Given tanA = -2/3, A is in quadrant II, find Sin2A

Given tanA = -2/3, A is in quadrant II, find Sin2A
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If tanA=-2/3 we can use a right-angled triangle to find sinA and cosA, then use 2sinAcosA=sin2A.

Hypotenuse=sqrt(2^2+3^2)=sqrt(13). sinA=2/sqrt(13), cosA=-3/sqrt(13), for A in the second quadrant, where sine is positive and cosine and tangent are negative, so sin2A=2(2/sqrt(13))(-3/sqrt(13))=-12/13. The angle A is in the second quadrant and has a value of about 146 degrees. 2A is therefore about 292 degrees in the fourth quadrant.

by Top Rated User (1.2m points)

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