①sin(-28π/3)
问题描述：
①sin(-28π/3)
②tan(-44π/3)
已知sin(π+a)=-1/2,计算:
①tan(π/2-a)
②cos(27π/2+a)
③sin(-a)-[cos(2π-a)] / [tan(a-π)]
答
①sin(-28π/3)=sin(-10π+2π/3)=sin(2π/3)=√3/2 ；
②tan(-44π/3)=tan(-15π+π/3)=tan(π/3)=√3；
由sin(π+a)=-1/2,得sina =1/2,所以,cosa = ±√3/2.
①tan(π/2-a)=cota=cosa/sina=±√3；
②cos(27π/2+a)=sina =1/2；
③sin(-a)- [cos(2π-a)] / [tan(a-π)]=-sina- (cosa / tana)=-sina- (cos²a / sina)=-1/2-3/2=-2.
注意运用“奇变偶不变,符号看象限”.如前①也可这样做：
sin(-28π/3)=sin(-9π-π/3)=sin(π/3)=√3/2.