2025年高考數(shù)學(xué)模擬檢測(cè)卷（新高考題型專(zhuān)項(xiàng)解析試題）考試時(shí)間：______分鐘總分：______分姓名：______一、選擇題（本大題共12小題，每小題5分，共60分。在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。）1.函數(shù)f(x)=|x-1|+|x+2|在區(qū)間[-3,3]上的最小值是（）A.1B.2C.3D.4解析：哎呀，同學(xué)們，這道題考的是咱們函數(shù)中的絕對(duì)值函數(shù)，這個(gè)可是咱們高中數(shù)學(xué)的難點(diǎn)之一呢！你們想想，絕對(duì)值函數(shù)就像一條條折線(xiàn)，這里有兩個(gè)折點(diǎn)，分別是x=1和x=-2。咱們要在區(qū)間[-3,3]上找最小值，那就要分情況討論啦！當(dāng)x在[-3,-2]區(qū)間時(shí)，f(x)=-(x-1)-(x+2)=-2x-1；當(dāng)x在[-2,1]區(qū)間時(shí)，f(x)=-(x-1)+(x+2)=3；當(dāng)x在[1,3]區(qū)間時(shí)，f(x)=(x-1)+(x+2)=2x+1。你們看，最小值就在x=-2的時(shí)候，這時(shí)候f(x)=5，但是咱們要找的是整個(gè)區(qū)間上的最小值，所以還要比較這三個(gè)分段函數(shù)的最小值，發(fā)現(xiàn)當(dāng)x=1時(shí)，f(x)最小，為2。所以答案是B。2.若復(fù)數(shù)z滿(mǎn)足z^2=1+i，則z的模長(zhǎng)為（）A.√2B.2C.1D.√5解析：好，咱們來(lái)看第二題，這道題考的是復(fù)數(shù)，復(fù)數(shù)啊，可是咱們數(shù)學(xué)里的一大亮點(diǎn)呢！你們想想，復(fù)數(shù)就像一個(gè)平面上的點(diǎn)，既有實(shí)部又有虛部。這道題給出了z的平方等于1+i，那咱們?cè)趺辞髗的模長(zhǎng)呢？首先，咱們知道復(fù)數(shù)z的模長(zhǎng)等于它平方的模長(zhǎng)的平方根。所以咱們先求1+i的模長(zhǎng)，1+i的模長(zhǎng)就是√(1^2+1^2)=√2。所以z的模長(zhǎng)就是√(√2)=√2/√2=1。所以答案是C。3.已知等差數(shù)列{a_n}的首項(xiàng)為1，公差為2，則前n項(xiàng)和S_n等于（）A.n(n+1)B.n^2C.n(n+3)D.n^2+1解析：第三題，等差數(shù)列，這個(gè)可是咱們數(shù)列里的基礎(chǔ)呢！首項(xiàng)是1，公差是2，那咱們求前n項(xiàng)和S_n，直接用等差數(shù)列求和公式S_n=n(a_1+a_n)/2，但是咱們知道a_n=a_1+(n-1)d，所以a_n=1+(n-1)×2=2n-1。所以S_n=n(1+2n-1)/2=n(2n)/2=n^2。所以答案是B。4.圓x^2+y^2-4x+6y-3=0的圓心坐標(biāo)是（）A.(2,-3)B.(-2,3)C.(2,3)D.(-2,-3)解析：第四題，圓的方程，這個(gè)咱們也要掌握??！圓的一般方程是x^2+y^2+Dx+Ey+F=0，圓心坐標(biāo)是(-D/2,-E/2)。咱們這個(gè)方程是x^2+y^2-4x+6y-3=0，所以D=-4，E=6，圓心坐標(biāo)就是(4/2,-6/2)=(2,-3)。所以答案是A。5.函數(shù)f(x)=sin(x+π/4)在區(qū)間[0,π]上的圖像與x軸的交點(diǎn)個(gè)數(shù)為（）A.1B.2C.3D.4解析：第五題，三角函數(shù)，咱們要畫(huà)出圖像來(lái)輔助理解。sin(x+π/4)就是sin函數(shù)向左平移π/4，咱們知道sin函數(shù)在一個(gè)周期內(nèi)與x軸相交兩次，所以sin(x+π/4)在[0,π]上也會(huì)與x軸相交兩次。具體來(lái)說(shuō)，當(dāng)x+π/4=kπ時(shí)，sin(x+π/4)=0，解得x=kπ-π/4，在[0,π]上，k=1時(shí)，x=3π/4；k=2時(shí)，x=7π/4，但是7π/4已經(jīng)超出[0,π]的范圍了，所以只有x=3π/4這一個(gè)交點(diǎn)。所以答案是A。6.不等式|x|+|y|≤1表示的平面區(qū)域是（）A.一個(gè)圓B.一個(gè)正方形C.一個(gè)矩形D.一個(gè)三角形解析：第六題，絕對(duì)值不等式，這個(gè)咱們也要掌握??！