f x sinx 根号3cosx
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(3/2)cosx-(√3/2)sinx=√3[(√3/2)cosx-(1/2)sinx]=√3[sin(π/3)cosx-cos(π/3)sinx]=√3sin(π/3-x)=-√3sin(x-π/
/>tanx=√3则x=kπ+π/3,k∈Z(1)k是偶数,则sinx=√3/2,cosx=1/2(2)k是奇数则sinx=√3/2,cosx=1/2(sinx+cosx)/(sinx-cosx)的值
用sinx2+cosx2=1再问:可以帮我把顺序写出吗。再问:谢谢。再答:就是3cos2x+cos2x=1再答:然后cosx=正负0.5就行了再问:就只有后面这两个步骤吧再问:还是前面那个也要再问:s
sinx+根号3cosx=22(sinx/2+根号3cosx/2)=2sinx/2+根号3cosx/2=1sinxcosπ/3+cosxsinπ/3=1sin(x+π/3)=1所以x+π/3=2kπ+
(1)3根号15sinx+3根号5cosx=6根号5((根号3)/2sinx+1/2cosx)=6根号5(cos10°sinx+sin30°cosx)=(6根号5)sin(x+30°)(2)根号3si
sinx+cosx=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)sinx-cosx=√2(sinxcosπ/4-cosxsinπ/4)=√2sin(x-π/4)sinx
=2*(根号3sinx/2+cosx/2)=2(sinx*cos30°+cosxsin30°)=2sin(x+30°)
√3sinx-cosx=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)
=(sinx)^2+√3sinxcosx=(1-cos2x)/2+√3/2(2sinxcosx)=1/2-1/2cosx+√3/2sin2x=1/2+sin(-π/6)cos2x+cos(-π/6)s
y=sinxcosx-√3(cosx)^2+√3/2=1/2*sin2x-√3*[(1+cos2x)/2]+√3/2=(sin2x)/2-(√3/2)*cos2x-√3/2+√3/2=√[(1/2)^
(1)解析:函数f(x)=√3cos²x+sinxcosx=√3(cos2x+1)/2+1/2sin2x=sin(2x+π/6)+√3则函数的最小正周期为π,图像的对称轴方程x=kπ+π/6
π/6+2kπ(k属于Z)
y=√3cosx+sinx=2[(√3/2)cosx+(1/2)sinx]=2(sinπ/3cosx+cosπ/3sinx)=2sin(x+π/3)所以它的最小正周期=2π;图像的对称轴为:x=kπ+
不用画图值域就是[-2,2]因为y=cosx的值域是[-1,1]
y=cosx(sinx-√3cosx)+√3/2=cosxsinx-√3(cosx)^2+√3/2=sin2x/2-√3cos2x/2=sin(2x-π/3)当x=-π/2时y=sin(-π-π/3)
Y=sinx*cosx*√3+cos^2x=√3/2*sin2x+1/2*cos2x+1/2=sin(2x+∏/6)+1/2.
不管怎么写,都是有意义的.它只是函数,不存在成立不成立的问题.
------a^2+c^2-b^2a^2+c^2-b^2a^2+c^2-accosB=-------------=-------------=-------------2ac2ac2ac2ac-ac1
没错,f(x)=2sin(2x+π/6)周期T=2π/2=π因为-1≤sin(2x+π/6)≤1f(x)max=2f(x)min=-2
cosx=根号3/2时x=2kπ+π/6或x=2kπ-π/6,k为整数所以不等式cosx>根号3/2解为:2kπ-π/6