求函数f(x)=2sin(x π 6)-2cosx的最大值

∵f（x）＝cos（2x－π/3）＋（sinx）^2－（cosx）^2＝cos（2x－π/3）－cos2x＝2sin（2x－π/6）sin（π/6）＝sin（2x－π/6）.∴g（x）＝［sin（2x

1、由于函数g(x)=sin(2(x－a)+π/3)为偶函数,所以g(x)的图像关于y轴对称,即函数g(x)当x=0时取得最值,所以g(0)=±1,解得sin(π/3－2a)=±1,sin(2a－π/

∵f(x)=2sin(π-x)cosx=2sinxcosx=sin2x1、最小正周期T=2π/2=π.2、∵-π/6≤x≤π/2∴-π/3≤2x≤π,∴-√3/2≤f(x)≤1,∴最大值1,最小值-√

f(x)=sin(2x+π/2)=cos2xg(x)=f(x)+f(π/4-x)=cos2x+cos(π/2-2x)=cos2x+sin2x=√2sin(2x+π/4)单增区间2x+π/4∈[2kπ-

你啊,要好好学习了!还没有悬赏分?把对称轴即x=∏／8代入原式子,即sin(∏／4+φ)=1或者-1,再用(-π

f(x)=cosx＋sinxf(x)=√2sin(x＋π/4)(1)递增区间：2kπ－π/2≤x＋π/4≤2kπ＋π/2得：2kπ－3/4π≤x≤2kπ＋π/4递增区间是：[2kπ－3π/4,2kπ＋

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin^2x=[sinx+√3cosx+sinx]cosx-√3sin^2x=2sinxcosx+√3cos^2x-√3sin^2x=sin

(1)f(x)=cos(-x/2)+sin(π-x/2)=cos(x/2)+sin(x/2)=√2[(√2/2)cos(x/2)+(√2/2)sin(x/2)]=√2[sin(π/4)cos(x/2)

f(x)=sin(π-x)sin(π/2-x)+cos²xf(x)=sinxcosx+cos²x=1/2sin2x+1/2+1/2cos2x=1/2(sin2x+cos2x)+1/

首先：定义域只有这一个,X+π/4≠2Kπ,所以X≠-π/4+2kπ..附上值域,化简原函数：f（X）=cos2X/[√2/2（sinX+cosX）]f(x)=(cos²X-sin²

因为f(x)=sinx+cosx=√2sin（x+π/4）第一题T=2π/1=2π第二题当sin（x+π/4）=1时,为最大值,即f(x)=√2sin（x+π/4）=-1时,为最小值,即f(x)=-√

因为f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)=根号3sin(2x-π/6)-（1-2sin的平方（x-π/12)）+1=根号3sin(2x-π/6)-cos(2x-π/6

f(x)=sin2(x+π)+根号3sin(x+π)sin(π-x)-1\2=sin2x－根号3sin²x-1/2=sin2x＋根号3/2cos2x－1=根号7/2sin（2x＋γ）－1co

f(x)=[2sin(x+π/3)+sinx]cosx-根号3sin^2x和差角公式展开f(x)=(sinx+根号3cosx+sinx)cosx-根号3sin2x=2sinx*cosx+根号3cosx

f(x)=2cosx*(1/2*sinx+√3/2*cosx)-√3sin²x+sinxcosx=sinxcosx+√3*cos²x-√3sin²x+sinxcosx=2

f(x)=根号2sin(π/2-x)sin(x+4/π)-1/2=√2cosx(√2/2×sinx+√2/2×cosx)-1/2=sinxcosx+cos²x-1/2=1/2sin2x+1/

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

f(x)=sin^2x+sinxcosx-sin^2x+cos^2x=sinxcosx+cos^2x=sin2x/2+(1+cos2x)/2=sin2x/2+cos2x/2+1/2(1)f(a)=si

sin(x+π/4)=sin(x-π/4+π/2)=cos(x-π/4)所以原式=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)=cos(2x-π/3)+sin(2x-π/2)=1

sin(___)的最大值是1,所以f(x)的最大值是2.