replies: 2
    Last Post: May 18th 2009, 12:51 PM
  4. Replies: 2
    Last Post: Apr 28th 2009, 06:42 AM
  5. Replies: 1
    Last Post: Oct 23rd 2008, 03:39 AM

Search Tags


/mathhelpforum @mathhelpforum