Question

化简根号下1+cos(3π-θ)/2 (3π/2<θ<2π)

Answer 1

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)

Answer 2

3π/2<θ<2π
π＜3π-θ＜3π/2
π/4＜(3π-θ)/4＜3π/8
cos(3π-θ)/4＞0
cos(3π-θ)/2=2cos^2[(3π-θ)/4]-1
根号{1+cos(3π-θ)/2）
=根号{1+2cos^2[(3π-θ)/4]-1}
=根号{2cos^2[(3π-θ)/4]}
=根号2 * cos[(3π-θ)/4
=根号2 cos[π-（π/4+θ）]
= -根号2 cos(θ+π/4)

Answer 3

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)

Answer 4

原式=√1+2cos^2[(3π-θ)/4]-1
=√2cos^2[(3π-θ)/4]
=√2 * cos[(3π-θ)/4
=√2 cos[π-（π/4+θ）]
= -√2 cos(θ+π/4)