已知sin(π+α)=-1/2,计算:(1)sin(5π-α); (2)cos(α-3π2).

问题描述:

已知sin(π+α)=-

1
2
,计算:
(1)sin(5π-α); 
(2)cos(α-
2
)

(1)∵sin(π+α)=-sinα=-

1
2

∴sinα=
1
2

则sin(5π-α)=sin(π-α)=sinα=
1
2

(2)∵sinα=
1
2

∴cos(α-
2
)=cos(
2
-α)=-sinα=-
1
2