Question 690928: If sin theta = 5/6, and theta is in quadrant 2, find the exact value of:
a) cos theta and b) sin(theta + pi/6)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If sin (t) = 5/6, and theta is in quadrant 2,
find the exact value of:
a) cos(t)
Since sin(t) = y/r, y = 5 and r = 6
Then x = sqrt[r^2-y^2] = sqrt(11)
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Note: Remember; cos is negative in the QII
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So cos(t)= x/r = -sqrt(11)/6
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b) sin(t + pi/6) = sin(t)*cos(pi/6) + cos(t)*sin(pi/6)
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= (5/6)(sqrt(3)/2) + (-sqrt(11)/6)(1/2)
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= (5sqrt(3) - sqrt(11))/12
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Cheers,
Stan H.
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