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Math Help - Trigonometry equation?

  1. #1
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    Trigonometry equation?

    Given the equation cos2x +7cosx-3 = 0 find the tan value of x

    I've tried using 2cos^2x-1 +7cosx-3=0 but I don't know what to do next??

    and one more: sin 105 - sin 15 = ???

    Please help. Thanks
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  2. #2
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    Hello, SodaCup!

    Given the equation: . \cos2x +7\cos x-3 \:=\: 0, find \tan x

    I've tried using: . 2\cos^2\!x-1 +7\cos x-3\:=\:0
    but I don't know what to do next.

    How about combining the -1 and -3?
    We have: . 2\cos^2\!x + 7\cos x - 3 \:=\:0

    Factor: . (\cos x + 4)(2\cos x - 1) \:=\:0


    And we have two equations to solve:

    . . \cos x + 4 \;=\:0 \quad\Rightarrow\quad \cos x \:=\:-4\quad\hdots\;\text{ no real roots}

    . . 2\cos x - 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad x \:=\:\pm\tfrac{\pi}{3}

    Therefore: . \tan\left(\pm\tfrac{\pi}{3}\right) \:=\:\pm\sqrt{3}




    Evaluate: . \sin105^o - \sin15^o
    We can use two identities: . \begin{array}{ccc}\sin\theta &=& \sqrt{\dfrac{1-\cos2\theta}{2}} \\ \\[-3mm] \cos\theta &=& \sqrt{\dfrac{1+\cos2\theta}{2}} \end{array}


    So that: . \begin{array}{ccccccccc}<br />
\sin105^o &=& \sqrt{\dfrac{1 - \cos210^o}{2}} &=& \sqrt{\dfrac{1-\left(-\frac{\sqrt{3}}{2}\right)}{2}} &=& \sqrt{\dfrac{2+\sqrt{3}}{4}} &=& \dfrac{\sqrt{2+\sqrt{3}}}{2} \\ \\[-3mm]<br /> <br />
\sin15^o &=& \sqrt{\dfrac{1-\cos30^o}{2}} &=& <br />
\sqrt{\dfrac{1 - \frac{\sqrt{3}}{2}}{4}} &=&  <br />
\sqrt{\dfrac{2-\sqrt{3}}{4}} &=& \dfrac{\sqrt{2-\sqrt{3}}}{2}<br />
\end{array}


    Then: . \sin105^o - \sin15^o \;=\; \frac{\sqrt{2+\sqrt{3}}}{2} - \frac{\sqrt{2-\sqrt{3}}}{2}


    Note that: . \begin{array}{ccc}2+\sqrt{3} &=& \dfrac{(\sqrt{3}+1)^2}{2} \\ \\[-3mm]<br />
2 - \sqrt{3} &=& \dfrac{(\sqrt{3}-1)^2}{2} \end{array}


    \text{Therefore: }\;\sin105^o - \sin15^o \;=\;\dfrac{\dfrac{\sqrt{3}+1}{\sqrt{2}} - \dfrac{\sqrt{3}-1}{\sqrt{2}}}{2} \;=\;\frac{1}{\sqrt{2}}

    Last edited by Soroban; November 13th 2009 at 06:12 AM.
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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, SodaCup!




    And we have two equations to solve:

    . . \cos x + 4 \;=\:0 \quad\Rightarrow\quad \cos x \:=\:-4\quad\hdots\;\text{ no real roots}

    . . 2\cos x - 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad x \:=\:\pm\tfrac{\pi}{3}

    Therefore: . \tan\left(\pm\tfrac{\pi}{3}\right) \:=\:\pm\sqrt{3}
    Um sorry I still don't get it, how do you find √3 from cosx = -4 and cosx = 1/2?

    Quote Originally Posted by Soroban View Post
    We can use two identities: . \begin{array}{ccc}\sin\theta &=& \sqrt{\dfrac{1-\cos2\theta}{2}} \\ \\[-3mm] \cos\theta &=& \sqrt{\dfrac{1+\cos2\theta}{2}} \end{array}


    So that: . \begin{array}{ccccccccc}<br />
\sin105^o &=& \sqrt{\dfrac{1 - \cos210^o}{2}} &=& \sqrt{\dfrac{1-\left(-\frac{\sqrt{3}}{2}\right)}{2}} &=& \sqrt{\dfrac{2+\sqrt{3}}{4}} &=& \dfrac{\sqrt{2+\sqrt{3}}}{2} \\ \\[-3mm]<br /> <br />
\sin15^o &=& \sqrt{\dfrac{1-\cos30^o}{2}} &=& <br />
\sqrt{\dfrac{1 - \frac{\sqrt{3}}{2}}{4}} &=&  <br />
\sqrt{\dfrac{2-\sqrt{3}}{4}} &=& \dfrac{\sqrt{2-\sqrt{3}}}{2}<br />
\end{array}


    Then: . \sin105^o - \sin15^o \;=\; \frac{\sqrt{2+\sqrt{3}}}{2} - \frac{\sqrt{2-\sqrt{3}}}{2}


    Note that: . \begin{array}{ccc}2+\sqrt{3} &=& \dfrac{(\sqrt{3}+1)^2}{2} \\ \\[-3mm]<br />
2 - \sqrt{3} &=& \dfrac{(\sqrt{3}-1)^2}{2} \end{array}


    \text{Therefore: }\;\sin105^o - \sin15^o \;=\;\dfrac{\dfrac{\sqrt{3}+1}{\sqrt{2}} - \dfrac{\sqrt{3}-1}{\sqrt{2}}}{2} \;=\;\frac{1}{\sqrt{2}}


    Very clear explanation. Thanks
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  4. #4
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    Quote Originally Posted by SodaCup View Post
    Um sorry I still don't get it, how do you find √3 from cosx = -4 and cosx = 1/2?
    Check the basic reference angles for the cosine ratio. For what angle value(s) is the cosine equal to 1/2?

    Also, check the properties of the cosine wave. What are the limits on the values of the cosine? Does -4 fall within this interval of values?

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  5. #5
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    Quote Originally Posted by stapel View Post
    Check the basic reference angles for the cosine ratio. For what angle value(s) is the cosine equal to 1/2?

    Also, check the properties of the cosine wave. What are the limits on the values of the cosine? Does -4 fall within this interval of values?

    1/2 = cos 60
    the limit is -2..? right?
    I still don't get it, how do you find √3 ??
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  6. #6
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    Quote Originally Posted by SodaCup View Post
    1/2 = cos 60
    And another value, as provided to you earlier. Then take the tangent of these angles to find the required values.

    Quote Originally Posted by SodaCup View Post
    the limit is -2..? right?
    Where are you taking limits...?
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  7. #7
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    Quote Originally Posted by stapel View Post
    And another value, as provided to you earlier. Then take the tangent of these angles to find the required values.


    Where are you taking limits...?
    I think I get it now..

    cos 60 = 1/2
    sin 60 = 1/2 √3

    sin/cos = √3

    anyway Thanks to all who helped.
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