ldcf.net
当前位置:首页 >> Cos3x=4Cos^3x<osx >>

Cos3x=4Cos^3x<osx

解:4cos³x-3cosx=cos(2x)4cos³x-3cosx-(2cos²x-1)=04cos³x-2cos²x-3cosx+1=04cos³x-4cos²x+2cos²x-2cosx-cosx+1=04cos²x(cosx-1)+2cosx(cosx-1)-(cosx-1)=0(cosx-1)(4cos²x+2cosx-1)=0(cosx...

见图

∫cos³xsinxdx =∫cos³xd(-cosx) =-∫cos³xd(cosx) =-¼cos⁴x +C 答案是-¼cos⁴x +C,前面有负号。

y=sin(3x-π/4) = cos[π/2-(3x-π/4)] = cos[3π/4-3x] = cos(3x-3π/4) = cos[3(x-π/4)] 将y=cos3x图象向右平移π/4个单位

分母是立方和的立方? 不是立方和的平方?

y=√(1+4^2)sin(3x+ψ)其中tanψ=1/4 所以最大值是√17

sin3x+cos3x =√2 [ sin3x.cos(π/4)+cos3x.sin(π/4) ] =√2 .sin(3x+ π/4)

(x→0)lim { (cosx-cos3x) / x³ } = (x→0)lim { (cosx-4cos³x+3cosx) / x³ } = (x→0)lim { (4cosx-4cos³x) / x³ } = (x→0)lim { 4cosx(1-cos²x) / x³ } = (x→0)lim { 4cosxsin²x / x³ } ...

一方面,(cos3x)'=(-sin3x)*(3x)'=-3sin3x; 另方面,[4(cosx)^3-3cosx]'=12(cosx)^2*(cosx)'-3(-sinx) =12[1-(sinx)^2]*(-sinx)+3sinx =12(sinx)^3-9sinx; 由已知,-3sin3x=12(sinx)^3-9sinx, 故 sin3x=-4(sinx)^3+3sinx.

cos4a =cos²2a-sin²2a =2cos²2a-1 =2[cos2a]²-1 =2[2cos²a-1]²-1 =2[4cos^4a-4cos²a+1]-1 =8cos^4a-8cos²a+1

网站首页 | 网站地图
All rights reserved Powered by www.ldcf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com