2 tan θ = 2; −180° < θ < −90° 求sin(θ/2) cos(θ/2) tan(θ/2) 求确切值,

2 tan θ = 2; −180° < θ < −90° 求sin(θ/2) cos(θ/2) tan(θ/2) 求确切值,
2 tan θ = 2; −180° < θ < −90°
求
sin(θ/2)
cos(θ/2)
tan(θ/2)
求确切值,

2 tan θ = 2; −180° < θ < −90° 求sin(θ/2) cos(θ/2) tan(θ/2) 求确切值,
见下图.

你好，这道题涉及了三角函数的半角公式，这是很常用的，希望你能记住它们。
公式如下：
半角公式
[sin(θ/2)]^2 = [1 - cos(θ)] / 2
[cos(θ/2)]^2 = [1 + cos(θ)] / 2
tan(θ/2) = [1 - cos(θ)] /sin(θ) = sin(θ) / [1 + cos(θ)]=sin(θ/2) / co...

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你好，这道题涉及了三角函数的半角公式，这是很常用的，希望你能记住它们。
公式如下：
半角公式
[sin(θ/2)]^2 = [1 - cos(θ)] / 2
[cos(θ/2)]^2 = [1 + cos(θ)] / 2
tan(θ/2) = [1 - cos(θ)] /sin(θ) = sin(θ) / [1 + cos(θ)]=sin(θ/2) / cos(θ/2)
利用公式求解，过程如下：
2 tan θ = 2; −180° < θ < −90°
所以 tan θ = 1;
因为 −180° < θ < −90°
所以 θ = -135°
所以 sin(θ)=sin(-135°)= -sin(180°-135°) = - sin(45°) = -√2/2
所以 cos(θ)=cos(-135°)= -cos(180°-135°) = - cos(45°) = -√2/2
然后利用公式：
[sin(θ/2)]^2 = [1 - cos(θ)] / 2 = (1+√2/2) / 2 = (2+√2) / 4
[cos(θ/2)]^2 = [1 + cos(θ)] / 2 = (1-√2/2) / 2 = (2-√2) / 4
tan(θ/2) = [1 - cos(θ)] /sin(θ)
= (1+√2/2) / (-√2/2) =[(1+√2/2) *√2] / [(-√2/2)*√2]= (√2+1)/(-1) = -(√2+1)
因为 θ = -135° 所以θ/2 = -67.5° 所以可知 sin(θ/2) < 0 且 cos (θ/2) > 0
所以
sin(θ/2) = -√[(sin(θ/2))^2] = - √ ( (2+√2) / 4) = - √(2+√2) / 2
cos(θ/2) = + √[(cos(θ/2))^2] = √ ( (2-√2) / 4) = √(2-√2) / 2
所以确切值为：
sin(θ/2) = - √(2+√2) / 2
cos(θ/2) = √(2-√2) / 2
tan(θ/2) = -(√2+1)
有不明白的还可以追问我或发消息给我，希望能帮到你。

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