二倍角的.
问题描述：
二倍角的.
cosπ/5乘cos2π/5的值?
y=cosx/(1-sinx)的单调递增区间?
答
cosπ/5*cos2π/5=(4sinπ/5*cosπ/5*cos2π/5)/(4sinπ/5) (用二倍角公式）=(2sin2π/5*cos2π/5)/(4sinπ/5)= (sin4π/5)/(4sinπ/5)=1/4y=cosx/(1-sinx)=[(cosx/2)^2-(sinx/2)^2]/(sinx/2-cosx/2)^2 (二倍角)=-(s...