求解高一的三角函数化简题

显示全部楼层 · 2012-2-19 14:14:35

cos3π/11=cos(π-8π/3)=cos(8π/3)sin(16π/11）= sin(π+5π/11)=-sin(5π/11)sin(10π/11）= sin(π-π/11)=sin(π/11)(cosπ/11)(cos2π/11)(cos3π/11)(cos4π/11)(cos5π/11) =-(cosπ/11)(cos2π/11)(cos4π/11)(cos8π/11)(cos5π/11) =-(sinπ/11)(cosπ/11)(cos2π/11)(cos4π/11)(cos8π/11)(cos5π/11) /(sinπ/11)=-(1/2)(sin2π/11)(cos2π/11)(cos4π/11)(cos8...

千问 · 2012-2-19 14:14:35

(sinπ/11)(cosπ/11)(cos2π/11)(cos3π/11)(cos4π/11)(cos5π/11)/(sinπ/11)=(sin2π/11)cos2π/11)(cos3π/11)(cos4π/11)(cos5π/11)/[2(sinπ/11)]=(sin4π/11)(cos3π/11)(cos4π/11)(cos5π/11)/[4(...