cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
=(cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11*2sinπ/11)/2sinπ/11
=(sin2π/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/2*sin4π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/8*sin6π/11*cos5π/11)/2sinπ/11
=(1/8*sin5π/11*cos5π/11)/2sinπ/11
=(1/16*sin10π/11)/2sinπ/11
=1/32
解題關(guān)鍵在于利用與構(gòu)造sin2x=2SinxCosx.如果不明白,等你追問.(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11得到(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11的理由是什么？sin(π-x)=sinx