已知cos^4a/cos^2β+sin^4a/sin2^β=1 求证cos^4β/cos^2a+sin^4β/sin^2
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/08/18 16:32:31
已知cos^4a/cos^2β+sin^4a/sin2^β=1 求证cos^4β/cos^2a+sin^4β/sin^2a=1
要证cos^4β/cos^2a+sin^4β/sin^2a=1
只需证cos^4βsin2α+sin^4βcos2α=cos2αsin2α
(1-sin2β)cos2βsin2α+(1-cos2β)sin2βcos2α=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2βsin2α-cos2βsin2βcos2α=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β(sin2α+cos2α)=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β-cos2αsin2α=0
cos2β(sin2α-sin2β)-cos2α(sin2α-sin2β)=0
(cos2β-cos2α)(sin2α-sin2β)=0
即证cos2β=cos2α,sin2α=sin2β然后···?
要证cos^4β/cos^2a+sin^4β/sin^2a=1
只需证cos^4βsin2α+sin^4βcos2α=cos2αsin2α
(1-sin2β)cos2βsin2α+(1-cos2β)sin2βcos2α=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2βsin2α-cos2βsin2βcos2α=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β(sin2α+cos2α)=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β-cos2αsin2α=0
cos2β(sin2α-sin2β)-cos2α(sin2α-sin2β)=0
(cos2β-cos2α)(sin2α-sin2β)=0
即证cos2β=cos2α,sin2α=sin2β然后···?
![已知cos^4a/cos^2β+sin^4a/sin2^β=1 求证cos^4β/cos^2a+sin^4β/sin^2](/uploads/image/z/17468214-6-4.jpg?t=%E5%B7%B2%E7%9F%A5cos%5E4a%2Fcos%5E2%CE%B2%2Bsin%5E4a%2Fsin2%5E%CE%B2%3D1+%E6%B1%82%E8%AF%81cos%5E4%CE%B2%2Fcos%5E2a%2Bsin%5E4%CE%B2%2Fsin%5E2)
反过来即可
若2sin(π/4+a)=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0.
若2sin(π/4+a)=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0
若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a
化简(1-sin^6 a-cos^6 a)/(cos^2 a-cos^4 a)==
证明sin(4A)sin(2A)(1-cos(2A)) cos(4A)cos(2A)(1 cos(2A))=cos(2A
已知tan a=-1/3,求4sin a-2cos a/5cos a+3sin a=
求证sin^4a+cos^4a=1-2sin²acos²a
已知cosa=1/4,求sin(2π+a)cos(-π+a)/cos(-a)tana
化简 1-sin^4a-cos^4a/cos^2a-cos^4a
已知(4sin a-2cos a)/(5cos a +3sin a)=6/11,求cos^4 a-sin^4 a
已知sin a-2cos a=0 求4sin^2 a-3sin a x cos a-5cos^2 a的值
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证:sin2θ+2cos2β=3