If cosA = 3/5, 0deg <A<90deg write sinA as a fraction. Hence write cos2A as a fraction. I've got the first part sinA as a fraction, 4/5 . The second part, cos2A as a fraction is getting me. Any help?
Follow Math Help Forum on Facebook and Google+
Originally Posted by zyx.gar If cosA = 3/5, 0deg <A<90deg write sinA as a fraction. Hence write cos2A as a fraction. I've got the first part sinA as a fraction, 4/5 . The second part, cos2A as a fraction is getting me. Any help? Hi zyx.gar, Try using the double angle identity: $\displaystyle \cos 2A = \cos^2A-\sin^2A$
View Tag Cloud