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2 Answers
Somebody N.
Mar 5, 2018

#sqrt(13)#

Explanation:

If we think of #sinA=2/3# in relation to a right triangle. Then the hypotenuse is #3# and the side opposite A is #2#.

The adjacent side of A is #sqrt((3)^2-(2)^2)=sqrt(13)#

This makes:

#cosA=sqrt(13)/3#

Using the identity:

#color(red)bb(sin(2A)=2sinAcosA)#

#sin(2A)=2(3/2)(sqrt(13)/3)=sqrt(13)#

Nghi N
Mar 6, 2018

#sin 2A = (4sqrt5)/9#

Explanation:

#sin A = 2/3#
#cos^2 A = 1 - sin^2 A = 1 - 4/9 = 5/9#
#cos A = +- sqrt5/3#
Since A is in Quadrant 1, so, cos A is positive
#cos A = sqrt5/3#
#sin 2A = 2sin A.cos A = 2(2/3)(sqrt5/3) = (4sqrt5)/9#

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