Question

高中数学三角函数一题。。在线等

已知函数F(X)=根号下<(1-X)/(1+X)>,若a属于（π/2，π），则F(COSa)+F(-COSa)可化为？
要有过程
答案是2/SINa，现在要过程

Answer 1

F(COSa)+F(-COSa)=根号下<(1-COSa)/(1+COSa)>+根号下<(1+COSa)/(1-COSa)>
（分子有理化）=(1-COSa)/|sina|+(1+COSa)/|sina|
=2/|sina|
∵ a∈(π/2，π)，原式=2/sina

Answer 2

∵ a∈（π/2,π）
∴a/2∈(π/4,π/2)
∴sin(a/2)>0,cos(a/2)>0

1-cosa=1-{1-2[sin(a/2)]^2}=2[sin(a/2)]^2
1+cosa=1+{2[cos(a/2)]^2-1}=2[cos(a/2)]^2

(1-cosa)/(1+cosa)
={2[sin(a/2)]^2}/{2[cos(a/2)]^2 }
=[tan(a/2)]^2

(1+cosa)/(1-cosa)
={2[cos(a/2)]^2}/{2[sin(a/2)]^2 }
=[cot(a/2)]^2

f(cosa)+f(-cosa)
=√[(1-cosa)/(1+cosa)]+√[(1+cosa)/(1-cosa)]
=tan(a/2)+cot(a/2)
=sin(a/2)/cos(a/2)+cos(a/2)/sin(a/2)
={[sin(a/2)]^2+[cos(a/2)]^2}/sin(a/2)cos(a/2)
=1/sin(a/2)cos(a/2)
=1/(sina/2)
=2/sina

Answer 3

cosa=2cos^2(a/2)-1=1-2sin^2(a/2)

(1-cosa)/(1+cosa)=sin^2(a/2)/cos^2(a/2)

a/2∈(π/4,π/2)

sin(a/2)>0 cos(a/2)>0
去根号
原式=sin(a/2)/cos(a/2)+cos(a/2)/sin(a/2)

sin(a/2)cos(a/2)=1/2sina
通分得 2/sina

Answer 4

sqrt((1-cosA)/(1+cosA))+sqrt((1+cosA)/(1-cosA))
=2/sqrt((1-cosA)(1+cosA)) //暴力通分
=2/sqrt(1-cos^2A)
=2/sinA

Answer 5

万能公式