Question

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

Answer 1

解:∵0＜α＜＜β＜π,

∴＜+α＜＜+β＜.

∵cos(+α)=- ＜0,

∴＜+α＜.

∴sin(+α)=.

∵sin(-β)=sin［π-(+β)］=sin(+β)=＞0,

∴＜+β＜π.

∴cos(+β)=.

∴cos(β-α)=cos［(+β)-(+α)］

=cos(+β)cos(+α)+sin(+β)sin(+α)

=-×()+ ×()=.