α∈(3π/2,2π),化简√[1/2+1/2√(1/2)-(1/2)COS 2α]
问题描述：
α∈(3π/2,2π),化简√[1/2+1/2√(1/2)-(1/2)COS 2α]
答
√[1/2+1/2√(1/2)-(1/2)COS 2α]
=√[1/2+1/2√(sina)^2
α∈(3π/2,2π),
sina=√[1/2-1/2(sina)]
=√(cosa/2-sina/2)^2
α/2∈(3π/4,π),
所以sina/2>0.cosa/2所以.√(cosa/2-sina/2)^2=sina/2-cosa/2=sin(a/2-π/4)