cosx大于2分之根号3
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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/
解析sinx+cosx=√2/2(sinx+cosx)2=1/2(sinx+cosx)2=1+sin2x=1/2sin2x=-1/2(sinx-cosx)2=1-sin2x=3/2sinx-cosx=
f(x)=sinxcosx-(√3/2)cos²x+(√3/2)=(1/2)sin2x-(√3/2)cos2x=sin(2x-π/3).最小正周期是T=2π/2=π
[2cos^2(x/2)]/[2cosx(-1/2-cos2x)]/[cosx(-1/2-cos2x)]-sinx-1=4cos^2(x/2)/[cos^2x(1+2cos2x)^2]-sinx-1c
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+
二分之一cosx-(根号下3分之2)sinx=√33/6sin(x-φ)其中φ=arctan(√6/4)
要详细过程,很急,明天要交,好心人帮帮忙~问题补充:求函数f(x)的最f(x)=2sinxcosx/2+√3[(cosx)^2+(√3/2)cos2x+1/2sin2x-√3/2=
y=√3/2sinx+1/2cosx+1=sin(x+30°)+1≥2所以最小值y=2
要证2根X>3/X(X>1)吗?这很显然嘛,该函数递增,大于F(1)
值域为(1,2],当x=6分之π是取最大值为2,当x=-6分之π或2分之π时取最小值为1..注意:1是开区间,2是闭区间.楼主不厚道,居然不给悬赏分!让我重新复习这么久之前的东西,我容易吗,学过数学的
a>0,b>0所以原式=√(ab)√b²-2√(ab/b²)=b√(ab)-(2/b)√(ab)=[(b²-2)/b]√(ab)
问题说清楚了么?2根号x大于3减负x分之1,x^(1/2)>3+1/x?
题主,你是不是把题目没有看全呀,我的肯定对着了.祝你心想事成再问:好吧
f(x)=向量om*向量on=2cos^2x+2倍的根号3(1/2*sin2x+根号3分之6a)=2cos^2+根号3*sin2x+12a=cos2x+根号3*sin2x+12a+1=2sin(2x+
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
可以这样考虑1/(根号n)=2/[2(根号n)]
1.原式=3√10(√2/2*sinx+√2/2cosx)=3√10(sinxcosπ/4+cosxsinπ/4)=3√10sin(x+π/4)2.2分之3cosx-2分之跟号3sinx=根号3(根号
想成整理后得f(x)=sin2x+√3cos2x=2sin(2x+3/π),周期为π单调增区间,2x+3/π∈[-π/2+2kπ,π/2+2kπ],x∈[-5π/12+kπ,π/12+kπ]单调减区间
(1)f(x)=sin(2x+π/6)+√3最小正周期为π,对称轴方程x=kπ+π/6,x=kπ-π/3(2)解析:f(-5π/12)=-sin(4π/6)+√3=√3/2,f(π/12)=sin(π