函数f(x)=4sin(2x π 6)图像的一个对称中心
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/25 06:33:18
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)=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+
2sin(x-π/4)sin(x+π/4)=cos(x-π/4-x-π/4)-cos(x-π/4+x+π/4)=-cos2xf(x)=cos(2x-π/3)-cos2x=cos(2x-π/6-π/6)
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)=-√
f(-x)=f(x)所以sin(-2x+a)=sin(2x+a)所以-2x+a=2kπ+2x+a或2x+a=2kπ+π-(2x+a)这是恒等式而-2x+a=2kπ+2x+a,2kπ+4x=0不是恒等式
f(x)=sin(x/2)-√3[1-cos(x/2)]+√3=2[(1/2)sin(x/2)+(√3/2)cos(x/2)]=2sin(x/2+π/3)(1)g(x)=f(x+π/3)=2sin[(
cos2x=sin(π/2-2x)=2sin(π/4-x)cos(π/4-x)cos2x/[sin(π/4-x)]=2sin(π/4-x)cos(π/4-x)/[sin(π/4-x)]=2cos(π/
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)=sin2(x+y/2)由于sin2x对称轴为π/4+kπ/2;故x+y/2=π/4+kπ/2x=π/4+kπ/2-y/2;将x=x=π/8代入,得y=π/4+kπ,根据y的范围可知:y=-3
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
1)f(x)=sin(2x+φ)一条对称轴是X=π/8则kπ+π/2=2*π/8+φ===>φ=kπ+π/4因为-π
1.由f(x)=sin(2x+φ)一条对称轴是直线x=π/2可得:在x=π/2时,函数取极值.则2*π/2+φ=kπ+π/2(k∈Z)φ=kπ-π/2又-π