Posted by Anonymous on Thursday, January 20, 2011 at 2:55am.
cos A = 4/5 for angle A in Quadrant II, find sin 2A

trig  Henry, Friday, January 21, 2011 at 7:13pm
cosA = 4/5 = 0.80,
A = 143.1deg.
sin(2A) = 2*sinA*cosA,
sin2A = 2*sin(143.1)*cos(143.1 =
0.9603.

trig  Henry, Friday, January 21, 2011 at 7:38pm
Solution using Exact values.
cosA = 4/5 = x/r.
(4)^2 + Y^2 = 5^2,
16 + Y^2 = 25,
Y^2 = 25  16 = 9,
Y = 3.
SinA = Y/r = 3/5.
sin2A = 2 * 3/5 * 4/5 = 24/25.
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