tana=2 求sin²+a+sinacosa

问题描述:

tana=2 求sin&sup2+a+sinacosa

姑且认为是sin²a+sinacosa
原式=(1-cos2a)/2+1/2sin2a
=1/2+1/2(sin2a-cos2a)
sin2a=2tana/(1+tan²a)=2×2/(1+2²)=4/5
cos2a=(1-tan²a)/(1+tan²a)=(1-4)/(1+4)=-3/5
那么原式=1/2+1/2×(4/5+3/5)=1/2+7/10=6/5