tanθ=(sina-cosa)/(sina+cosa) 求sina-cosa/sinθ
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/08 13:37:04
tanθ=(sina-cosa)/(sina+cosa) 求sina-cosa/sinθ
![tanθ=(sina-cosa)/(sina+cosa) 求sina-cosa/sinθ](/uploads/image/z/19375209-9-9.jpg?t=tan%CE%B8%3D%28sina-cosa%29%2F%28sina%2Bcosa%29+%E6%B1%82sina-cosa%2Fsin%CE%B8)
tanθ=(sina-cosa)/(sina+cosa) =(tana-1)/(tana+1)=tan(a-45°)
θ=α-45°或者θ=180°+α-45°=α+135°
当θ=α-45°,sina-cosa/sinθ=sina-cosa/1/根号2(sinα-cosα)=根号2
当θ=α+135°,sina-cosa/sinθ=sina-cosa/1/根号2(-sinα+cosα)=-根号2
θ=α-45°或者θ=180°+α-45°=α+135°
当θ=α-45°,sina-cosa/sinθ=sina-cosa/1/根号2(sinα-cosα)=根号2
当θ=α+135°,sina-cosa/sinθ=sina-cosa/1/根号2(-sinα+cosα)=-根号2
已知sin=-3/5,那么(cosa+sina/cosa-sina)+(cosa-sina/cosa+sina)
若θ,a为锐角且tanθ=(sina-cosa)/(sina+cosa),求证:sina-cosa=√2sinθ.
已知(2sinA+cosA)/(sinA-cosA)=-5 求1、(sinA+cosA)/(sinA-cosA) 2、3
若|cosa|/cosa+|sina|/sinθ=0,试判断sin(cosa)cos(sina)的符号
( sina*sina-cosa*cosa)/cos2a*sina+cos2a*cosa)=
证明(1-cos^2a)/(sina+cosa)-(sina+cosa)/(tan^2a-1)=sina+cosa
证明(1-cos^2a)/(sina-cosa)-(sina+cosa)/(tan^2-1)=sina+cosa
[cosa -sina
求证:1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina
已知tana=3,求①sina+cosa/2sina-cosa ②sin²a+sina·cosa+3cos
若|cosa|/cosa+sina/|sina|=0,试判断sin(cosa)·cos(sina)的符号
tanA=2,求sinA+cosA/sinA-cosA=