函数f(x)=sin(2x-1 4π)-2√2(sinx的平方)的最小正周期

①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c

f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T

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-√

先用tanx=sinx/cosx、倍角公式、诱导公式化简原函数：f（x）=sin²x+sinxcosx-sin[2（x+π/4）]=（1-cos2x）/2+1/2sin2x-sin（2x+π

f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,

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

f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x=2sin2xc

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

答：f(x)=(cosx)^2-√3sinxcosx+2(sinx)^2-1/2f(x)=(1/2)*cos(2x)-(√3/2)sin2x+1-cos(2x)f(x)=-(√3/2)sin2x-(1

f(x)=sin2x＋cos2x－1=√2sin(2x＋π/4)－1.1、最小正周期是π,最大值时2x＋π/4=2kπ＋π/2,即x=kπ＋π/4,k是整数.再问：已知函数f(x)=2sin(∏-X)

f(x)=2sinx*sin(π/2+x)-2sin^2x+1＝2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)因为f(x0/2)=根2/3所以sin(x0+π/4)

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)=cos(x-π/3)-sin(π/2-x)=(1/2)cosx+(√3/2)sinx-cosx=(√3/2)sinx-(1/2)cosx=sin(x-π/6）,它的最小值=-1.

http://zhidao.baidu.com/question/7920192.html这里有所有的需要的公式.我帮你算了半天,没算完,估计给你这些公式你自己算好一些.

f(x)=2cosx*sinx-2cosx^2+1f(x)=sin2x-cos2xf(x)=根号2*sin(2x-45)周期T=π

f（x）=(1+cotx)sinx^2-2sin(x+π/4)sin(x-π/4)=（1+cosx/sinx)*sinx^2-(sinx^2-cosx^2)=cosx^2+sinxcosx1、若tan