Question 590934: If cos x = -(2/3) and pi/2 < x < pi, what is the exact value of sin2x, cos2x, and tan2x
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! If cos x = -(2/3) and pi/2 < x < pi, what is the exact value of sin2x, cos2x, and tan2x
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If cos x = -(2/3) (adjacent/hypotenuse in quadrant II where cos<0)
sinx=√5/3 (opposite/hypotenuse in quadrant II where sin>0)
tanx=-√5/2(opposite/adjacent in quadrant II where tan<0)
..
sin2x=2sinxcosx=2*√5/3*-2/3=-4√5/9
cos 2x=cos^2x-sin^2x=(-2/3)^2-(√5/3)^2=4/9-5/9=-1/9
tan2x=2tanx/(1-tan^2x)=2*-√5/2/(1-5/4)=-√5/-1/4=4√5
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