已知sin(α+π/12)＝1/3,求cos(α+7π/12)的值

已知sin(α+π/12)＝1/3,求cos(α+7π/12)的值
已知sin(α+π/12)＝1/3,求cos(α+7π/12)的值

已知sin(α+π/12)＝1/3,求cos(α+7π/12)的值
令x=α+π/12
则α=x-π/12
且sinx=1/3
所以原式=cos(x-π/12+7π/12)
=cos(x+π/2)
=*-sinx
=-1/3

cos(α+7π/12)
=sin(-α+-π/12)
=-sin(α+π/12)
=-1/3

7π/12-π/12=6π/12=π/2；
∵cos(π/2+α)=-sinα
∴cos(α+7π/12)=cos[π/2+(α+π/12)]=-sin(α+π/12)=-1/3；