cos(x-π)

乌马田2961已知cos(x - π 6 )= - 3 3 ,则cosx+cos(x - π 3 )= -
酆斩庾15850413691 ______ ∵cos(x- π 6 )=- 3 3 ,∴cosx+cos(x- π 3 )=cosx+cosxcos π 3 +sinxsin π 3 = 3 2 cosx+ 3 2 sinx= 3 ( 3 2 cosx+ 1 2 sinx)= 3 cos(x- π 6 )= 3 *(- 3 3 )=-1. 故答案为:-1.

乌马田2961函数y=cos(x - π/3)cos(x+2π/3)的最大值是? -
酆斩庾15850413691 ______[答案] ∵y=cos(x-π/3)cos(x+2π/3)=cos(x-π/3)﹛﹣cos[π-(x+2π/3)]﹜=cos²(x-π/3) =1/2*[cos(2x-2π/3)+1]=1/2*[cos(2x-π/3)+1] ∴最大值=1

乌马田2961cos(x - π/4)=根号2/10,则sin(x+π/4)=?? -
酆斩庾15850413691 ______ π/2<x<3π/4 π/4<x-π/4<π/2 因为cos(x-π/4)=√2/10 所以sin(x-π/4)=√98/10 那么tan(x-π/4)=√(98/2)=7 (tanx-1)/(1+tanx)=7 解得 tanx=-4/3 sin2x=2tanx/(1+tan²x)=-24/25 cos2x=(1-tan²x)/(1+tan²x)=-7/25 sin(2x+π/3)=sin2xcosπ/3+cos2xsinπ/3=(-24/25)*(1/2)+(-7/25)*(√3/2)=-(24+7√3)/50 过程

乌马田2961函数y=cos(x - π6)(x∈[π6,23π])的最小值是______. -
酆斩庾15850413691 ______[答案] ∵x∈[ π 6, 2 3π],可得x- π 6∈[0, π 2] ∴当x- π 6= π 2时,即x= 2 3π时,函数y=cos(x- π 6)的最小值是0 故答案为:0

乌马田2961cos(x - π/3)怎么得到sin(π/2 - (x - π/3))为什么不能得sin(π/2+(x - π/3))诱导公式cosa=sin(π/2 - a)可也有cosa=sin(π/2+a)嘛 -
酆斩庾15850413691 ______[答案] cos(x-π/3)=sin(π/2-(x-π/3))=sin(π/2+(x-π/3))

乌马田2961函数y= - 2cos(x - π\3)的最大值是?过程 -
酆斩庾15850413691 ______[答案] 对于Y=-2cos(x-π/3) 因为cosX在为cos0+2kπ(cos0+2kπ代表比如cos0、cos2π、cos-2π这些都是为余弦的最大值1)有原函数y=AcosX,A为正数 的最大值 但是本题的A为负数 当然要取余弦函数为最小值-1的时候Y=-2cos(x-π/3)才会有最大值!cosX在为...

乌马田2961cos(x - 派+四分之派)怎么化简,正玄.余弦都化下 -
酆斩庾15850413691 ______[答案] cos(x-π+π/4)=cos[(x+π/4)-π]=-cos(x+π/4) 又=cos(x-3π/4)=cos[(x-π/4)-π/2]=sin(x-π/4) 关键是诱导公式,不懂追问,有助请采纳.

乌马田2961已知函数πf(x)=2cosxcos(x - π/6) - √3(sinx)^2+sinxcosx, -
酆斩庾15850413691 ______ f(x)=2cosx(cosx*√3/2+sinx*1/2)-√3(sinx)^2+sinxcosx=√3(cosx)^2+sinxcosx-√3(sinx)^2+sinxcosx=√3[(cosx)^2-(sinx)^2]+2sinxcosx=sin2x+√3cos2x=√[1^2+(√3)^2]sin(2x+z)=2sin(2x+z),其中tanz=√3=tanπ/3 所以f(x)=2sin(2x+π/6) 所以T=2π/2=π f(a)=2sin(2a+π/6)=1 sin(2a+π/6)=1/2 所以2a+π/6=2kπ+π/6或2kπ+5π/6 a=kπ或kπ+π/30<=a<=π 所以a=0,a=π,a=π/3

乌马田2961cos(x - π∕2)化简 在线急 -
酆斩庾15850413691 ______ cos(x-π∕2)=cos(π∕2-x)=cosx 因为余弦函数是偶函数,cosx=cos(-x)

乌马田2961若cos(x - π/4)= - 3/4,则sin2x的值为多少? -
酆斩庾15850413691 ______[答案] cos(x-π/4)=-3/4 cos(2x-π/2)=2[cos(x-π/4)]^2-1 =1/8 sin2x=cos(2x-π/2)=1/8