化简根号下1+cos(3π-θ)/2 (3π/2＜θ＜2π)

化简根号下1+cos(3π-θ)/2 (3π/2＜θ＜2π)
化简根号下1+cos(3π-θ)/2 (3π/2＜θ＜2π)

化简根号下1+cos(3π-θ)/2 (3π/2＜θ＜2π)
我们先计算根号里面的
1+cos(3π/2-θ/2)=1-cos(π/2-θ/2)=1-sin(θ/2)=1-2sin(θ/4)cos(θ/4)=[sin(θ/4)-cos(θ/4)]²
因为3π/2＜θ＜2π
所以3π/8＜θ/4＜π/2
所以sin(θ/4)>cos(θ/4)
所以1+cos(3π/2-θ/2)=√[sin(θ/4)-cos(θ/4)]²=sin(θ/4)-cos(θ/4)