Question 834741: if cscθ=4 and θ is in quadrant II, find the exact value of sin2θ
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! if cscθ=4 and θ is in quadrant II, find the exact value of sin2θ
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With theta in QII, x is negative and y is positive.
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Since csc = r/y, r = 4 and y = 1
Therefore -x = -sqrt[4^2-1^2] = -sqrt(15)
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Your Problem::
sin(2t) = 2*sin(t)cos(t) = 2(y/r)(x/r) = 2(1/4)(-sqrt(15)/4) = -sqrt(15)/8
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Cheers,
Stan H.
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