已知sina+cosa=3√5/5,a属于(0,π/4),sin(β-π/4)=3/5,β属于（π/4,π/2).(1)求sina和tan2a的值
问题描述：
已知sina+cosa=3√5/5,a属于(0,π/4),sin(β-π/4)=3/5,β属于（π/4,π/2).(1)求sina和tan2a的值
答
sina=3√5/5-cosa ,sin^2a=9/5-6√5/5cosa+cos^2a=1-cos^2a ,4/5-6√5/5cosa+2cos^2a=0 ,(2√5/5-cosa)(√5/5-cosa)=0 ,cosa= 2√5/5 sina=√5/5 ,tana=1/2tan2a=2tana/(1-tan^2a)=1/(1-1/4)=4/3那解一下第二问（2）求cos(a+2β)的值sin(β-π/4)=√2/2(sinβ-cosβ)=3/5 , sinβ-cosβ=3√2/5 , 1 -2sinβcosβ=18/25 , sin2β=7/25cos2β=-24/25cos(a+2β)=cosacos2β-sinasin2β= 2√5/5 *(-24/25)-√5/5*7/25=-11√5/25