函数f(x)=2sin(2x π 3),g(x)=mcos(2x-
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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-√
(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间:π/2+2
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(π-x)sin(π/2-x)=2sinxcosx=sin2x1)最小正周期=2π/2=π2)在区间[-派/6,派/2]上x=π/4时,有最大值=sinπ/2=1x=-π/6时,有最
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)=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)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T
fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3(cos*2x-sin*2x)=sin2x+根号3cos2x=2sin(2x+派
f(x)=cos(-x/2)+sin(π-x/2)=cosx/2+sinx/2f(a)=cos(a/2)+sin(a/2)=(2√10)/5cos(a/2)+sin(a/2)=(2√10)/5平方1+
(Ⅰ)f(x)=sinx•cosx+12cos2x+12=12sin2x+12cos2x+12=22sin(2x+π4)+12∴函数f(x)的最小正周期T=2π2=π(Ⅱ)当x∈[−π8,3π8]时,
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)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+
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)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-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
由1,3作为条件,可以得到2,由2,3作为条件,可以得到1,由1,3得到2,证明:由3可知w=2或-2,设定w=2时,由1可以得到2*π/12+t=kπ/2,k为不等于0的整数.得到t=kπ/2-π/