Question 610303
Find the exact values of sin(2theta), and cos(2theta) if sin(theta)= -9/10 and theta is between 180 degrees and 270 degrees.
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use x in place of theta.
sinx=-9/10=opposite side/hypotenuse (in quadrant III where sin is <0)
Adjacent side=&#8730;(10^2-9^2)=&#8730;(100-81)=&#8730;19
cosx=-&#8730;19/10  (in quadrant III where cos is also<0)
..
sin2x=2sinxcosx
=2*(-9/10)*(-&#8730;19/10)
=18&#8730;19/100
=(9&#8730;19)/50 (in quadrant II where sin>0)
..
cos2x=cos^2x-sin^2x
=(-&#8730;19/10)^2-(-9/10)^2
=19/100-81/100
=-62/100
=-31/50  (in quadrant II where cos<0)