tanx+sinx+x大小比较

江咳邰4676如何比较sinx,cosx,tanx的大小? -
欧贱待18074818671 ______ 1)sinx-cosx=√2[sin(x-π/4)],x-π/4∈(-π/4,3π/4),显然当x∈(0,π/4)时,√2sin(x-π/4)<0,sinx<cosxx=π/4时,,√2sin(x-π/4)=0,sinx=cosxx∈(π/4,π)时,√2sin(x-π/4)>0,sinx>cosx2)tanx-cosx=[sinx-(cosx)^2]/cosx=[(sinx)^2+sinx-1]/cosx.x∈(0,π/2)∪(π/2,π),...

江咳邰4676tanx.sinx/tanx - sinx=tanx+sinx/tanxsinx -
欧贱待18074818671 ______ 1-cos ^2x=sin^2x 方程两边同时乘以sinx^2/cosx^2 sin^2x/cos^2x - sin^2x =sin^2x sin^2x/cos^2x tan^2x -sin^2x =sin^2 x tan^2 x(tanx-sinx)(tanx+sinx)=(sinx tanx) (sinx tanx) divide both sides by sinxtanx/(tanx+sinx) tanx-sinx/tanx sinx=tanxsinx/tanx+...

江咳邰4676求证:tanxsinx/(tanx - sinx)=(tanx+sinx)/(tanxsinx) -
欧贱待18074818671 ______ 两边同时乘以(tan(x)sin(x))/(tan(x)+sin(x)),可得原命题即证明 tan(x)tan(x)sin(x)sin(x)/( tan(x)tan(x)-sin(x)sin(x))=1 tan(x)tan(x)sin(x)sin(x)/( tan(x)tan(x)-sin(x)sin(x))分子分母同时除以 sin(x)sin(x)可得: tan(x)tan(x)/( 1/(cos(x)cos(x))-1)=tan(x)tan(x)/tan(x)tan(x)=1=右边,证毕

江咳邰4676当函数y=sin(π∕3+x)cos(π∕3 - x)取最大值时,tanx的值为 -
欧贱待18074818671 ______ 解:y=sin(π∕3+x)cos(π∕3-x) = (sinπ∕3•cosx+cosπ∕3•sinx)(cosπ∕3•cosx+sinπ∕3•sinx) =(√3/2•cosx+1/2•sinx)(1/2•cosx+√3/2•sinx) =√3/4•(cosx)^2+√3/4•(sinx)^2+3/4sinxcosx+1/4sinxcosx =√3/4+sinxcosx =√3/4+1/2sin2x 所以当y=sin(π...

江咳邰4676求证:函数Y=(sinx+tanx)/(cos x+cot x)在定义域内恒大于零 -
欧贱待18074818671 ______ 函数的定义域不包含x=kpi+pi/2,也不包含x=kpi 所以sinx不等于0且不等于-1,cosx不等于-1且不等于0 y=(sinx+tanx)/(cos x+cot x)=[(sinxcosx+sinx)/cosx]/[(sinxcosx+cosx)/sinx]=(sinxcosx+sinx)sinx/[(sinxcosx+cosx)cosx]=sinx^2(1+cosx)/[cosx^2(1+sinx)]1+cosx>01+sinx>0 sinx^2(1+cosx)/[cosx^2(1+sinx)]>0 所以函数在定义域内恒大于零

江咳邰4676已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x - π4),当m=0时, -
欧贱待18074818671 ______ 1、将m=0带入f(x),f(x)=(1+1/tanx)sin^2x利用半角公式化简整理得f(x)=1+√2*sin(2x-π/4)/2 x属于[π/8,3π/4],2x-π/4属于[0,5π/4],sin(2x-π/4)∈[-√2/2,1],带入可得f(x)的取值范围[0,1+√2/2]2、tana=sina/cosa=2,sin^2a+cos^2a=1,sina=2√5/5,cos=√5/5,带入求得m=-2

江咳邰4676若函数f(x)=(1+tanx)cosx,0≤x<π/2,则f(x)的最大值和最小值分别为,要过程 -
欧贱待18074818671 ______ f(x)=(1+tanx)cosx=cosx+sinx=√2((√2/2)cosx+(√2/2)sinx)=√2sin(x+π/4) f(x)在[0,π/4]内增,在[π/4,π/2]内减 因此最大值为:当x=π/4时,f(π/4)=√2 最小值为:当x=0时,f(0)=√2/2

江咳邰4676tanx+sinx=1 有一道题tanx+sinx=1 求x -
欧贱待18074818671 ______[答案] tanx+sinx=sinx/cosx +sinx=1(sinx+sinxcosx)/cosx =1sinx+sinxcosx=cosxsinxcosx=cosx-sinx=√2cos(π/4 +x)1/2 sin2x =√2cos(π/4 +x)sin2x =2√2cos(π/4 +x)两边平方 sin²2x =8cos²(π/4 +x...

江咳邰4676当函数y=sin(π3+x)cos(π3 - x)取得最大值时,tanx的值为______. -
欧贱待18074818671 ______[答案] y=sin( π 3+x)cos( π 3-x) =( 3 2cosx+ 1 2sinx)( 1 2cosx+ 3 2sinx) = 3 4+sinxcosx = 3 4+ 1 2sin2x 当函数y=sin( π 3+x)cos( π 3-x)取得最大值时,sin2x=1,即有x=kπ+ π 4,k∈Z. 此时有tanx=1. 故答案为:1.

江咳邰4676已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x - π4),当m=0时,求f(x)在区间[π/8,3π/4]上的取值范围 -
欧贱待18074818671 ______ m=0,所以f(x)=)=(1+1/tanx)sin^2x=sin^2x+sinxcosx=1-cos^2x+1/2sin2x=1-(cos2x+1)/2+1/2sin2x=-1/2cos2x+1/2sin2x+1/2=二分之根号二sin(2x-π/4)+1/2,下面的应该就好做了吧