f(x)=cos(2x-x/3)+2sin(x-π/4)sin(x+π/4)sin(x+π/4)化简.
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 02:53:54
x)KӨдM/0Э76*Ө=ߠo fj3L{A&H`}03l(vē={{6c=9@U
kí(
d֠bVa4E%PaLaQ )0 0T&Z-C}#`a
LqrrP{!jj_\gA ͒
f(x)=cos(2x-x/3)+2sin(x-π/4)sin(x+π/4)sin(x+π/4)化简.
f(x)=cos(2x-x/3)+2sin(x-π/4)sin(x+π/4)sin(x+π/4)
化简.
f(x)=cos(2x-x/3)+2sin(x-π/4)sin(x+π/4)sin(x+π/4)化简.
似乎是f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=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/4)cos(x-π/4)
=cos(2x-π/3)+sin[2(x-π/4)]
=cos2xcosπ/3+sin2xsinπ/3+sin(2x-π/2)
=cos2x*1/2+sin2xsinπ/3-cos2x
=sin2xsinπ/3-cos2x*1/2
=sin2xsinπ/3-cos2x*cosπ/3
=-cos(2x+π/3)
f(sin x)=3-cos 2x,求f(cos x)=?
设f(sin x)=3-cos 2x,则f(cos x)=?
若F(sin x)=3-cos 2x 则F(cos x)=?
f(sin x)=3-cos 2x求f(cos x)
设f(x)=4cos(ωx-π/6)sinωx-cos(2ωx+π),其中ω>0,求函数y=f(x)的值域,请看问题补充f(x)=4cos(ωx-π/6)sinωx-cos(2ωx+π) =4(coswxcosπ/6+sinwxsinπ6)sinwx+cos2wx =2√3sinwxcoswx+2sin²wx+cos2wx =√3si
f(x)为奇函数,x>0,f(x)=sin 2x+cos x,则x
证明f(x)=cos^2x+cos^2(x+∏/3)+cos^2(x-∏/3)是常数函数
化简f(x)=2cos^3 x+sin^2 (360-x)-cos(180-x)-3 / 2+2cos^2(180+x)+cos(-x)求化简图片
化简f(x)=cos(2x-π/3)-cos2x
h(x)=根号 2x=1 / 3x f(x)=cos²x+cos(x^2) 的一阶导数 h(x)=根号 2x+1 / 3xf(x)=cos²x+cos(x^2)
f(x)=cos(2x)+2sin(x)化简
若f(Sinx)=3-COS^2X求f(COS^X)表达式
f'(sin^2x)=cos^2x,求f(x)f'((sin^x)^2)=(cos^x)^2
一道 三角题 急求解题过程设f(x)满足2f(sin x)+3f(cos x)=2cos x,则f(x)=______.
f(cos)=-cos(2x),则f(sinx)
f(x)=2(cosx)^3+sin(360°-x)^2-cos(180°-x)-3/2+2cos(180°+x)+cos(-x)求f(π/3)
f ' (sinx)=cos^2x,求f(x)
f(sinx)=1+cos(2x),求f(x),