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Thread: Finding an angle in radians and finding exact values

  1. #1
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    Finding an angle in radians and finding exact values

    Hi

    I'm really stuck with these two questions would anyone be able to help?

    a) Given that sec theta=-sqrt2, tan theta=-1 and -pi<theta<pi, find the exact value of the angle theta in radians. Justify your answer.

    b) Given that -pi<theta<-1/2pi and cos theta =-1/6 use appropriate trigonometric formulas to find the exact values of the following:

    • Cos (2theta)
    • sin theta
    • sin (2theta)


    I don't have a clue where to start for part a but for part b i've worked out the following. Is any of it correct?


    • Cos (2theta) = -1/6 x 2 = -1/3




    • Sin theta: Because cos2theta + sin2theta =
    • Sin2theta = 1 + 1/62
    • Sin2theta = 37/36
    • Sin theta = sqrt 37/36
    • Sin theta = -16 sqrt 37 I've put negative answer because of the interval it lies in but not sure if that's correct




    • Sin (2theta) = 2 sin theta x cos theta
    • = 2 x (-1/6 sqrt 37) x -1/6
    • = 1/18 sqrt 37



    I would be incredibly grateful if someone could help me out i've been looking at these and trying to work them out for hours
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  2. #2
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    Re: Finding an angle in radians and finding exact values

    (a) \sec{t} = -\sqrt{2} \implies \cos{t} = -\frac{\sqrt{2}}{2}

    from this piece of info, we can surmise that t = \frac{3\pi}{4} \, or \, -\frac{3\pi}{4}, a quad III and quad III angle, respectively.

    now ... in what quadrant is \tan{t} = -1 ?
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  3. #3
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    Re: Finding an angle in radians and finding exact values

    Quote Originally Posted by soapqueen2011 View Post
    Hi

    b) Given that -pi<theta<-1/2pi and cos theta =-1/6 use appropriate trigonometric formulas to find the exact values of the following:

    • Cos (2theta)
    • sin theta
    • sin (2theta)


    I don't have a clue where to start for part a but for part b i've worked out the following. Is any of it correct?


    • Cos (2theta) = -1/6 x 2 = -1/3


    no ... cos(2t) = 2cos2t - 1

    cos(2t) = 1 - 2sin2t



    • Sin theta: Because cos2theta + sin2theta =
    • Sin2theta = 1 + 1/62
    • Sin2theta = 37/36
    • Sin theta = sqrt 37/36
    • Sin theta = -16 sqrt 37 I've put negative answer because of the interval it lies in but not sure if that's correct


    sin2t = 1 - cos2t

    sin2t = 1 - (-1/6)^2 = 35/36



    • Sin (2theta) = 2 sin theta x cos theta
    • = 2 x (-1/6 sqrt 37) x -1/6
    • = 1/18 sqrt 37


    now use what you were given and the correct value of sin(t)

    ...
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