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Thread: [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 $\displaystyle \pi\leq y\leq2\pi$. Note that you will only have one solution.

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

    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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