Question

若sin(+α)=,cos(-β)=,且0＜α＜＜β＜,求cos(α+β)的值.

Answer 1

解:∵0＜α＜＜β＜,

∴＜+α＜π,-＜-β＜0.

又已知sin(+α)=,cos(-β)=,∴cos(+α)=-,sin(-β)=-

∴cos(α+β)=sin［+(α+β)］=sin［(+α)-(-β)］

=sin(+α)cos(-β)-cos(+α)sin(-β)

=×-(-)×(-)=-.