Question 522875: If tanx=-1/3 ; cosx>0 then
sin2x=
cos2x=
tan2x=
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! If tanx=-1/3 ; cosx>0
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Since tan is negative and cos is positive, x is in the 4th quadrant.
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tan = y/x = -1/3, so y = -1 and x = 3
Therefore r = sqrt[1^2 + 3^2] = sqrt(10)
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sin(x) = y/r = -1/sqrt(10) ; cos(x) = x/r = 3/sqrt(10)
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then
sin2x= 2sin(x)*cos(x) = 2*(-1/sqrt(10))*(3/sqrt(10)) = -6/10 = -3/5
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cos2x= cos^2 - sin^2 = (9/10) - (1/10) = 8/10 = 4/5
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tan2x= sin(2x)/cos(2x) = [-3/5]/[4/5] = -3/4
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Cheers,
Stan H.
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