sina=(-2根5)/5,(3π/2

问题描述:

sina=(-2根5)/5,(3π/2

a在第四象限所以cosa>0sin²a+cos²a=1所以cosa=√5/5所以sin2a=2sinacosa=4/5cos2a=2cos²a-1=-3/5cos[(2a-π)/3]错了吧,应该是cos[2a-(π/3)]原式=cos2acos(π/3)+sin2asin(π/3)=(√5-2√15)/10...