求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/15 13:25:29
求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)
求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3
求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3
求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3
是否是:(-1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3
证明:
左边=[-1/sina-(-sina)]/[1/(-cosa)+cosa]
=[(-1+sin^2a)/sina]/[(cos^2-1)/cosa]
=-cos^2a/sina*cosa/(-sin^2a)
=cos^3a/sin^3a
=1/(tana)^3
=右边.
求证:sinα
求证:sinα
求证:sinα/(1+cosα)+(1+cosα)/sinα=2/sinα
求证:2(1-Sinα)(1+Sinα)=(1-Sinα+Cosα)的平方
求证:2(1-Sinα)(1+Sinα)=(1-Sinα+Cosα)的平方
求证 sinαcosβ=1/2[sin(α+β)+sin(α-β)]
求证:sinα+cosα>1
求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1
求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3
求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
三角函数证明(sinα+sinθ)*(sinα-sinθ)=sin(α+θ)*sin(α-θ)求证(sinα+sinθ)*(sinα-sinθ)=sin(α+θ)*sin(α-θ)
求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos
sinα-cosα+1/sinα+cosα-1=1+sinα/cosα 求证 sinα-cosα+1/sinα+cosα-1=1+sinα/cosα 求证
求证 1+sinα+cosα+2sinαcosα/求证 (1+sinα+cosα+2sinαcosα)/(1+sinα+cosα)=sinα+cosα
求证(1+sinα+cosα+2sincosα)/(1+sinα+cosα)=sinα+cosα