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Math Help - [SOLVED] trig problem

  1. #1
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    [SOLVED] trig problem

    cosY= -sqrt3/2

    and

    cosY= -sqrt3/2 for y in [pi,2pi]

    The question ask to solve these equations
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by john doe View Post
    cosY= -sqrt3/2

    and

    cosY= -sqrt3/2 for y in [pi,2pi]

    The question ask to solve
    Keep in mind that cosine is negative in the second and third quadrants.

    The second question is asking for particular angles within the interval \pi\leq y\leq2\pi. Note that you will only have one solution.

    keep in mind that \cos\left(\tfrac{\pi}{6}\right)=\tfrac{\sqrt{3}}{2  } Change the angle to get \cos(?)=-\tfrac{\sqrt{3}}{2}

    The first question wants you to generalize the solution. This can be a little tricky. Find several solutions to the equation, and then try to find a pattern given in the coefficients of the angles. Try this one and let us know where you get stuck...if you get stuck...

    I hope this helps.

    --Chris
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  3. #3
    Senior Member nikhil's Avatar
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    Lightbulb Check this out

    Quote Originally Posted by john doe View Post
    cosY= -sqrt3/2

    and

    cosY= -sqrt3/2 for y in [pi,2pi]

    The question ask to solve
    cosY= -sqrt3/2 for y in [pi,2pi]
    cosY is negative in third quadrant that is in interval [pi,3pi/2]
    if cosY=sqrt3/2 then
    Y=pi/6 (taking principle value)
    so value of Y for which cosY=-sqrt3/2 will be
    =pi+pi/6=7pi/6 (for third quadrant)
    therfor y=7pi/6
    hope this helps
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