Results 1 to 15 of 15

Thread: Trig equation, is this correct?

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    631

    Trig equation, is this correct?

    Solve the following equations in the given intervals.

    $\displaystyle sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $

    $\displaystyle 0\leq \theta \leq 2\pi $
    $\displaystyle sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $

    $\displaystyle 1 + tan^2 \theta = sec^2 \theta $

    $\displaystyle 1 + tan^2 \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $

    $\displaystyle tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0 $

    $\displaystyle tan^2 \theta - tan \theta - \sqrt{3}tan \theta = -\sqrt{3} $

    $\displaystyle tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3} $

    so

    $\displaystyle tan \theta = -\sqrt{3} $

    or

    $\displaystyle tan \theta = 1 $

    is this method correct? thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Tweety View Post
    $\displaystyle sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $

    $\displaystyle 1 + tan^2 \theta = sec^2 \theta $

    $\displaystyle 1 + tan^2 \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1 $

    $\displaystyle tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0 $

    $\displaystyle tan^2 \theta - tan \theta - \sqrt{3}tan \theta = -\sqrt{3} $

    $\displaystyle tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3} $

    so

    $\displaystyle tan \theta = -\sqrt{3} $

    or

    $\displaystyle tan \theta = 1 $

    is this method correct? thanks!
    it is wrong
    let $\displaystyle tan(\theta)=t$ and solve it by $\displaystyle t=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ in your question a=1 , b=-(1+sqrt(3)) , c=sqrt(3)

    then

    $\displaystyle tan(\theta)=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$

    $\displaystyle \theta = \tan^{-1}\left(\frac{-b\mp \sqrt{b^2-4ac}}{2a}\right)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by Amer View Post
    it is wrong
    let $\displaystyle tan(\theta)=t$ and solve it by $\displaystyle t=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ in your question a=1 , b=-(1+sqrt(3)) , c=sqrt(3)

    then

    $\displaystyle tan(\theta)=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$

    $\displaystyle \theta = \tan^{-1}\left(\frac{-b\mp \sqrt{b^2-4ac}}{2a}\right)$
    Hi,

    I dont understand how you got to the quadratic equation?

    the correct answers are, $\displaystyle \frac{\pi}{4} , \frac{\pi}{3} , \frac{5\pi}{4} , \frac{4\pi}{3} $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    $\displaystyle tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3}
    $

    You can't factorise when it's like that, you can only factorise when one side is equal to 0

    $\displaystyle
    tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0
    $

    You can use the quadratic equation from there. If it helps let $\displaystyle u = tan \theta$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Tweety View Post
    Hi,

    I dont understand how you got to the quadratic equation?

    the correct answers are, $\displaystyle \frac{\pi}{4} , \frac{\pi}{3} , \frac{5\pi}{4} , \frac{4\pi}{3} $
    $\displaystyle tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0$

    let t=tan(theta)

    $\displaystyle t^2 -(1+\sqrt{3})t+\sqrt{3}=0$

    $\displaystyle t=\frac{1+\sqrt{3}\mp \sqrt{1+2\sqrt{3}+3-4(\sqrt{3})}}{2}$

    $\displaystyle t=\frac{1+\sqrt{3}\mp \sqrt{1-2\sqrt{3}+3}}{2}$

    $\displaystyle t=\frac{1+\sqrt{3}\mp \sqrt{(1-\sqrt{3})^2}}{2}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by Amer View Post
    $\displaystyle tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0$

    let t=tan(theta)

    $\displaystyle t^2 -(1+\sqrt{3})t+\sqrt{3}=0$

    $\displaystyle t=\frac{1+\sqrt{3}\mp \sqrt{1+2\sqrt{3}+3-4(\sqrt{3})}}{2}$

    $\displaystyle t=\frac{1+\sqrt{3}\mp \sqrt{1-2\sqrt{3}+3}}{2}$

    $\displaystyle t=\frac{1+\sqrt{3}\mp \sqrt{(1-\sqrt{3})^2}}{2}$
    Oh right, I get it now thanks
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    $\displaystyle \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 $

    $\displaystyle (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$

    $\displaystyle \tan{\theta} = \sqrt{3}$ ... $\displaystyle \theta = \frac{\pi}{3}$ , $\displaystyle \frac{4\pi}{3}
    $

    $\displaystyle \tan{\theta} = 1$ ... $\displaystyle \theta = \frac{\pi}{4}$ , $\displaystyle \frac{5\pi}{4}$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by skeeter View Post
    $\displaystyle \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 $

