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Math Help - Trig Exam Question

  1. #1
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    Trig Exam Question

    I have had a go at this question but i feel i am on the wrong course as i end up with a far too complicated equation.

    This is the question:

    Solve, for 0<= θ<180, the equation
    4cosec ²2θ º=3+5cot2θ º


    Thanks for anyone who can show me how to do this in the most simple way.

    P.S. can someone please tell me how to insert the special maths images.

    P.P.S. Could do with knowing this by tommorow
    Last edited by Big_Joe; December 16th 2008 at 11:09 AM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Big_Joe View Post
    I have had a go at this question but i feel i am on the wrong course as i end up with a far too complicated equation.

    This is the question:

    Solve, for 0<= θ<180, the equation
    4cosec ²2θ º=3+5cot2θ º


    Thanks for anyone who can show me how to do this in the most simple way.
    4\csc^2\left(2\vartheta\right)=3+5\cot\left(2\vart  heta\right)

    Recall that \cot^2\varphi+1=\csc^2\varphi

    Thus, we get 4\left[\cot^2\left(2\vartheta\right)+1\right]=3+5\cot\left(2\vartheta\right)\implies 4\cot^2\left(2\vartheta\right)-5\cot\left(2\vartheta\right)+1=0

    Making the substitution z=\cot\left(2\vartheta\right) gives us the quadratic equation 4z^2-5z+1=0.

    This factors to \left(4z-1\right)\left(z-1\right)=0.

    This produces two different solutions:

    4z-1=0\implies z=\tfrac{1}{4} or z-1=0\implies z=1

    Since we let z=\cot\left(2\vartheta\right), this now means that:

    \cot\left(2\vartheta\right)=\tfrac{1}{4} or \cot\left(2\vartheta\right)=1

    Can you take it from here?

    P.S. can someone please tell me how to insert the special maths images.
    See the LaTeX help section here on the forums. You can generate the images by putting the LaTeX code in [tex][/tex] tags.
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  3. #3
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    Thanks a lot for that, I was on completely the wrong course trying to use double angle formula. I think i can complete this now. If i get the wrong answer i will post back tomorrow.

    Thanks for your help!
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  4. #4
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    I tried to continue the question but again i feel i am on the wrong track. I tried to find θ by using the double angle formula and saying:

    cot2θ=¼
    so 1/tan2θ=¼
    1/2tanθ/1-tan²θ=¼
    1-tan²θ/2tanθ=¼
    4(1-tan²θ) = 2tanθ
    so 4tan²θ + 2tanθ - 4 = 0
    2(2tanθ+2)(tanθ-1)=0
    therefore 2tanθ+2 = 0
    tanθ = -2/2
    θ=tan ‾¹(-1)
    θ=-45

    and tanθ-1 = 0
    θ=tan‾¹(1)
    θ=45

    Using my method i would also have to do this for the other equation for 2θ that i obtained earlier but the values i have got here are incorrect.
    Can someone please tell me where i went wrong!

    Edit i have tried to write my calculations using the Latex script:

    \cot{(2\theta)}=\frac{1}{4}

    so \frac{1}{\tan{(2\theta)}}=\frac{1}{4}

    \frac{1}{\frac{2\tan\theta}{1-\tan^2\theta}}=\frac{1}{4}

    \frac{1-\tan^2\theta}{2\tan\theta}=\frac{1}{4}

    4(1-\tan^2\theta)=2\tan\theta

    so 4\tan^2\theta+2\tan\theta-4=0

    2(2\tan\theta+2)(\tan\theta-1)=0

    therefore 2\tan\theta+2=0

    \tan\theta=\frac{-2}{2}

    \theta=\arctan{-1}

    \theta=-45

    and \tan\theta-1=0

    \theta=\arctan{1}

    \theta=45
    Last edited by Big_Joe; December 17th 2008 at 06:43 AM.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Big_Joe View Post
    I tried to continue the question but again i feel i am on the wrong track. I tried to find θ by using the double angle formula and saying:

    cot2θ=¼
    so 1/tan2θ=¼
    1/2tanθ/1-tan²θ=¼
    1-tan²θ/2tanθ=¼
    4(1-tan²θ) = 2tanθ
    so 4tan²θ + 2tanθ - 4 = 0
    2(2tanθ+2)(tanθ-1)=0
    therefore 2tanθ+2 = 0
    tanθ = -2/2
    θ=tan ‾¹(-1)
    θ=-45

    and tanθ-1 = 0
    θ=tan‾¹(1)
    θ=45

    Using my method i would also have to do this for the other equation for 2θ that i obtained earlier but the values i have got here are incorrect.
    Can someone please tell me where i went wrong!

    Edit i have tried to write my calculations using the Latex script:

    \cot{(2\theta)}=\frac{1}{4}

    so \frac{1}{\tan{(2\theta)}}=\frac{1}{4}

    \frac{1}{\frac{2\tan\theta}{1-\tan^2\theta}}=\frac{1}{4}

    \frac{1-\tan^2\theta}{2\tan\theta}=\frac{1}{4}

    4(1-\tan^2\theta)=2\tan\theta

    so 4\tan^2\theta+2\tan\theta-4=0

    2(2\tan\theta+2)(\tan\theta-1)=0

    therefore 2\tan\theta+2=0

    \tan\theta=\frac{-2}{2}

    \theta=\arctan{-1}

    \theta=-45

    and \tan\theta-1=0
    \theta=\arctan{1}

    \theta=45
    It looks alright...but one of your angles doesn't fall in the given interval 0\leq\vartheta\leq180!

    When you take \tan^{-1}\left(-1\right) you get one angle that is restricted by 0\leq\vartheta\leq180. The angle is in the second quadrant. In particular, \vartheta=135.

    When you take \tan^{-1}\left(1\right) you get one angle that is restricted by 0\leq\vartheta\leq180. The angle is in the first quadrant. In particular, \vartheta=45

    So your solution set would be \color{red}\boxed{\vartheta=45,~135}

    Does this make sense?
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  6. #6
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    Yes but the actual answers to the question are 22.5,112.5,38 and 128. Something must be wrong.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Big_Joe View Post
    Yes but the actual answers to the question are 22.5,112.5,38 and 128. Something must be wrong.
    Oh..

    They went ahead and directly evaluate it...

    \cot\left(2\vartheta\right)=1\implies \vartheta=\tfrac{1}{2}\cot^{-1}\left(1\right)

    But \cot^{-1}\left(1\right)=45 or \cot^{-1}\left(1\right)=225

    Thus, \vartheta=\tfrac{1}{2}\left(45\right)=\color{red}\  boxed{22.5} or \vartheta=\tfrac{1}{2}\left(225\right)=\color{red}  \boxed{112.5}

    But you will have to approximate the other two using a calculator...

    \cot\left(2\vartheta\right)=\tfrac{1}{4}\implies \vartheta=\tfrac{1}{2}\cot^{-1}\left(\tfrac{1}{4}\right)

    But \cot^{-1}\left(\tfrac{1}{4}\right)\approx 76 or \cot^{-1}\left(\tfrac{1}{4}\right)\approx 256

    Therefore, \vartheta=\tfrac{1}{2}\left(76\right)=\color{red}\  boxed{38} or \vartheta=\tfrac{1}{2}\left(256\right)=\color{red}  \boxed{128}

    So our solution set is \color{red}\boxed{\vartheta=22.5,~38,~112.5,~128}
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