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