given: cosa=(-2/3),(pi/2)<a<pi, Find the exact value for: sin2a cos2a and tan2a

given: cosa=(-2/3),(pi/2)<a<pi, Find the exact value for: sin2a
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cos(a)=-⅔, sin2(a)=1-cos2(a)=1-4/9=5/9, sin(a)=±√5/3. But π/2<a<π. In this quadrant sine is positive, so sin(a)=√5/3.

sin(2a)=2sin(a)cos(a)=2(√5/3)(-⅔)=-4√5/3.

cos(2a)=2cos2(a)-1=8/9-1=-1/9.

tan(2a)=sin(2a)/cos(2a)=-4√5/3/(-1/9)=9(4√5/3)=12√5.

by Top Rated User (1.2m points)

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