Question

数学求过程 谢谢

2.sin(π\6+α)=4\5,sin（5π\6-α）= cos(2π\3-2α)=

Answer 1

sin（5π\6-α）=sin（π-π\6-α）
= sin[π-(π\6+α)]
=sin(π\6+α)
=4\5
cos(2π\3-2α)=cos(π-π\3-2α)
=cos[π-(π\3+2α)]
=-cos(π\3+2α)
=-cos2(π\6+α)
=-{1-2[sin(π\6+α)]^2}
=7/25

Answer 2

sin（5π\6-α）=sin(π-(π\6+α))=sin(π\6+α)
cos(2π\3-2α)= cos(π-(π\3+2α))=-cos(π\3+2α)=-cos(2*(π\6+α))
=-(1-2*sin(π\6+α)*sin(π\6+α))=2*sin(π\6+α)*sin(π\6+α)-1