题目答案

参考题1、（1）f(x)=A[1-cos(2ωx+2φ)+/2=A[-1/2*cos(2ωx+2φ)+1/2+最大=A(1/2+1/2)=A=2所以f(x)=-cos(2ωx+2φ)+1相邻两对称轴间距是半个周期所以T/2=2T=4T=2π/2ω=4ω=π/4所以f(x)=-cos(πx/2+2φ)+1把点代入2=-cos(π/2+2φ)+1cos(π/2+2φ)=-1-sin(2φ)=-1sin(2φ)=12φ=π/2，φ=π/4（2）易知A=2,f(x)=2sin^2(wx+r)=2*0.5*(1-cos(2wx+2r)),周期为4，即每四个的和为4，f(1)+…+f(2008)=20082、y=2cos(x+π/4)*cos(x-π/4)+(√3)sin2x=2sin(π/4-x)*cos(π/4-x)+(√3)sin2x=sin(π/2-2x)+(√3)sin2x=cos2x+(√3)sin2x=2[cosπ/3cos2x+sinπ/3sin2x]=2cos(2x-π/3)所以,值域[-2,2],周期为π,3、4、解：(1)因为函数图象过点(0,1),所以2sinφ=1,即sinφ=12.因为0≤φ≤,所以φ=.(2)由函数y=2sin(πx+)及其图象,得M(-16,0),P(13,2),N(56,0).所以=(-12,-2), =(12,-2).从而cos〈,〉==,故〈,〉=arccos.5、（1）cosx≠0.x≠kπ+π/2.k∈Z.∴f(x)的定义域为{x│x≠kπ+π/2.k∈Z（2）α是第四象限的角，tanα=-4/3.sinα=-4/5, cosα=3/5.sin2α=2sinαcosα=-24/25.cos2α=2cos^2-1=-7/25.√2sin(2x-π/4）=√2(sin2xcosπ/4-cos2xsinπ/4)=sin2x-cos2x=-17/25.∴f(α)=1-(-17/25)/(3/5)=32/15.6、解：（1）当f(x)=(1-cos2x0/2+sin2x+3(1+cos2x)/2=1+sin2x+cos2x=2+sqr2sin(2x+pai/4)当2x+pai/4=2kpai+pai/2,即x=kpai+pai/8 时, f(x) 取得最大值2+sqr2 .函数f(x) 的取得最大值的自变量x 的集合为{x=kpai+pai/8 k∈Z}（2）*kπ-π/8, kπ+π/8]7、8、cosA+2cosB+C/2=cosA+2sinA/2=1-sin^2(A/2)+sinA/2=-[sin(A/2)-1/2]^2+5/4sin(A/2)=1/2最大A=60９、（１）f(x)=向量a*(b+c)由题f(x)=(sinx,-cosx)*(sinx-cosx,-3cosx+sinx)f(x)=sinx(sinx-cosx)-cosx(-3cosx+sinx)=sinxsinx-sinxcosx+3cosxcosx-sinxcosx=sinxsins+3cosxcosx-2sinxcosx=sinxsinx+cosxcosx+2cosxcosx-2sinxcosx=cos2x-sin2x=根号2/2 sin(2x+45度)最大值2/2，最小正周期π（２）f(x)=a*(b+c)=(sinx,-cosx)（sinx-cosx,3cosx-sinx）=1-2sinx cosx+2cos^2x=-(sin2x-cos2x)+2 =-根号2*sin(2x-π/4)+2 知f(x)是由-sin2x 先向右平移π/8，再向上平移2得到的，所以逆着可知d=(-π/8,-2)１０、（１）a⊥b，则a*b=sina＋cosa=0即：tana=sina/cosa=－1因－π/2<a<π/2，则：a=－π/4（２）|a＋b|²=|a|²＋2a*b＋|b|²=(sin²a＋1)＋2(sina＋cosa)＋(1＋cos²a)=3＋2(sina＋cosa)=3＋2√2sin(a＋π/4)因－π/2<a<π/2，则sin(a＋π/4)的最大值是1则：|a＋b|²的最大值是3＋2√2则：|a＋b|的最大值是√2＋111、（１）根据余弦定理c^2= a^2 + b^2- 2abcosC =1+4-2*1*2*3/4=2∴AB=c=√2（2）∵cosC=3/4∴sinC=√（1-cos²C)=√7/4正弦定理a/sinA=c/sinC,∴sinA=asinC/c=(√7/4)/√2=√14/8∵a<b ∴A是锐角∴cosA=√（1-sin²A)=5√2 /8∴sin2A=2sinAcosA=2×√14/8×5√2/8=5√7/16cos2A=2cos²A-1=9/16sin(2A+C)=sin2AcosC+sinCcos2A=5√7/16×3/4+√7/4×9/16=3√7/812、（1）.