求证:tan^2+cot^2=[2(3+cos4)]/(1-cos4x)
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/06 00:28:50
xM@3#!
e.!AWnz/A{+8T /q<~"Co3ST{VF[POFGL;a*qu8 P$[n(fgJOOpe|e8BFtX ej+Mo!\v'-y2{ʼH,M
求证:tan^2+cot^2=[2(3+cos4)]/(1-cos4x)
求证:tan^2+cot^2=[2(3+cos4)]/(1-cos4x)
求证:tan^2+cot^2=[2(3+cos4)]/(1-cos4x)
tan^2x+cot^2x
=sin^2x/cos^2x+cos^2x/sin^2x
=(sin^4x+cos^4x)/(cos^2xsin^2x)
=(sin^4x+cos^4x+2cos^2xsin^2x-2cos^2xsin^2x)/(cos^2xsin^2x)
=[(sin^2x+cos^2x)^2-2cos^2xsin^2x]/(cos^2xsin^2x)
=[1-2cos^2xsin^2x]/(cos^2xsin^2x)
=[1-2cos^2xsin^2x]/(cosxsinx)^2
=[1-2cos^2xsin^2x]/(1/2*sin2x)^2
=4[1-2cos^2xsin^2x]/(sin2x)^2
=4[1-2cos^2xsin^2x]/[(1-cos4x)/2]
=8[1-2cos^2xsin^2x]/(1-cos4x)
=4[2-4cos^2xsin^2x]/(1-cos4x)
=4[2-(sin2x)^2]/(1-cos4x)
=4[2-(1-cos4x)/2]/(1-cos4x)
=4[2-1/2+cos4x)/2]/(1-cos4x)
=4[3/2+cos4x/2]/(1-cos4x)
=2(3+cos4x)/(1-cos4x)
求证sin^2tan+cos^2cot+2sincos=tan+cot
求证tanα+cotα=2/sin2α
求证tan^2α+cot^2α+1=(tan^2α+tanα+1)(cot^2α-cotα+1)
求证:tan^2x+cot^2x=2(3+cos4x)/(1-cos4x)
求证:tan^2+cot^2=[2(3+cos4)]/(1-cos4x)
tan(3/2π-α)=?;cot(-α-π)=?
请教几道高一三角函数恒等变换题求证:tan ^2 A- cot^2 A=-2 sin4A/ sin^3A
sin(2a)[cot(a/2)-tan(a/2)]=4cos^2(a)求证,
求证(1+tan^2A)/(1+cot^2A)=(1-tanA/1-cotA)^2
求证:cos^2a/[cot(a/2)-tan (a/2)]=1/4sin2a
数学三角函数证明求证2cotA=cot(A/2)-tan(A/2)
求证(cos^α)/(cotα/2-tanα/2)=1/4sin2α
求证:sin^2θtanθ+cos^2θcotθ+2sinθcosθ=tanθ+cotθ
求证4cos^2α/cotα-tanα=sin4α
cot^-1(x)=π/2-tan^-1(x) 求证过程
求证该恒等式2tan2α+tanα=cotα-4cot4α急,please
设α为锐角,tanα+cotα=3,则tan^2α+cot^2α=
诱导公式推广tan(3/2π+α)=cotα还是tan(1/2π+α)=cotα