lnsinx (pai-2x)^-搜答案-拍题作业网

洛必达法则上下求导得答案是1/2再答：把cot2x写成cos2x/sin2x

1）因为π

诱导公式f(x)=(1+2cos²x-1)/(4cosx)+asin(x/2)cos(x/2)=(cosx)/2+a/2*sinx=(a/2)sinx+(1/2)cosx=√[(a/2)&s

当x趋于0时(lnsinx/x)~sinx/x-1lim((lnsinx/x)/x^2),=lim((sinx/x-1)/x^2)=lim(（sinx-x)/x^3)用洛比达法则=lim(cosx-1

(1)因为x∈[π/4,π/2]所以2x-π/3∈[π/6,2π/3]所以x=5π/12时有最大值3,x=π/4时有最小值2（2）若不等式-2

记积分值为I,I=积分（0到pi/2)(ln2+lnsinx/2+lncosx/2)dx=（第三项做变换x=pi-t）pi/2ln2+积分（0到pi/2）lnsinx/2dx+积分（pi/2到pi）l

cos(x-π/4)=cosxcosπ/4+sinxsinπ/4=√2/2*(cosx+sinx)=√2/10sinx+cosx=1/5cosx=1/5-sinx平方cos²x=1-sin&

f(x)=sin(π-x)cos(3π/2+x)+sin(π+x)sin(3π/2-x)=(sinx)(sinx)+(-sinx)(-cosx)=sinx(sinx+cosx)f'(x)=cosx(s

cos(2(x-pai/4))=2(cos(x-pai/4))^2-1=2×(√2/10)^2-1=-24/25cos(2(x-/4))=cos(2x-pai/2)=sin2xsin2

y=sin(π/2+x)*cos(π/6-x)=cosx*(cosπ/6*cosx+sinπ/6*sinx)=√3/2(cosx)^2+1/2sinx*cosx=√3/4cos2x+√3/4+1/4s

f（x）=sin（πx/4-π/6）-2cos²πx/8+1={sinπx/4*cosπ/6-cosπx/4*sinπ/6}-{2cos²πx/8-1}={sinπx/4*cosπ

f(x)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/4)cos(π/4-x)=cos(2x-π/3)+2sin(x-π/

cos(2x+pai/4)+sin(2x+pai/4)=√2*[√2/2cos(2x+pai/4)+√2/2sin(2x+pai/4)]=√2*[sin45*cos(2x+pai/4)+cos45si

-1/8用罗比达法则

函数y=log2底sin（2x+π/4）的单调增区间就是sin（2x+π/4）的单调增-pai/8+kpai<x<pai/8+kpai同理可求减区间pai/8+kpai<x<3

因为sin(pai/4+x)=sin[pai/2-(pai/4-x)](诱导公式)=cos(pai/4-x)(cosx是偶函数)=cos(x-pai/4)=根号2/10由x属于[pai/2,3pai/

用cos2x=1-2倍sin方x.cos(2π/3+2x)=1-2倍sin(π/6-x)方=1-2乘1/9=7/9

f(x)=sin(π/2+x)cos(π/6-x)=cosxcos(π/6-x)=cosx(1/2cosx+√3/2sinx)=1/2cos^2x+√3/2sinxcosx=1/4+1/4cos2x+

Kpai-pai/2