二倍角的.cosπ/5乘cos2π/5的值?y=cosx/(1-sinx)的单调递增区间?

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)
=1/4
y=cosx/(1-sinx)
=[(cosx/2)^2-(sinx/2)^2]/(sinx/2-cosx/2)^2 (二倍角)
=-(sinx/2+cosx/2)/(sinx/2-cosx/2) （分子分母同除以cosx/2）
=(tanx/2+1)/(1-tanx/2) (用1=tanπ/4配正切两角和)
=tan(x/2+π/4)
-π/2+kπ