已知cos(α-π/6)+sinα=(4倍根号3)/5，则sin(α+7π/6)=

已知cos(α-π/6)+sinα=(4倍根号3)/5，则sin(α+7π/6)=

cos(α-π/6)+sinα=4√3/5
cos(α-π/6)+cos(π/2-α)=4√3/5
2cos{[(α-π/6)+(π/2-α)]/2}*cos{[(α-π/6)-(π/2-α)]/2}=4√3/5
2cosπ/6*cos(α-π/3)=4√3/5
2√3/2*sin(α+π/6)=4√3/5
sin(α+π/6)=4/5
sin(α+7π/6)=-sin(α+π/6)=-4/5

cos(α-π/6)+sinα
=cosα*√3/2+1/2 *sinα+sinα
=cosα*√3/2+3/2 *sinα
=√3(sinπ/6cosα+cosπ/6sinα)
=√3sin(π/6+α)=4/5√3;

sin(α+7π/6)=-sin(α+π/6+π)=-sin(α+π/6)=-4/5 
修改了一下楼上的