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Math Help - Solving for sin(x)

  1. #1
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    Solving for sin(x)

    Given that  \tan(x) = \frac{2}{3}\mbox{ and } \pi<x<\frac{3\pi}{2} \mbox{, find }\sin(x)\\

    I know that \pi<x<\frac{3\pi}{2} puts x in Quadrant III in the unit circle, but I don't see how that helps me, other than the fact that \sin(x) will be negative. I solved for \sin(x), but I'm pretty sure it isn't the right answer.

    \tan(x) = \frac{2}{3}\\\\\frac{\sin(x)}{cos(x)} = \frac{2}{3}\\\\\sin(x) = \frac{2cos(x)}{3}\\

    also, it gave a,b,c,d,e choices.

    a) \frac{\sqrt{10}}{10}

    b) \frac{-\sqrt{13}}{2}

    c) \frac{-\sqrt{10}}{10}

    d) \frac{-2\sqrt{13}}{13}

    e) \frac{2\sqrt{13}}{13}
    Last edited by Fenixx09; July 23rd 2011 at 03:36 PM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Solving for sin(x)

    If you know:
    \tan(x)=\frac{2}{3} \Leftrightarrow x=\arctan\left(\frac{2}{3}\right).
    and you know \tan(x)=\tan(\pi+x)

    Can you now complete? ...
    Last edited by Siron; July 23rd 2011 at 02:45 PM.
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  3. #3
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    Re: Solving for sin(x)

    Quote Originally Posted by Fenixx09 View Post
    Given that  \tan(x) = \frac{2}{3}\mbox{ and } \pi<x<\frac{3\pi}{2} \mbox{, find }\sin(x)
    I know that \pi<x<\frac{3\pi}{2} puts x in Quadrant III in the unit circle.
    Recall that \tan(t)=\frac{y}{x}\text{ and }x^2+y^2=r^2.
    Also \sin(t)=\frac{y}{r}.

    Don't forget that in III, y<0~\&~r>0.
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  4. #4
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    Re: Solving for sin(x)

    Quote Originally Posted by Fenixx09 View Post
    Given that  \tan(x) = \frac{2}{3}\mbox{ and } \pi<x<\frac{3\pi}{2} \mbox{, find }\sin(x)
    reference triangle in quad III ... note the tangent value of the reference angle is \frac{-2}{-3} = \frac{2}{3}

    calculate the value of the hypotenuse and use your basic definition of sine for the reference angle.

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    Re: Solving for sin(x)

    Thanks guys, I really wanted to know how to do this without a calculator! And I couldn't get off of the unit circle until skeeter posted that graph :/

    (-3)^{2}+(-2)^{2}=(\mbox{hypotenuse})^{2}\\\\9+4=(\mbox{hypot  enuse})^{2}\\\\13=(\mbox{hypotenuse})^{2}\\\\\sqrt  {13}=(\mbox{hypotenuse})\\\\sin(x)=\frac{\mbox{(\o  pposite)}}{(\mbox{hypotenuse})}\\\\sin(x)=\frac{-2}{\sqrt{13}}\\\\sin(x)=\frac{-2\sqrt{13}}{13}
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Solving for sin(x)

    My solution would be:
    \tan(x)=\frac{2}{3} \Leftrightarrow x=\arctan\left(\frac{2}{3}\right) and x+\pi=\arctan\left(\frac{2}{3}\right) \Leftrightarrow x=\arctan\left(\frac{2}{3}\right) - \pi

    Now you just have to calculate:
    \sin\left[\arctan\left(\frac{2}{3}\right)-\pi \right]=\sin \left[\arctan\left(\frac{2}{3}\right)\right]\cdot \cos(\pi)-\sin(\pi)\cdot \cos\left[\arctan\left(\frac{2}{3}\right)\right]

    =-\frac{\frac{2}{3}}{\sqrt{1+\left(\frac{2}{3}\right  )^2}}
    =-\frac{\frac{2}{3}}{\sqrt{\frac{13}{9}}}
    =-\frac{2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}
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  7. #7
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    Re: Solving for sin(x)

    Quote Originally Posted by Siron View Post
    My solution would be:
    \tan(x)=\frac{2}{3} \Leftrightarrow x=\arctan\left(\frac{2}{3}\right) and x+\pi=\arctan\left(\frac{2}{3}\right) \Leftrightarrow x=\arctan\left(\frac{2}{3}\right) - \pi
    Now you just have to calculate:
    \sin\left[\arctan\left(\frac{2}{3}\right)-\pi \right]=\sin \left[\arctan\left(\frac{2}{3}\right)\right]\cdot \cos(\pi)-\sin(\pi)\cdot \cos\left[\arctan\left(\frac{2}{3}\right)\right]
    =-\frac{\frac{2}{3}}{\sqrt{1+\left(\frac{2}{3}\right  )^2}}
    =-\frac{\frac{2}{3}}{\sqrt{\frac{13}{9}}}
    =-\frac{2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}
    I must say that approach seems over-the-top to me.
    Here are some ideas you may consider, I realize are the product of a European system. This forum is titled pre-university trigonometry. Now if this poster is in North America he/she may not ever have seen the arctangent function. In fact he/she may well only 9th or 10th year student. But looking at his/her reply, it appears that skeeter’s reply was the one understood. That tells me that this poster is not even up to the analytic trigonometry that I posted.
    My point is that pre-university trigonometry has many different meanings.
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: Solving for sin(x)

    You're absolutely right (what do you mean with an European product?) and probably my solution doens't interest Fenixx09 but maybe there are other members interested.
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