Question

cos2Θ/sin（Θ＋π/4）＝_√2/2，则sinΘ－cosΘ的值为

Answer 1

cos2Θ/sin（Θ＋π/4）＝-√2/2
cos2Θ = --√2/2sin（Θ＋π/4）
cos2Θ = --√2/2(sinΘcosπ/4+cosΘsinπ/4)
cos2Θ = --√2/2【(√2/2)sinΘ+(√2/2)cosΘ】
cos²Θ-sin²Θ = -(1/2)(sinΘ+cosΘ)
（cosΘ-sinΘ）（sinΘ+cosΘ）+(1/2)（sinΘ+cosΘ）=0
（cosΘ-sinΘ+1/2）（sinΘ+cosΘ）=0
cosΘ-sinΘ = -1/2

Answer 2

你好

cos2Θ/sin（Θ＋π/4）＝-√2/2
cos2Θ=-√2/2sin（Θ＋π/4）=-√2/2（sinΘcosπ/4+cosΘsinπ/4）=-（sinΘ+cosΘ）
cos² Θ-sin² Θ=-（sinΘ+cosΘ）
cos Θ-sinΘ=-1
sinΘ-cos Θ=1

Answer 3

解：
cos2θ/sin（θ+π/4）=-√2/2
(cos²θ-sin²θ)/(sinθcosπ/4+cosθsinπ/4)= -√2/2
[(cosθ+sinθ)(cosθ-sinθ)] / [√2/2(sinθ+cosθ)] = -√2/2
√2(cosθ+sinθ)(cosθ-sinθ) / (cosθ+sinθ) = -√2/2
√2(cosθ-sinθ) = -√2/2
cosθ-sinθ = -1/2