Question

三角函数高手进！！！ 有答案给你 但是要很清晰的过程！！

Answer 1

解法一：
cos2x=(cosx)^2-(sinx)^2
∴cosθ=(cosθ/2)^2-(sinθ/2)^2
∴1+cosθ=[(cosθ/2)^2+(sinθ/2)^2]+[(cosθ/2)^2-(sinθ/2)^2]=2(cosθ/2)^2
1-cosθ=[(cosθ/2)^2+(sinθ/2)^2]-[(cosθ/2)^2-(sinθ/2)^2]=2(sinθ/2)^2
√[(1+cosθ)/(1+cosθ)]=cos(θ/2)/sin(θ/2) 分子分母同乘以2cos(θ/2)
=2[cos(θ/2)]^2/[2sin(θ/2)cos(θ/2)]
=(1+cosθ)/sinθ
解法二：
(cosθ)^2+(sinθ)^2=1
∴(sinθ)^2=1-(cosθ)^2
√[(1+cosθ)/(1+cosθ)]
=√[(1+cosθ)(1+cosθ)/(1+cosθ)(1+cosθ)] (分子分母同乘以(1+cosθ))
=√{(1+cosθ)^2/[1-(cosθ)^2]}
=√[(1+cosθ)^2/(sinθ)^2]
=(1+cosθ)/sinθ

Answer 2

根号里上下乘以（1+COSθ)分母用平方差公式解为SIN²θ