    $\displaystyle (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$

    $\displaystyle \tan{\theta} = \sqrt{3}$ ... $\displaystyle \theta = \frac{\pi}{3}$ , $\displaystyle \frac{4\pi}{3}
    $

    $\displaystyle \tan{\theta} = 1$ ... $\displaystyle \theta = \frac{\pi}{4}$ , $\displaystyle \frac{5\pi}{4}$
    how did you get this expression $\displaystyle (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$
    from
    $\displaystyle \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 $

    ? is it a difference of two square??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003
    Quote Originally Posted by Tweety View Post
    how did you get this expression $\displaystyle (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$
    from
    $\displaystyle \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 $

    ? is it a difference of two square??
    Suppose you had $\displaystyle x^2- (1+ a)x+ a$. Could you see that it factors as (x-1)(x-a)? Do you see that a factors as (-1-)(a) and those add to -(1+ a), the coefficient of x?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Tweety View Post
    how did you get this expression $\displaystyle (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$
    from
    $\displaystyle \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 $

    ? is it a difference of two square??
    look at this

    $\displaystyle x^2-5x+6 \Rightarrow (x-2)(x-3)$ -2,-3 the multiply of them is 6 and the sum is -5 . 6 is the constant term and -5 is the factor of x

    $\displaystyle x^2-4x+3\Rightarrow (x-1)(x-3)$ -1,-3 multiply of them is 3 and the sum is -4 .3 is the constant term and -5 is the factor of x

    like this or use

    $\displaystyle x=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ if you do not understand it
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Difference of 2 squares here is not needed.

    Skeeter has used the fact that $\displaystyle -\sqrt{3}\times -1 = \sqrt{3}$

    and that $\displaystyle -\sqrt{3}+ -1 = -(\sqrt{3}+1)$ like what you would do when solving an ordinary quadratic that has 1 as the coeffecient for the $\displaystyle x^2$ term
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by HallsofIvy View Post
    Suppose you had $\displaystyle x^2- (1+ a)x+ a$. Could you see that it factors as (x-1)(x-a)? Do you see that a factors as (-1-)(a) and those add to -(1+ a), the coefficient of x?

    if you had the expression $\displaystyle (x-1)(x-a) $ this expands out to $\displaystyle x^2-ax-x +a $

    right?

    Than you can't really factor it any further to get back to the original expresssion. which is why I can't understand how it turn outs to be (x-1)(x-a)
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by Amer View Post
    look at this
    Quote Originally Posted by Amer View Post

    $\displaystyle x^2-5x+6 \Rightarrow (x-2)(x-3)$ -2,-3 the multiply of them is 6 and the sum is -5 . 6 is the constant term and -5 is the factor of x

    $\displaystyle x^2-4x+3\Rightarrow (x-1)(x-3)$ -1,-3 multiply of them is 3 and the sum is -4 .3 is the constant term and -5 is the factor of x

    like this or use

    $\displaystyle x=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ if you do not understand it
    I understand your examples but how does this relate to the trig expression, are you saying that I can use the same method to factorize
    $\displaystyle \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 $ As I would with $\displaystyle x^2-5x+6 $ ?

    thanks.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    Quote Originally Posted by Tweety View Post

    I understand your examples but how does this relate to the trig expression, are you saying that I can use the same method to factorize
    $\displaystyle \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 $ As I would with $\displaystyle x^2-5x+6 $ ?

    thanks.
    yes ...

    $\displaystyle x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x^2 - (3+2)x + 6$

    the only difference here is that $\displaystyle 3$ and $\displaystyle 2$ can be combined, wheras the $\displaystyle 1$ and $\displaystyle \sqrt{3}$ cannot.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by skeeter View Post
    yes ...

    $\displaystyle x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x^2 - (3+2)x + 6$

    the only difference here is that $\displaystyle 3$ and $\displaystyle 2$ can be combined, wheras the $\displaystyle 1$ and $\displaystyle \sqrt{3}$ cannot.
    oh right finally I get it, that makes so much sense now!!

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving a trig equation - Is my work correct?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Nov 28th 2009, 09:26 AM
  2. Is this equation correct?
    Posted in the Algebra Forum
    Replies: 10
    Last Post: Aug 13th 2009, 09:04 AM
  3. Is this tensor equation correct?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Jul 11th 2009, 10:42 PM
  4. trig question (correct me) [part 1]
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 26th 2009, 09:05 AM
  5. Log equation is this correct?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Dec 16th 2008, 11:20 PM

Search Tags


/mathhelpforum @mathhelpforum