If sin A = -4/5 and cos A < 0, then the terminal side of angle (2A) lies in quadrant

Answer 1

A in quadrant III; sin2A = 2(-4/5)(-3/5) = 24/25 > 0; cos2A = 1 - 2(-4/5)² = - 7/25 < 0

180° < A < 270°; 360° < 2A < 540°; but 2A in quadrant II, so 450° < 2A < 540°

Answer 2

A is in the 2nd Quadrant, work the problem several different ways, and it comes up A=106 degrees in Quadrant II

sinA = -4/5

cosA =-3/5,

A = about180+53 degrees = 233 A

=466 degrees in quadrant II

466 -360 = 106 degrees is in quadrant IL

sin2A = 2sinAcosA = 2(-4/5)(-3/5) =24/25

2A = sin^-1(24/25) = about 74 degrees in quadrant I or 106 degrees in quadrant II, but that's misleading, as only 106 degrees is correct

it helps to draw a unit circle and plot the points

A = 233

sin233=-4/5

cos233 =-3/5

sin106 = .96

cos233= -.6