Question

当θ为第二象限角,且sin(θ/2+π/2)=1/3时[√(1-sinθ)]/[cos(θ/2)-sin(θ/2)]的值是

主要是[√(1-sinθ)]/[cos(θ/2)-sin(θ/2)]的化简写详细点，快~

Answer 1

sin(θ/2+π/2)=1/3
cos(θ/2)=1/3
根据 θ为第二象限角 和 cos（θ/2）为正值
可得 sin（θ/2）为正值
那么 sin（θ/2）=2√2/3
cos(θ/2)-sin(θ/2)=(1-2√2)/3<0

[√(1-sinθ)]/[cos(θ/2)-sin(θ/2)]
=[√(sin^2(θ/2)+cos^2(θ/2)-2sin(θ/2)cos(θ/2))]/[cos(θ/2)-sin(θ/2)]
= [√(sin(θ/2)-cos(θ/2))^2]/[cos(θ/2)-sin(θ/2)]
=|cos(θ/2)-sin(θ/2)|/[cos(θ/2)-sin(θ/2)]
=-1

欢迎追问！

Answer 2

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