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若tanθ+1/tanθ=3则sinθ·cosθ=若tanθ+1/tanθ=3则sinθ·cosθ=若tanθ+1/tanθ=3则sinθ·cosθ=有什么不懂的可以随时追问

求值：2(sin^6θ+cos^6θ)-3(sin^4θ+cos^4θ)求值：2(sin^6θ+cos^6θ)-3(sin^4θ+cos^4θ)求值：2(sin^6θ+cos^6θ)-3(sin^4θ

求值tan(π／6-θ)+tan(π／6+θ)+√3tan(π／6-θ)tan(π／6+θ)求值tan(π／6-θ)+tan(π／6+θ)+√3tan(π／6-θ)tan(π／6+θ)求值tan(π／

已知sin2θ+4/cosθ+1=2,则(cosθ+3)·(sinθ+1)的值为已知sin2θ+4/cosθ+1=2,则(cosθ+3)·(sinθ+1)的值为已知sin2θ+4/cosθ+1=2,则

|4cosθ+3sinθ-6|的最小值|4cosθ+3sinθ-6|的最小值|4cosθ+3sinθ-6|的最小值1前两项值在-5到5当等于5是总试最小为1

化简:{cot(θ+4π)·cos(θ+π)·sin2次方(θ+3π)}/{tan(π+θ)·cos3次方(-π-化简：{cot（θ+4π）·cos（θ+π）·sin2次方（θ+3π）}/{tan（π

已知cos(75°+θ)=1/3,θ为第三象限角,求cos(-225°-θ)+sin(435°+θ)的值.·······若有大神路过,顺便帮个忙~灰常灰常滴感谢!已知cos(75°+θ)=1/3,θ为

设f(θ)=[sin2(6π+θ)+cosθ-2cos3(3π+θ)-3]/2+2cos2(θ-4π)-cos(-θ),求f(π/3)设f(θ)=[sin2(6π+θ)+cosθ-2cos3(3π+θ

设f(θ)=[sin2(6π+θ)+cosθ-2cos3(3π+θ)-3]/2+2cos2(θ-4π)-cos(-θ),求f(π/3)设f(θ)=[sin2(6π+θ)+cosθ-2cos3(3π+θ

tan(π/6-θ)+tan(π/6+θ)+根号3tan(π/6-θ)tan(π/6+θ)化简这个式子,tan[(π/6-θ)+(π/6+θ)]*[1-tan(π-θ)tan(π/6+θ)]+根号3t

已知三角形ABC的面积为3,且满足0≤向量AB·向量AC≤6,设向量AB、AC的夹角为θ1.求θ的取值范围2.求函数ƒ（θ）=2sin^2（π/4+θ)-√3cos2θ的最大值和最小值已知三

若tanθ=根号3/3,则cos^2θ-sin(θ+π/6)cosθ的值若tanθ=根号3/3,则cos^2θ-sin(θ+π/6)cosθ的值若tanθ=根号3/3,则cos^2θ-sin(θ+π/

已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcosθ-3cos^2θ)已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=

∫（sin∧3θ／cos∧6θ）dθ∫（sin∧3θ／cos∧6θ）dθ∫（sin∧3θ／cos∧6θ）dθ∫(sinθ)^3/(cosθ)^6dθ=-∫(sinθ)^2/(cosθ)^6dcosθ=

[cot（θ+4π）·cos（θ+π）·sin^2（θ+3π）]/[tan（π+θ）·cos^3（-π-θ）]如题,望尽快予以解答[cot（θ+4π）·cos（θ+π）·sin^2（θ+3π）]/[t

1+cos(2θ)+cos(4θ)+cos(6θ)=4cosθcos(2θ)cos(3θ)证明相等.1+cos(2θ)+cos(4θ)+cos(6θ)=4cosθcos(2θ)cos(3θ)证明相等.

x除以4······3x除以5······4x除以6······5x除以4······3x除以5······4x除以6······5求xx除以4······3x除以5······4x除以6······

已知函数f(x)=x^2+2x·tanθ-1,x∈[-根号3,根号3]1`当θ=-π/6时,求f(x)的最大值最小值2·求使f(x)在区间[-1,根号3]上是单调函数的θ的取值范围已知函数f(x)=x

f（x）=sinΘ/3·x³+√3cosΘ/2·x²+tanΘ这个函数的导函数是?f（x）=sinΘ/3·x³+√3cosΘ/2·x²+tanΘ这个函数的导函数

已知cos(θ-π/6)＝3/5,θ∈(π/6,2π/3),求sinθ已知cos(θ-π/6)＝3/5,θ∈(π/6,2π/3),求sinθ已知cos(θ-π/6)＝3/5,θ∈(π/6,2π/3),