|x|+|y|≤1表示的是以原點(diǎn)為中心，邊長(zhǎng)為√2的正方形內(nèi)部及邊界。你們可以想象一下，|x|+|y|=1是一個(gè)正方形，|x|+|y|≤1就是這個(gè)正方形內(nèi)部及邊界。所以答案是B。7.已知向量a=(1,2)，b=(3,-4)，則向量a×b等于（）A.-10B.10C.-14D.14解析：第七題，向量積，這個(gè)咱們也要掌握啊！向量a=(1,2)，b=(3,-4)，向量積a×b=1×(-4)-2×3=-4-6=-10。所以答案是A。8.函數(shù)f(x)=e^x-x在區(qū)間(-∞,+∞)上的單調(diào)性是（）A.單調(diào)遞增B.單調(diào)遞減C.先增后減D.先減后增解析：第八題，指數(shù)函數(shù)，咱們要利用導(dǎo)數(shù)來(lái)研究單調(diào)性。f(x)=e^x-x，f'(x)=e^x-1。當(dāng)x>0時(shí)，e^x>1，f'(x)>0，所以f(x)在(0,+∞)上單調(diào)遞增；當(dāng)x<0時(shí)，e^x<1，f'(x)<0，所以f(x)在(-∞,0)上單調(diào)遞減。所以答案是C。9.已知直線(xiàn)l1:ax+3y-6=0與直線(xiàn)l2:3x+by-9=0平行，則a的值為（）A.1B.2C.3D.4解析：第九題，直線(xiàn)方程，咱們要利用平行直線(xiàn)的性質(zhì)。兩條直線(xiàn)平行，它們的斜率相等。l1的斜率是-a/3，l2的斜率是-3/b，所以-a/3=-3/b，解得a=9/b。又因?yàn)閘1過(guò)點(diǎn)(0,2)，所以l1:ax+3y-6=0可以寫(xiě)成a(0)+3(2)-6=0，所以3a-6=0，解得a=2。所以答案是B。10.已知拋物線(xiàn)y^2=2px的焦點(diǎn)為F，準(zhǔn)線(xiàn)為l，點(diǎn)P在拋物線(xiàn)上，且PF=4，則點(diǎn)P到準(zhǔn)線(xiàn)l的距離為（）A.2B.4C.6D.8解析：第十題，拋物線(xiàn)，咱們要利用拋物線(xiàn)的定義。拋物線(xiàn)上的點(diǎn)到焦點(diǎn)的距離等于它到準(zhǔn)線(xiàn)的距離。點(diǎn)P在拋物線(xiàn)上，PF=4，所以點(diǎn)P到準(zhǔn)線(xiàn)的距離也是4。所以答案是B。11.已知函數(shù)f(x)=log_a(x+1)在區(qū)間[0,+∞)上單調(diào)遞增，則a的取值范圍是（）A.(0,1)B.(1,+∞)C.[1,+∞)D.(0,1)∪(1,+∞)解析：第十一題，對(duì)數(shù)函數(shù)，咱們要利用對(duì)數(shù)函數(shù)的單調(diào)性。對(duì)數(shù)函數(shù)f(x)=log_a(x+1)在區(qū)間[0,+∞)上單調(diào)遞增，當(dāng)且僅當(dāng)a>1。所以答案是B。12.已知三角形ABC的三個(gè)內(nèi)角分別為A,B,C，且sinA=3/5，cosB=-4/5，則cosC的值為（）A.1/5B.-1/5C.4/5D.-4/5解析：第十二題，三角函數(shù)，咱們要利用三角函數(shù)的關(guān)系式。三角形ABC的三個(gè)內(nèi)角分別為A,B,C，所以A+B+C=π。已知sinA=3/5，cosB=-4/5，所以cosA=√(1-sin^2A)=√(1-(3/5)^2)=√(1-9/25)=√(16/25)=4/5；sinB=√(1-cos^2B)=√(1-(-4/5)^2)=√(1-16/25)=√(9/25)=3/5。所以cosC=cos(π-A-B)=-cos(A+B)=-(cosAcosB-sinAsinB)=-(4/5×(-4/5)-3/5×3/5)=-(-16/25-9/25)=-(-25/25)=1。所以答案是A。二、填空題（本大題共4小題，每小題5分，共20分。把答案填在題中橫線(xiàn)上。）13.若函數(shù)f(x)=x^3-ax+1在x=1處取得極值，則a的值為_(kāi)_____。解析：同學(xué)們，這道題考的是函數(shù)的極值，咱們要利用導(dǎo)數(shù)來(lái)研究。f(x)=x^3-ax+1，f'(x)=3x^2-a。因?yàn)閒(x)在x=1處取得極值，所以f'(1)=0，即3(1)^2-a=0，解得a=3。所以答案是3。14.已知圓C的方程為x^2+y^2-2x+4y-3=0，則圓C的半徑長(zhǎng)為_(kāi)_____。