三道高一三角函数证明题(1)tanX*cosX/tanX-sinX=1+cosX/sinX(2)1/sin^2X+1/cos^2X-1/tan^2X=2+tan^2X(3)1-cos^4X-sin^4X/1-cos^6X-sin^6X=2/3
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![三道高一三角函数证明题(1)tanX*cosX/tanX-sinX=1+cosX/sinX(2)1/sin^2X+1/cos^2X-1/tan^2X=2+tan^2X(3)1-cos^4X-sin^4X/1-cos^6X-sin^6X=2/3](/uploads/image/z/4330583-71-3.jpg?t=%E4%B8%89%E9%81%93%E9%AB%98%E4%B8%80%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E8%AF%81%E6%98%8E%E9%A2%98%281%29tanX%2AcosX%2FtanX-sinX%3D1%2BcosX%2FsinX%282%291%2Fsin%5E2X%2B1%2Fcos%5E2X-1%2Ftan%5E2X%3D2%2Btan%5E2X%283%291-cos%5E4X-sin%5E4X%2F1-cos%5E6X-sin%5E6X%3D2%2F3)
三道高一三角函数证明题(1)tanX*cosX/tanX-sinX=1+cosX/sinX(2)1/sin^2X+1/cos^2X-1/tan^2X=2+tan^2X(3)1-cos^4X-sin^4X/1-cos^6X-sin^6X=2/3
三道高一三角函数证明题
(1)tanX*cosX/tanX-sinX=1+cosX/sinX
(2)1/sin^2X+1/cos^2X-1/tan^2X=2+tan^2X
(3)1-cos^4X-sin^4X/1-cos^6X-sin^6X=2/3
三道高一三角函数证明题(1)tanX*cosX/tanX-sinX=1+cosX/sinX(2)1/sin^2X+1/cos^2X-1/tan^2X=2+tan^2X(3)1-cos^4X-sin^4X/1-cos^6X-sin^6X=2/3
1,你的题目抄错,左边应是tanx*sinx/tanx-sinx
证明(化弦法):左边=(sin^2x/cosx)/[(sinx-sinx*cosx)/cosx]
=sinx/(1-cosx)
=[sinx(1+cosx)]/[(1+cosx)(1-cosx)]
=(1+cosx)sinx/sin^2x
=(1+cosx)/sinx=右边
2,证明(使用平方关系):左边=csc^2x+sec^2x-cot^2x
=1+sec^2x
=1+(1+tan^2x)
=2+tan^2x=右边
3,证明(使用配方+平方关系,包括立方和公式)
:左边=[1-(cos^2x+sin^2x)^2+2sin^2x*cos^2x]/{1-(cos^2x+sin^2x)(sin^4x+cos^4x-sin^2x*cos^2x)]
=[1-(cos^2x+sin^2x)^2+2sin^2x*cos^2x]/{1-(cos^2x+sin^2x)[(sin^2x+cos^2x)^2-3sin^2x*cos^2x)]}
=[1-1+2sin^2x*cos^2x]/{1-1*[1-3sin^2x*cos^2x]}
=(2sin^2x*cos^2x)/(3sin^2x*cos^2x)
2/3=右边.