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Math Help - A little help with reference angles.

  1. #1
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    A little help with reference angles.

    So this is from a calculus problem but I'm not sure how to achieve the other angle (240 degrees). I know you get the (120 degrees from reference angle (180-60 = 120) but for 240 I'm not sure. Thanks and heres the problem.

    2\cos t+1=0\Longrightarrow \cos t = -\frac{1}{2}\Longrightarrow<br />
t=\left\{\begin{array}{c}\frac{2\pi}{3}\\{}\\\frac  {4\pi}{3}\end{array}\right.\,+2\pi<br />
k\,,\,\,k\in\mathbb{Z}<br />


    How would I go about achieving the  \frac {4 \pi} 3 ?
    Thanks.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by elitenewbz View Post
    so this is from a calculus problem but i'm not sure how to achieve the other angle (240 degrees). I know you get the (120 degrees from reference angle (180-60 = 120) but for 240 i'm not sure. Thanks and heres the problem.

    2\cos t+1=0\longrightarrow \cos t = -\frac{1}{2}\longrightarrow t=\left\{\begin{array}{c}\frac{2\pi}{3}\\{}\\\frac {4\pi}{3}\end{array}\right.\,+2\pi
    k\,,\,\,k\in\mathbb{z}
    " alt="
    t=\left\{\begin{array}{c}\frac{2\pi}{3}\\{}\\\frac {4\pi}{3}\end{array}\right.\,+2\pi
    k\,,\,\,k\in\mathbb{z}
    " />

    how would i go about achieving the  \frac {4 \pi} 3 ?
    Thanks.
    180+60=240
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  3. #3
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    So does that mean 240 = coterminal with the 60 degrees? I'm not sure how it fits... Von, which equation is that?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by EliteNewbz View Post
    So does that mean 240 = coterminal with the 60 degrees? I'm not sure how it fits... Von, which equation is that?
    I'm not quite sure what you are asking me. It looks like you have done a great job at finding the solution in you first post.

    I was only pointing out that \cos{t}=\frac{-1}{2} implies that t lies in QII or QIII. So that this means that we have the two solutions

    1. The one that you pointed out, namely t=\pi-\frac{\pi}{3}=\frac{2\pi}{3}

    And

    2. \pi+\frac{\pi}{3}=\frac{4\pi}{3}

    for t\in[0,2\pi]

    But, you even GENERALIZED the solution to include ALL values of t. So, what's your question?
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  5. #5
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    Oh I get it now. You've answered it. You've added  \frac \pi 3 to  \pi to include it into Quadrant III correct? (The pi/3 comes from the cos(t) = -1/2). So lets say the answers are in QUAD 4, you would add  \frac {5 \pi} 6 to  \frac \pi 3 . Correct?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by EliteNewbz View Post
    Oh I get it now. You've answered it. You've added  \frac \pi 3 to  \pi to include it into Quadrant III correct? (The pi/3 comes from the cos(t) = -1/2). So lets say the answers are in QUAD 4, you would add  \frac {5 \pi} 6 to  \frac \pi 3 . Correct?
    If \cos\theta is negative, then there will be 2 solutions (excluding \theta=-1). They are given by \theta=\pi+\varphi and \theta=\pi-\varphi, where \varphi is a refrence angle.

    Likewise, if \cos\theta is positive then the solutions will be given by \theta=\varphi and 2\pi-\varphi.
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