sin(π-x)

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sin(x+π/4)

sin(x+π/4)首先你写的答案不对,是2kπ-3π/4<x+π/4<2kπ-π/4对于这种题需要知道sinx的图像很清楚看到满足sinx<-√2/2的部分是2kπ-3π/4<x<2kπ-π/4∴sin(x+

sin平方x-sinxcosx-2cos平方x 求sin(x-π/3)X为锐角 sin平方x-sin

sin平方x-sinxcosx-2cos平方x求sin(x-π/3)X为锐角sin平方x-sinxcosx-2cos平方x=0求sin(x-π/3)请看图片

sin^2(π/3-x)+sin^2(π/6+x)

sin^2(π/3-x)+sin^2(π/6+x)解sin²(π/3-x)+sin²(π/6+x)=sin²[π/2-(π/6+x)]+sin²(π/6+x)=cos²(π/6+x)+sin

2sin(x+π/4)sin(x-π/4)=?

2sin(x+π/4)sin(x-π/4)=?2sin(x+π/4)sin(x-π/4)=2sin(x+π/4)sin[-π/2+(π/4+x)]=-2sin(x+π/4)cos(π/4+x)=-sin(2x+π/2)=-cos2x2sin

sin(x+π/4)sin(x-π/4)等于什么?

sin(x+π/4)sin(x-π/4)等于什么?

sinπ(x-1)=-sinπx如何证明

sinπ(x-1)=-sinπx如何证明sinπ(x-1)=sin(πx-π)=-sinπx

sin x + sin πx 的周期是多少?

sinx+sinπx的周期是多少?无周期y=sinx+sinπx=2sin[(1+π)x/2]cos[(1-π)x/2]

sin^2(π/4 +x)

sin^2(π/4+x)sin^2(π/4+x)=[1-cos2(π/4+x)]/2=[1-cos(π/2+2x)]/2=1/2+sin2x/2运用公式Cos2A=CosA^2-SinA^2=1-2SinA^2sin^2(π/4+x)=[1

sinπx的导数

sinπx的导数用链式法则:y=sin(πx)dy/dx=dsin(πx)/d(πx)*d(πx)/dx=cos(πx)*π(dx/dx)=cos(πx)*π=πcos(πx)πcosπx

sin(πx/2)求导

sin(πx/2)求导y=sin(πx/2)则y'=cos(πx/2)*(πx/2)'=π/2*cos(πx/2)

-sin(π+x)等于多少?

-sin(π+x)等于多少?-sin(π+x)=-[-sin(x)]=sin(x)

化简 sin²x+sin²(x+2π/3)+sin²(x-2π/3)

化简sin²x+sin²(x+2π/3)+sin²(x-2π/3)sin²x+sin²(x+2π/3)+sin²(x-2π/3) 

已知tanx=3,则[sin(π+x)-2sin(π/2+x)]/[sin(7π-x)+sin(7π

已知tanx=3,则[sin(π+x)-2sin(π/2+x)]/[sin(7π-x)+sin(7π/2-x)]已知tanx=3,则[sin(π+x)-2sin(π/2+x)]/[sin(7π-x)+sin(7π/2-x)]的值是tanx=

sin x

sinx-1/2π(0,π/2)0到1/2π

计算sin(π/10)x sin (13π/10)

计算sin(π/10)xsin(13π/10)sin(π/10)xsin(13π/10)=-1/2[cos(π/10+13π/10)-cos(π/10-13π/10)]=-1/2[cos(7π/5)-cos(-6π/5)]=-1/2[cos

已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(

已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于(

已知函数f(x)=sin^2 x+2根号3sinxcosx+sin(x+π/4)sin(x-π/4)

已知函数f(x)=sin^2x+2根号3sinxcosx+sin(x+π/4)sin(x-π/4),x属于R,求f(x)的最小正周期和值域若x=x00≤x0小于等于π/2为f(x)的一个零点,求sin2x0的值f(x)=sin^2x+2√3

f(x)=(1+cotx)sin^2x-2sin(x+π/4)sin(x-π/4)还有一题,

f(x)=(1+cotx)sin^2x-2sin(x+π/4)sin(x-π/4)还有一题,f(x)=(1+cotx)sin^2x-2sin(x+π/4)sin(x-π/4)=(sinx+cosx)sinx+cos((x+π/4)+(x-π

已知函数f(x)=(1+1\tanx)sin^2x+m sin(x+π/4)sin(x-π/4)

已知函数f(x)=(1+1\tanx)sin^2x+msin(x+π/4)sin(x-π/4)sin(x+π/4)sin(x-π/4)=[cos(π/2)-cos(2x)]/2=-cos(2x)/2(1+1\tanx)sin^2x=[(si

求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π

求证:(sin2x/(1-cos2x))·(sinx/(1+sinx))=tan(π/4-x/2).[sin2x/(1-cos2x)]·[sinx/(1+sinx)]=2sinx*cosx*sinx/[2(sinx)^2*(1+sinx)]