设α为锐角,若cos（α+π/6）=4/5,则sin（2α+π/12）=?

α∈(0,π/2)
α+π/6∈(π/6,2π/3)
cos(α+π/6)=4/5>0
∴α+π/6∈(π/6,π/2)
∴2α+π/3∈(π/3,π)
cos(2α+π/3)
=2cos²(α+π/6)-1
=2*(4/5)²-1
=32/25-1
=7/25>0
∴2α+π/3∈(π/3,π/2)
sin(2α+π/3)=24/25
sin(2α+π/12)
=sin(2α+π/3-π/4)
=sin(2α+π/3)cosπ/4-cos(2α+π/3)sinπ/4
=24/25*√2/2-7/25*√2/2
=24√2/50-7√2/50
=17√2/50

sinα咋求啊？

α∈(0,π/2)α+π/6∈(π/6,2π/3)cos(α+π/6)=4/5>0∴α+π/6∈(π/6,π/2)∴2α+π/3∈(π/3,π)cos(2α+π/3)=2cos²(α+π/6)-1=2*(4/5)²-1=32/25-1=7/25>0∴2α+π/3∈(π/3,π/2)sin(2α+π/3)=24/...

首先你先判断取值范围，α+π/6小于45°，那么2α+π/6也是锐角
sin(α+π/6)=1-(cos(α+π/6))^(1/2)=3/5
cos(2α+π/3)=2cos(α+π/6)^2-1=7/25
sin(2α+π/3)=2sin(α+π/6）cos(α+π/6)=2*3/5*4*5=24/25
∴sin（2α+π/12）
=sin[（2α+π/3）-π/4]
=sin（2α+π/3）cosπ/4-cos（2α+π/3）sinπ/4
=24/25*（1/2)^(0.5)-7/25*(1/2)^0.5
=50分之17倍根号2