题目内容
已知0<β<<α<π,cos(-α)=,sin(+β)=,求sin(α+β)的值.
∵<α<,∴ -<-α<-,∴ -<-α<0.
又cos(-α)=,∴ sin(-α)=-.
∵ 0<β<,∴ <+β<π.
又sin(+β)=,∴ cos(+β)=-.
∴sin(α+β)=-cos =-cos[(+β)-(-α)]=
-cos cos-sin(+β)·sin
=
又cos(-α)=,∴ sin(-α)=-.
∵ 0<β<,∴ <+β<π.
又sin(+β)=,∴ cos(+β)=-.
∴sin(α+β)=-cos =-cos[(+β)-(-α)]=
-cos cos-sin(+β)·sin
=
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