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Thread: terminal side

  1. #1
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    Hello. I am having difficulty with this problem:
    Find the point on the terminal side of θ = ​-3pi/4 that has an x coordinate of -1

    I was hoping that someone here would be able to help me out? I am actually trying to figure out what I am doing so if you could explain how you found the answer, I would really appreciate it. Thank you so much guys.
    Last edited by RiderMind; Oct 3rd 2012 at 01:57 PM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: terminal side

    θ = ?
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  3. #3
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    Re: terminal side

    oops, sorry.

    θ=-3pi/4

    Thanks
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: terminal side

    From the given information, we know:

    $\displaystyle r\cos\left(-\frac{3\pi}{4} \right)=-1$

    $\displaystyle r=\sqrt{2}$

    The $\displaystyle y$-coordinate is then:

    $\displaystyle r\sin\left(-\frac{3\pi}{4} \right)=?$
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  5. #5
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    Re: terminal side

    0?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: terminal side

    No, what is $\displaystyle \sin\left(-\frac{3\pi}{4} \right)=-\sin\left(\frac{3\pi}{4} \right)=-\sin\left(\pi-\frac{3\pi}{4} \right)=-\sin\left(\frac{\pi}{4} \right)$ ?
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  7. #7
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    Re: terminal side

    -45 degrees?

    or sqrt2
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: terminal side

    No, the sine of an angle will not return an angle. $\displaystyle \frac{\pi}{4}$ is a special angle for which you should know the trig. functions at that angle.

    $\displaystyle -\sin\left(\frac{\pi}{4} \right)=-\frac{1}{\sqrt{2}}$

    So, the $\displaystyle y$-coordinate of the point is:

    $\displaystyle y=r\cdot\left(-\frac{1}{\sqrt{2}} \right)$

    Recall we found $\displaystyle r=\sqrt{2}$ hence:

    $\displaystyle y=\sqrt{2}\cdot\left(-\frac{1}{\sqrt{2}} \right)=-1$

    And so, the point in question is (-1,-1). Try drawing a diagram, and you will easily see that the $\displaystyle y$-coordinate has to be equal to the $\displaystyle x$-coordinate, as the given angle lies along the line $\displaystyle y=x$.
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  9. #9
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    Re: terminal side

    I see what you mean now. Thanks for telling me to draw a diagram. That made it easier for me to understand. So -pi/4 is one of those I just need to memorize then right?

    Thanks so much bro.
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  10. #10
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    Re: terminal side

    Well, it's a lot easier to memorize if you understand it. Imagine a right triangle having one angle of $\displaystyle \pi/4$ and one leg of length 1. Since $\displaystyle \pi/2= 2(\pi/4)$, so that the other angle is also $\displaystyle \pi/4$ which means that the other leg also has length 1. By the Pythagorean theorem, the length of the hypotenuse is given by $\displaystyle c^2= 1^2+ 1^2= 2$ so that $\displaystyle c= \sqrt{2}$. That gives $\displaystyle sin(\pi/4)= 1/\sqrt{2}= \frac{\sqrt{2}}{2}$.
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  11. #11
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    Re: terminal side

    Could someone send me a link to teach me how to do this? I'm afraid I don't even understand it...
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