求Am·n=(-1)*cosA+√3*sinA=√3sinA-cosA=[(√3)^+1^]*sin[A-arctan(1/√3)]=2sin(A-π/6)已知m·n=1<=>sin(A-π/6)=1/2∵A为△ABC的内角∴0<A<π<=>-π/6<A-π/6<5π/6结合基本正弦函数y=sinx的图像，当自变量x∈(-π/6,5π/6)时，能使y=sinx=1/2的x值只有一个：x=π/6∴A-π/6=π/6<=>A=π/3（2）.由已知：(1+sin2B)/(cos^B*sin^B)=3<=>1+sin2B=3*sin^B*cos^B=(3/4)*4*sin^B*cos^B=(3/4)*(2sinBcosB)^=(3/4)*sin^2B<>3sin^2B-4sin2B-4=0<=>(3sin2B+2)*(sin2B-2)=0<=>sin2B=-2/3设tanB=t,由万能公式，有：sin2B=2t/(1+t^)=-2/3<=>t^+3t+1=0<=>t=(-3/2)±√5/2即：tanB1=(√5-3)/2≈-0.38或tanB=-(√5+3)/2≈-2.62１３、（１）f(x)=2 (sin2xcos π/6 + cos2xsin π/6) （正弦两角和）=2*(2sinxcosx)* (√3/2) + 2*(cos^2 x- sin^2 x)* (1/2）(cos2x，sin2x 二倍角展开,cosπ/6,sinπ/6值代入）X∈（0,π/2）所以cosx>0sinx = 4/5. cosx = √(1-sin平方x) = 3/5代入上面的f(x) = √3 * 2* 4/5* 3/5 + 9/25- 16/25 = （24√3-7）/25 赞同0| 评论（２）先展开f（x）=2sin（x+π/6）-2cosx=2sinxcosπ/6+2cosxsinπ/6-2cosx=2sinxcosπ/6-2cosxsin π/6=2sin（x-π/6）当sinx=4/5时，cosx=3/5，代入f（x）=2sinxcosπ/6-2cosxsinπ/6即可至于值域也很简单x∈[π/2，π] 时x-π/6∈[π/3，5π/6]f(x)∈[-根号3，2]14、（1）由cos2α=1-2sin²α，把2sin^2(x-π/12）表示成1-cos2(x-π/12)=1-cos(2x-π/6)，这样f(x)=根号3sin(2x-6/π)-cos(2x-π/6）+1.然后用辅助角公式把sinα和cosα结合起来就可以了。

f(x)=2cos2x+1.所以最小正周期是π，（2）当2x-π/3=2kπ+π/2,即x=kπ+5π/12时,f(x)max=3１５、tanα+cotα=-10/33(tanα)^2+10tanα+3=0(3tanα+1)(tanα+3)=0tanα=-1/3,或tanα=-3而：3π/4<α<π0<π-α<π/4tanα=-tan(π-α)>-1所以：tanα=-3（２）16、１７、（１）a,b,c成等比数列,所以a*c=b^2根据正弦定理，a/sinA=b/sinB=c/sinC所以sinA=a/b*sinB,sinC=c/b*sinC1/tanA+1/tanC=cotA+cotC=cosA/sinA+cosC/sinC=(cosA*sinC+sinA*cosC)/sinA*sinC=sin(A+C)/[(a/b*sinB)*(c/b*sinC)]=sinB/[(a/b*sinB)*(c/b*sinC)]=1/sinB=4/(根号7）（２）a,b,c成等比数列,设公比为q,则b=a*q,c=a*q^2cosB=(a^2+c^2-b^2)/2*a*c=(a^2+a^2*q^4-a^2*q^2)/2*a*a*q^2=(1+q^4-q^2)/2*q^2=3/4化简为：2*q^4-5*q^2+2=0解得：q=1/（根号2）,或者q=根号2向量BA点乘向量BC＝a*c*cosB=a*a*q^2*cosB=3/2将cosB和q代入，解得：a