f(π/12)
=cos²(π/12-π/6)-sin²(π/12)
=cos²(-π/12)-sin²(π/12)
=cos²(π/12)-sin²(π/12) 两倍角公式
=cos[2(π/12)]
=cosπ/6
=√3/2
f(x)=cos^2(x-π/6)-sin^2x
=（2cos^2(x-π/6)-1+1）/2+(1-2sin^2x-1)/2
=[cos(2x-π/3)+cos2x]/2
=(cos2xcosπ/3+sin2xsinπ/3+cos2x)/2
=根号3（2分之根号3倍的cos2x+2分之sin2x)/2
=根号3/2sin(2x+π/3)
0<=x<=Pai/2,则有Pai/3<=2x+Pai/3<=4Pai/3
故有-根号3/2<=sin(2x+Pai/3)<=1
故最大值是:根号3/2*1=根号3/2.
已知函数f(x)=cos^2(x-π\/6)-sin^2x
=cos²(-π\/12)-sin²(π\/12)=cos²(π\/12)-sin²(π\/12) 两倍角公式 =cos[2(π\/12)]=cosπ\/6 =√3\/2 f(x)=cos^2(x-π\/6)-sin^2x =（2cos^2(x-π\/6)-1+1）\/2+(1-2sin^2x-1)\/2 =[cos(2x-π\/3)+cos2x]\/2 =(cos2xcosπ\/3+sin...
设函数f(x)=2cos^2(x+π\/6)-cos^2x
1)f(x)=1+cos(2x+π\/3)-(1+cos2x)\/2=1\/2-sin2x根号3\/2 最小值1\/2-根号3\/2 最小正周期π 2）c带入得sinC=根号3\/2 C=π\/3 A=π-B-C=2π\/3-arccosB
已知函数fx=cos^2(x-pai\/6)-sin^2
(2)f(x) =[cos(x-π\/6)]^2 - (sinx)^2 = [(√3\/2)cosx + (1\/2)sinx]^2 - (sinx)^2 = (1\/4)( 3(cosx)^2 + 2√3sinxcosx -3(sinx)^2 )= (1\/4) ( 3cos2x + √3sin2x )= (√3\/2)((√3\/2)cos2x+ (1\/2)sin2x)= (√3\/2)sin(2x+π\/3)max f(...
已知函数f(x)=cosxsin(x-π\/6)-sin平方x求f(x)的最小正周期
f(x)=cosxsin(x-π\/6)-sin平方x =cosx(√3\/2*sinx-1\/2*cosx)-(sinx^2）=√3\/2*sinxcosx-1\/2*(cosx)^2-[1-(cosx)^2]=√3\/4*sin2x+1\/4*(1+cos2x)-1 =√3\/4*sin2x+1\/4*cos2x)-3\/4 =1\/2*sin(2x+π\/6)-3\/4 最小正周期T=2π\/2=π ...
已知函数f(x)=cos^2x-sin^2x
解:1.∵f(x)=cos^2x-sin^2x=cos2x ∴f(x)最小正周期T=2π\/2=π 2.∵f(x)=cos2x在[0，π\/2]上单调递减 ∴f(x)的最大植为f(0)=cos0=1 f(x)的最小植为f(π\/2)=cosπ=-1
已知函数f(x)=sincos(x+π\/6)-sin∧2x
答：f(x)=sinxcos(x+π\/6)-(sinx)^2 f(x)=sinx*[ cosxcos(π\/6)-sinxsin(π\/6) ]-(sinx)^2 f(x)=(√3\/2)sinxcosx-(1\/2)(sinx)^2-(sinx)^2 f(x)=(√3\/4)sin2x-(3\/2)(sinx)^2 f(x)=(√3\/4)sin2x-(3\/4)(1-cos2x)f(x)=(√3\/2)*[(1\/2)sin2x+(...
已知函数f(x)=cos²(x-π\/6)-sin²x,求f(x)的最大值
f(x)＝cos²(x-π\/6)-sin²x ＝[1+cos(2x-π\/3)]\/2-(1-cos2x)\/2 ＝[cos(2x-π\/3)+cos2x]\/2 ＝cos(2x-π\/6)cos(-π\/6)＝(1\/2)cos(2x-π\/6).∴cos(2x-π\/6)＝1,即2x-π\/6＝2kπ→x＝kπ+π\/12时，所求最大值：f(x)|max＝1\/2。
已知函数f(x)= cos^2x+ sin^2x,求f(x)
∫cosx\/(sinx+cosx) dx=(1\/2)(x+ln|sinx+cosx|) + C。（C为积分常数）解答过程如下：∫cosx\/(sinx+cosx) dx = (1\/2)∫[(cosx+sinx)+(cosx-sinx)]\/(sinx+cos)] dx = (1\/2)∫ dx + (1\/2)∫(cosx-sinx)\/(sinx+cosx) dx = x\/2 + (1\/2)∫d(sinx+cosx)\/(sinx+...
f(x)=cos方(x+π\/6)+1\/2cos(2x-π\/6)
=-1\/2(1\/2cos2x+√3\/2sin2x)=-1\/2sin(2x+π\/6)所以,最小正周期为：T=π；因为sin2x的对称轴为：x=π\/4+kπ\/2 所以f(x)图像对称轴的方程为：x=(π\/4-π\/6)+kπ\/2=π\/12+kπ\/2 (2) 因为 -π\/12≤x≦π\/12 所以-π\/6≤2x≤π\/6 所以0≦2x+π\/6≦π\/3 由正弦...
已知函数f(x)=sin(2x-π\/6)-2cos^2x-1(x∈R),(1)求函数f(x)的单调递...
见图 解：（I）f（x）==sin2x+cos2x=sin（2x+）．令 2kπ-≤（2x+）≤2kπ+，可得 kπ-≤x≤kπ+，k∈z．即f（x）的单调递增区间为[kπ-，kπ+]，k∈z．（II）在△ABC中，由，可得sin（2A+）=，∵＜2A+＜2π+，∴＜2A+= 或，∴A= （或A=0 舍去）．∵b，a，c...
