若sin(π/3+a)=1/3,且a属于(-π/2,0),则sina=

∵a属于(-π/2,0),则π/3+a属于(-π/6,π/3)∴cos(π/3+a)>0∵sin(π/3+a)=1/3∴cos(π/3+a)=√[1-(sin(π/3+a))^2]=2√2/3故sina=sin[(π/3+a)-π/3]=sin(π/3+a)cos(π/3)-cos(π/3+a)sin(π/3) (应用差角公式)=(1/3)(1/2)-(2√2/3)(√3/2)=(1-2√6)/6。