=2，此时q=1/（根号2）,c=1,a+c=3或者a=1，此时q=根号2,c=2,a+c=3综合以上a+c=3１８、cos*(6k+1)π/3+2x+=cos*2kπ+π/3+2x+=cos*π/3+2x+cos[(6k-1)π/3-2x+=cos*2kπ-π/3-2x+=cos*π/3+2x+那么原式=2cos[π/3+2x]+2√3sin(π/6-2x)2√3sin(π/6-2x)=2√3sin[π/2-(π/3+2x)]=2√3cos(π/3+2x)即可求出函数f(x)的值域和最小正周期１９、tan(α+π/4)=(tanα+tanπ/4)/(1-tanαtanπ/4)=(tanα+1)/(1-tanα)=,2(tanα/2)/*1-(tanα/2)^2++1-/,1-2(tanα/2)/*1-(tanα/2)^2+-={2*2/[1-2^2]+1}/{1-2*2/[1-2^2]={-4/3+1]/{1+4/3]=(-1/3)/(7/3)=-1/7（２）(6sinα+cosα)/(3sinα-2cosα)上下同除cosα得(6sinα+cosα)/(3sinα-2cosα)=(6tanα+1)/(3tanα-2)=(6*(-4/3)+1)/(3*(-4/3)-2)=7/6２０、sin（α-π/4）=√2/2(sinα-cosα)=7√2/10sinα-cosα=7/5 (1)cos2α=-(sinα+cosα)(sinα-cosα)=7/25 (2)由（1）（2）解得sinα=3/5,cosα=-4/5第二问直接计算就行２１、余弦定理cosA=(b^2+c^2-a^2)/2bc=1/2 A=60°正弦定理c/b=sinC/sinB=sin(120°-B)/sinB=(√3/2cosB-1/2sinB)/sinB=(√3/2)/tanB-1/2=1/2+√3√3/2tanB=1+√3tanB=√3/2(1+√3)=(3-√3)/4２２、f(25π/6)=sin(2*25π/6-π/3)+√3/2=sin(8π+π/3-π/3)+√3/2=sin0+√3/2=√3/2设a∈(0,π)f(a/2)= sin( a+π/3) -(√3)/2＝1/4-(√3)/2sin( a+π/3)＝1/4<(√3)/2=sinπ/3a+π/3∈（π/3，2π/3）∴a+π/3=π-arc sin(1/4)a=2π/3-arc sin(1/4)sina=sin(2π/3-arc sin(1/4))=sin(π/3+ ar c sin(1/4)=(√3)/2*√15/4+1/2*1/4=1/8*(3√5+1)23、（1）.f(π/4)=√2sin(2×π/4+π/4)=√2×√2/2=1（2）.由题意，sin(a+π/4)=1/2所以a=7π/12所以sina=(√6+√2)/4２４、f(x)=2cosx/2×(√2sin(x/2+π/4）+ tan(x/2+π/4)×tan(x/2－π/4))=√2[sin(x+π/4)+sin(π/4)] + [1+tan(x/2)]/[1－tan(x/2)]×[tan(x/2)－1]/[1+tan(x/2)] =√2sin(x+π/4)最大值=√2最小正周期=2πsinx的增区间是：-π/2+2kπ≤x≤π/2+2kπ带入-π/2+2kπ≤x+π/4≤π/2+2kπ所以增区间-3π/4+2kπ≤x≤π/4+2kπ减区间同理25、sinAsinB+sinAcosB=sinC=sin(pi-A-B)=sin(A+B)=sinAcosB+sinBcosA那么sinAsinB=sinBcosAsinB不可能为0,所以sinA=cosA若cosA=0则sinA不为0,矛盾!所以cosA不为0所以tanA=1得到A=45SinB+Cos2C=0sinB=cos(180-2c)=sin(2C-90)sinB>0,所以sin(2C-90)>0,故2C>90,C>45因为sinB=cos(180-2c)=sin(2C-90)所以B=2c-90或B=180-(2C-90)又B+C=180-45=135分别解出B=60,C=75B=0,C=135(舍)综上A=45,B=60,C=75２６、设高为hh=√[54-(cosC*3√6)²]=√52S=1/2*tgB+1/2*3√6/cosC*√52=1/6+√26=﹙1+6√26﹚／6２７、延长BD至E，使BD=DE，连结AE 。