sin(π 6-x)=1 3
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(Ⅰ)f(x)=sin(x+π6)-cos(x+π3)+cosx=32sinx+12cosx-(12cosx-32sinx)+cosx=3sinx+cosx=2sin(x+π6),∵ω=1,∴T=2π
(Ⅰ)∵sin(2x+π6)=sin2xcosπ6+cos2xsinπ6,sin(2x−π6)=sin2xcosπ6−cos2xsinπ6,cos2x=12(cos2x+1)∴f(x)=sin(2x+
由于f(3α+π2)=2sinα=1013,∴sinα=513,再由α∈[0,π2],可得cosα=1213.再由f(3β+2π)=2sin(β+π2)=2cosβ=65,∴cosβ=35,再由β∈[
在同一个坐标系中作出y=sinπx和y=14x的图象,如图所示:由于y=sinπx和y=14x的图象在[-4,4]上有7个交点,而当x<-4,或 x>4时,两个曲线不会有交点,方程sinπx
sin(PI/6+2x)=cos(PI/2-PI/6-2x)=cos(PI/3-2x)=cos(2*(PI/6-x))=1-2*sin(PI/6-x)^2=1-2*(1/4)^2=7/8tan70*c
2sinπ/6x=1-cosπ/3xf(x)=2sinπ/6x+sin(π/3x+π/6)—1=(1-cosπ/3x)+sinπ/3xcosπ/6+cosπ/3xsinπ/6-1=√3/2sinπx/
(1)∵f(x)=sin(2x+π6)+sin(2x−π6)+2cos2x=sin2xcosπ6+cos2xsinπ6+sin2xcosπ6−cos2xsinπ6+2cos2x+1=3sin2x+co
(1)∵f(x)=sin(2x-π6),∴最小正周期T=2π2=π;…(3分)f(0)=sin(-π6)=-12,…(6分)(3)由f(α+π3)=35得sin(2α+π2)=35,…(7分),∴co
y=sin(x+π3)sin(x+π2)=(sinxcosπ3+cosxsinπ3)cosx=12sinxcosx+32cos2x=14sin2x+32•1+cos2x2=34+12sin(2x+π3
(1)f(0)=2sin(−π6)=−1…(3分)(2)f(3α+π2)=2sin[13(3α+π2)−π6]=2sinα=1013,即sinα=513…(5分)f(3β+2π)=2sin[13(3β
sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π
f(x)=√3sin(2x-π/6)+2sin^2(x-π/12)=√3sin(2x-π/6)-cos(2x-π/6)+1=2sin[(2x-π/6)-π/6]+1=2sin(2x-π/3)+1(1)
sin(x+11π/6)=sin(x+2π-π/6),因为2π是一个周期,可以消去,于是得到答案sin(x-π/6)
令2kπ-π2≤2x-π6≤2kπ+π2,k∈z,解得kπ-π6≤x≤kπ+π3,故函数y=2sin(2x−π6)的单调递增区间是[kπ-π6,kπ+π3],k∈z,故答案为[kπ-π6,kπ+π3]
把函数y=sin(2x−π6)的图象向左平移φ(φ>0)个单位得到函数y=sin(2(x-π12+φ))的图象,因为函数y=sin(2(x-π12+φ))为奇函数,故-π12+φ=kπ,故φ的最小值是
y=32sin2x+12cos2x+12cos2x−32sin2x=cos2x.所以函数的最大值为1故答案为:1
∵0≤x≤π2,∴π6≤x+π6≤2π3;∴当x+π6=π2时,函数取得最大值是y=sin(x+π6)=1;当x+π6=π6时,函数取得最小值是y=sin(x+π6)=12;∴函数y=sin(x+π6
∵sin(x+π6)=14∴sin(x−5π6)+ sin2(π3−x)=sin[π−(x+π6)]+sin2[π2−( π6+x)]=sin(x+π6)+cos2(x+π6)=1
sin(7派/6X)=1/4,cos(11派/6-X)=(根号下15)/4,这两式相加等于(1根号15)/4
f(x)=sin(2x+π/6)-cos2x+1所以为2π/2=πf(x)=根号3/2sin2x-(cos2x)/2+1=sin(2x-π/6)+1所以最大值为2,x=π/2+2kπ-π/6=π/3+