sinπ–x等于sinx

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/29 22:30:12
函数y=sin(π/2-x)+sinx的最大值

cosx+sinx=√2(√2/2cosx)+√2(√2/2sinx)=√2(√2/2cosx+√2/2sinx)=√2(sinπ/4cosx+cosπ/4sinx)=√2sin(x+π/4)划一公式

sinx*sin(x+y)+cosx*cos(x+y)等于?

∵cos(A-B)=cosA*cosB+sinA*sinB∴逆用以上公式得sinx*sin(x+y)+cosx*cos(x+y)=cos[(x+y)-x]=cosy∴sinx*sin(x+y)+cos

化简sin(x+y)sinx+cos(x+y)cosx等于?

sin(x+y)sinx+cos(x+y)cosx=(sinxcosy+sinycosx)sinx+(cosxcosy-sinxsiny)cosx=(sinx)^2*cosy+(cosx)^2*cos

已知函数f(x)=sin(π/2-x)+sinx

f(x)=cosx+sinxf(x)=√2sin(x+π/4)(1)递增区间:2kπ-π/2≤x+π/4≤2kπ+π/2得:2kπ-3/4π≤x≤2kπ+π/4递增区间是:[2kπ-3π/4,2kπ+

求证sinx-cosx=根号2sin(x-π/4)

√2sin(x-π/4)=√2(sinxcosπ/4-cosxsinπ/4)=√2[(1/√2)sinx-(1/√2)cosx]=sinx-cosx

函数f(x)=sinx-sin(x-π3

f(x)=sinx-sin(x-π3)=12sinx+32cosx=sin(x+π3)∴函数f(x)=sinx-sin(x-π3)的最大值为1故答案为:1

函数f(x)=3sinx+sin(π2+x)

由f(x)=3sinx+cosx=2sin(x+π6)⇒f(x)max=2.故答案为:2

已知函数f(x)=2sinx*sin(π/2+x)-2sin^2x+1

f(x)=2sinx*sin(π/2+x)-2sin^2x+1=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)因为f(x0/2)=根2/3所以sin(x0+π/4)

已知函数f(x)=sinx+sin(x+π/2) ,

因为f(x)=sinx+cosx=√2sin(x+π/4)第一题T=2π/1=2π第二题当sin(x+π/4)=1时,为最大值,即f(x)=√2sin(x+π/4)=-1时,为最小值,即f(x)=-√

设函数f(x)=sinx+sin(x+π/3)

1)由三角函数和差化积公式:f(x)=2sin(x+x+π/3)/2cos(x-x-π/3)/2=2sin(x+π/6)cos(π/6)=√3sin(x+π/6)f(x)的最小值为-√3.当x+π/6

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si

sinx+cosx=√2sin(x+π/4)

sinx+cosx=√2(cos45°sinx+sin45°cosx)=√2sin(x+45°)==√2sin(x+π/4)

求X解集:sinx=2sin(π/3-x)

这类题目的一般解法是先化成asinx+bcosx=0,再化成√(a^2+b^2)sin(x+φ)=0,即可求出解集.

limx趋近0 {【sinx---sin(sinx)】sinx}/(x^4)

利用罗必塔法则limx趋近0{【sinx---sin(sinx)】sinx}/(x^4)=limx趋近0{(sinx)的平方---sin(sinx)乘以sinx}/(x^4)=limx趋近0{sinx

化简{sin(π+x)}/1+cos(3π-x)根号下(tanx-sinx)/(tanx+sinx)

=-sinx/(1-cosx)*√[(1/cosx-1)/(1/cosx+1)]]=-sinx/(1-cosx)*[(1-cosx)/|sinx|]sinx>0=-1sinx再问:化简,不用求值再答:

sin²x+sinx=1则sinx等于多少?

sin²x+sinx=1sin²x+sinx+(1/2)²=1+1/4(sinx+1/2)²=5/4sinx+1/2=±√5/2sinx=-1/2±√5/2si

|sin(x+π/2)|-|sinx|求导过程?

(1)当x∈(0,π/2)时:y=|sin(x+π/2)|-|sinx|=sin(x+π/2)-sinxy′=cos(x+π/2)-cosx(2)当x∈(π/2,π)时:y=-sin(x+π/2)-s

a*sinx+b*sin(x+c)等于什么表达式?

展开两式,x的正余弦系数,分别对应相等,得:A*cosB=a+b*cosc;AsinB=b*sinc;

sin(pai/3-x)等于sinx?

不等于sin(2kπ+x)=sinxsin(π-x)=sinx再问:那是多少再答:sin(π/3-x)=sinπ/3cosx-cosπ/3sinx=√3/2cosx-1/2sinx