已知cosα=1/7,cos（α-β）=13/14,且0＜ β ＜α＜π/2
问题描述：
已知cosα=1/7,cos（α-β）=13/14,且0＜ β ＜α＜π/2
cos(π+2α)tan(π-2α)sin(π/2-2α)
(1)求 ---------------------------------------- 的值
cos(π/2+2α)
(2)求角β
答
（1）cos(π+2α)tan(π-2α)sin(π/2-2α)/cos(π/2+2α)
=[-cos(2α)][-tan(2α)]cos(2α)/[-sin(2α)] (应用诱导公式)
=-cos²(2α)[sin(2α)/cos(2α)]/sin(2α)
=-cos(2α)
=1-2cos²α(应用倍角公式)
=1-2*(1/7) (∵cosα=1/7)
=5/7；
（2）∵cosα=1/7,cos（α-β）=13/14,且0＜ β ＜α＜π/2
∴sinα=√(1-cos²α)=4√3/7,sin(α-β)=√(1-cos²(α-β))=3√3/14
∵sinβ=sin(α-(α-β))
=sinαcos(α-β)-cosαsin(α-β)
=(4√3/7)(13/14)-(1/7)(3√3/14)
=√3/2
∴β=π/3.