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Math Help - How do I find negative radians on the unit circle?

  1. #1
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    How do I find negative radians on the unit circle?

    Hello all. For the last 2 hours I've been working on sum and difference problems and cannot seem to get answers that match the book's answers. Here's what i did:

    sin(-7pi/12)

    = sin(2pi/12 - 9pi/12)

    = sin(pi/6 - 3pi/4)

    = sin(pi/6)cos(3pi/4) - cos(pi/6)sin(3pi/4)

    = (1/2)(-rad2/2) - (rad3/2)(rad2/2)

    = -rad2/4 - rad6/4

    = (-rad2-rad6)/4

    The books says the answer is supposed to come out to be (-rad2+rad6)/4

    I know i'm missing some key concept here and hours of scouring the internet has failed me. I don't know how to find the sin or cos of a negative radian measure on the unit circle; Like if i need to find the sin of -pi/4 or cos of -pi/3. I first attempted this problem using sin(-pi/4 + -pi/3) but switched to using the difference formula as to avoid using negative radians. I get a different wrong answer each way. This is my first post so i hope i have provided adequate information for you to answer my question and i hope i haven't written too much.. Thank you so very much in advance, this is driving me absolutely bonkers!!

    -Nick
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  2. #2
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    Re: How do I find negative radians on the unit circle?

    Quote Originally Posted by guitargeek70 View Post
    Hello all. For the last 2 hours I've been working on sum and difference problems and cannot seem to get answers that match the book's answers. Here's what i did:

    sin(-7pi/12)

    = sin(2pi/12 - 9pi/12)

    = sin(pi/6 - 3pi/4)

    = sin(pi/6)cos(3pi/4) - cos(pi/6)sin(3pi/4)

    = (1/2)(-rad2/2) - (rad3/2)(rad2/2)

    = -rad2/4 - rad6/4

    = (-rad2-rad6)/4

    The books says the answer is supposed to come out to be (-rad2+rad6)/4

    I know i'm missing some key concept here and hours of scouring the internet has failed me. I don't know how to find the sin or cos of a negative radian measure on the unit circle; Like if i need to find the sin of -pi/4 or cos of -pi/3. I first attempted this problem using sin(-pi/4 + -pi/3) but switched to using the difference formula as to avoid using negative radians. I get a different wrong answer each way. This is my first post so i hope i have provided adequate information for you to answer my question and i hope i haven't written too much.. Thank you so very much in advance, this is driving me absolutely bonkers!!

    -Nick
    Your answer is correct. It must be a typo in the book.

    An alternative method is to remember that \displaystyle \begin{align*} \sin{(-\theta)} \equiv -\sin{(\theta)} \end{align*}, so

    \displaystyle \begin{align*} \sin{\left( -\frac{7\pi}{12} \right)} &= -\sin{\left( \frac{7\pi}{12} \right) } \\ &= -\sin{\left( \frac{\pi}{4} + \frac{\pi}{3} \right)} \\ &= -\left[ \sin{\left( \frac{\pi}{4} \right)}\cos{\left( \frac{\pi}{3} \right) } + \cos{\left( \frac{\pi}{4} \right)}\sin{ \left( \frac{\pi}{3} \right) } \right] \\ &= -\left[ \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) \right] \\ &= -\left( \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} \right) \\ &= -\frac{ \left( \sqrt{2} + \sqrt{6} \right) }{ 4 } \end{align*}
    Last edited by Prove It; February 25th 2013 at 10:22 PM.
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    Re: How do I find negative radians on the unit circle?

    oh wow... lol.. that makes me feel a lot better. I'm in an math course which uses the online service CourseCompass. There's another problem on here which is giving an answer with signs different from mine.

    The problem is sin(-13pi/12)

    here's what I did:

    = sin(-5pi/6 + -pi/4)
    = sin(-5pi/6)cos(-pi/4) + cos(-5pi/6)sin(-pi/4)
    = (-1/2)(rad/2) + (-rad3/2)(-rad2/2)
    = (-rad2+rad6)/4

    They say it should equal (-rad2-ra6)/4

    So what gives? Should i just disregard their answers for these problems? I can't even begin to tell you the amount of time i've wasted looking for errors in my work... Thank you very much for your help, im amazed in the speed with which you responded.
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    Re: How do I find negative radians on the unit circle?

    Again I agree with your answer. Again, I would have simplified at the beginning.

    \displaystyle \begin{align*} \sin{\left( -\frac{13\pi}{12} \right)} &= -\sin{\left( \frac{13\pi}{12} \right)} \\ &= -\sin{\left( \frac{5\pi}{6} + \frac{\pi}{4} \right)} \\ &= -\left[ \sin{\left( \frac{5\pi}{6} \right)} \cos{\left( \frac{\pi}{4} \right)} + \cos{\left( \frac{5\pi}{6} \right)} \sin{\left( \frac{\pi}{4} \right)} \right] \\ &= -\left[ \left( \frac{1}{2} \right) \left( \frac{\sqrt{2}}{2} \right) + \left( -\frac{\sqrt{3}}{2} \right) \left( \frac{\sqrt{2}}{2} \right) \right] \\ &= -\left( \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \right) \\ &= \frac{\sqrt{6} - \sqrt{2}}{4} \end{align*}

    I wouldn't throw out the answers you have been given. Just keep in mind that they may have mistakes in them. I find it more effective to check my answers using technology such as a CAS calculator or Wolfram Alpha.
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    Re: How do I find negative radians on the unit circle?

    I can't even begin to tell you the amount of time i've wasted looking for errors in my work... Thank you very much for your help, im amazed in the speed with which you responded.


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