Question 517738: If cos x = -2/3 and x terminates in quadrant 2 find the exact value of cos 2x, sin 2x, and tan 2x...
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! If cos x = -2/3 and x terminates in quadrant 2 find the exact value of cos 2x, sin 2x, and tan 2x
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Cos x terminates in quadrant II where sin>0, cos<0 and tan<0
You are working with a right triangle where the adjacent side=-2, hypotenuse=3, and opposite side
=√[(3^2)-(-2)^2]=√(9-4)=√5
cos x=-2/3 (given)
sin x=√5/3
tan x=sin/cos=(√5/3)/(-2/3)=-√5/2
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cos 2x=cos^2x-sin^2x (identity)
cos 2x=(-2/3)^2-(√5/3)^2=4/9-5/9=-1/9
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sin 2x=2sinx cosx (identity)
sin 2x=2*√5/3*2/3=4√5/9
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tan 2x=2tanx/(1-tan^2x) (identity)
tan 2x=(2*-√5/2)/(1-(-√5/2)^2))=-√5/(1-5/4)=√5(-1/4)=-√5/4
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