若3π/4<θ<5π/4,化简,根号下[cosπ/4乘sin(3π/4--θ)][sin(π--θ)--sin(θ--π/2)],除sin（θ+π/4）

3π/4<θ<5π/4,化简,根号下[cosπ/4乘sin(3π/4--θ)][sin(π--θ)--sin(θ--π/2)],除sin（θ+π/4）
解：[cosπ/4乘sin(3π/4--θ)= cosπ/4乘sin(π-θ-π/4)=√2/2[sin(π-θ)cosπ/4-cos(π-θ)sinπ/4]
=1/2[sin(π-θ)+cos(π-θ)
[sin(π--θ)--sin(θ--π/2)]= sin(π--θ)- sin(θ-π+π/2)= sin(π--θ)-[ sin(θ--π)cosπ/2+cos(θ-π) sinπ/2]
= sin(π--θ)- cos(π--θ)
sin（θ+π/4）=sin(θ-π+5π/4)= sin(θ--π)cos(5π/4)-cos (θ--π)sin(5π/4)
=√2/2[sin(π--θ)+cos (π--θ)]
综上三式，
[cosπ/4乘sin(3π/4--θ)][sin(π--θ)--sin(θ--π/2)],除sin（θ+π/4）=√2/2[sin(π--θ)-cos(π--θ)
后面你再化简一下，化成单角θ的单一三角函数形式就完成了