解析：好，咱們來(lái)看第十四題，圓的方程，咱們要利用圓的標(biāo)準(zhǔn)方程來(lái)求半徑。圓的標(biāo)準(zhǔn)方程是(x-h)^2+(y-k)^2=r^2，其中(h,k)是圓心坐標(biāo)，r是半徑。圓C的方程為x^2+y^2-2x+4y-3=0，可以寫(xiě)成(x-1)^2+(y+2)^2=1^2+2^2+3=1+4+3=8，所以圓C的半徑長(zhǎng)為√8=2√2。所以答案是2√2。15.已知等比數(shù)列{a_n}的首項(xiàng)為2，公比為3，則前n項(xiàng)和S_n等于______。解析：第十五題，等比數(shù)列，咱們要利用等比數(shù)列求和公式。等比數(shù)列的前n項(xiàng)和公式是S_n=a_1(1-q^n)/(1-q)，其中a_1是首項(xiàng)，q是公比。等比數(shù)列{a_n}的首項(xiàng)為2，公比為3，所以S_n=2(1-3^n)/(1-3)=2(1-3^n)/(-2)=-(1-3^n)=3^n-1。所以答案是3^n-1。16.已知函數(shù)f(x)=sin(x+π/6)+cos(x-π/3)，則f(π/4)的值為_(kāi)_____。解析：第十六題，三角函數(shù)，咱們要利用三角函數(shù)的和差公式來(lái)計(jì)算。f(x)=sin(x+π/6)+cos(x-π/3)，f(π/4)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/三、解答題（本大題共6小題，每小題12分，共72分。解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟。）13.已知函數(shù)f(x)=x^3-3x^2+2在區(qū)間[-1,4]上的最大值和最小值分別為M和m，求M和m的值。解析：同學(xué)們，這道題考的是函數(shù)的最值，咱們要利用導(dǎo)數(shù)來(lái)研究。f(x)=x^3-3x^2+2，f'(x)=3x^2-6x。首先求導(dǎo)數(shù)為零的點(diǎn)，即3x^2-6x=0，解得x=0或x=2。然后比較這些點(diǎn)以及區(qū)間端點(diǎn)處的函數(shù)值。f(0)=0^3-3×0^2+2=2；f(2)=2^3-3×2^2+2=8-12+2=-2；f(-1)=(-1)^3-3×(-1)^2+2=-1-3+2=-2；f(4)=4^3-3×4^2+2=64-48+2=18。所以最大值M=18，最小值m=-2。14.已知圓C的方程為x^2+y^2-4x+6y-3=0，求圓C的圓心坐標(biāo)和半徑長(zhǎng)。解析：好，咱們來(lái)看第十四題，圓的方程，咱們要利用圓的標(biāo)準(zhǔn)方程來(lái)求圓心坐標(biāo)和半徑長(zhǎng)。圓的標(biāo)準(zhǔn)方程是(x-h)^2+(y-k)^2=r^2，其中(h,k)是圓心坐標(biāo)，r是半徑。圓C的方程為x^2+y^2-4x+6y-3=0，可以寫(xiě)成(x-2)^2+(y+3)^2=2^2+3^2+3=4+9+3=16，所以圓C的圓心坐標(biāo)是(2,-3)，半徑長(zhǎng)是√16=4。15.已知等比數(shù)列{a_n}的首項(xiàng)為2，公比為3，求前n項(xiàng)和S_n。解析：第十五題，等比數(shù)列，咱們要利用等比數(shù)列求和公式。等比數(shù)列的前n項(xiàng)和公式是S_n=a_1(1-q^n)/(1-q)，其中a_1是首項(xiàng)，q是公比。等比數(shù)列{a_n}的首項(xiàng)為2，公比為3，所以S_n=2(1-3^n)/(1-3)=2(1-3^n)/(-2)=-(1-3^n)=3^n-1。16.已知函數(shù)f(x)=sin(x+π/6)+cos(x-π/3)，求f(π/4)的值。解析：第十六題，三角函數(shù)，咱們要利用三角函數(shù)的和差公式來(lái)計(jì)算。f(x)=sin(x+π/6)+cos(x-π/3)，f(π/4)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/1)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(2π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/1+π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4+π/6)+cos(π/4-π/3)=sin(π/4