求sin2xsin3x的极限

单钧汤4217如何求sinx+sin2xsin3x+.+sinnx -
殷姚寿18452911528 ______ 设S=sinx+sin2x+sin3x+……+sinnx 两边同乘以2sin(x/2)(x≠2kπ,k∈Z) 得2sin(x/2)S=cos(x/2)-cos[(2n+1)x/2]=2sin(nx/2)sin[(n+1)x/2] 所以S=sin(nx/2)sin[(n+1)x/2]÷sin(x/2)

单钧汤4217sin2xsin3x=cos2xcos3x,则x的值可能为 -
殷姚寿18452911528 ______ sin2xsin3x=cos2xcos3x 所以 cos2xcos3x-sin2xsin3x=0 即 cos5x=05x=k*pi+pi/2 k属于整数 x=k*pi/5+pi/10 所以选A

单钧汤4217f(x)=sinxsin2xsin3x 求其n阶导数 -
殷姚寿18452911528 ______ lim(tanx-sinx)/sin^3x =limsinx(1/cosx-1)/sin^3x =lim(1-cosx)/[sin^2x*cosx] =lim2(sinx/2)^2/[4(sinx/2)^2(cosx/2)^2*cosx] =lim1/[2(cosx/2)^2*cosx] 当x→2π时 cosx/2→ -1 cosx→ 1 原极限=1/2

单钧汤4217高阶导数y=sinxsin2xsin3x的20阶怎么求 -
殷姚寿18452911528 ______ y=sinxsin2xsin3x =-[cos4x-cos(-2x)]sin2x/2 =(sin2xcos2x-sin2xcos4x)/2 ={sin4x/2-[sin6x+sin(-2x)]/2}/2 =-sin2x/4+sin4x/4-sin6x/4 y'=-cos2x/2+cos4x-3cos6x/2 y''=sin2x-4sin4x+9sin6x y<3>=2cos2x-16cos4x+54cos6x y<4>=-4sin2x+64sin4x-324sin6...

单钧汤4217高数问题 求积分sin(2x)cos(3x)怎么求...求高手 -
殷姚寿18452911528 ______ ∫sin2xcos3xdx=∫(1/2)[sin(2x+3x)+sin(2x-3x) ]dx=(1/2)∫sin5xdx+(-1/2)∫sinxdx=(-1/10)cos5x+(1/2)cosx+C

单钧汤4217满足等式cos2xcos3x=sin2xsin3x的角是 -
殷姚寿18452911528 ______[答案] 原式即化为cos2xcos3x-sin2xsin3x=0 cos(2x+3x)=0 所以5x=kπ+π/2 其实这道题稍稍思考就行咯,你应该有学过公式:cos(a+b)=cosacosb-sinasinb cos(a-b)=cosacosb+sinasinb

单钧汤4217y 等于s i n 2x s i n 3x 求导 -
殷姚寿18452911528 ______ y=sin2xsin3x y'=(sin2x)'sin3x+sin2x(sin3x)'=cos2x(2x)'sin3x+sin2xcos3x(3x)'=2cos2xsin3x+3sin2xcos3x

单钧汤4217sinxsin2xsin3x怎么推导出(sin4x)/4 - (xin6x - sin2x)/4的 -
殷姚寿18452911528 ______ 两次用积化和差公式: sinxsin2xsin3x =-sin2x(cos4x-cos2x)/2 =-(sin2xcos4x)/2+(sin2xcos2x)/2 =-(sin6x-sin2x)/4+(sin4x)/4

单钧汤4217求一道积分题,sinxsin2xsin3xdx的积分 -
殷姚寿18452911528 ______[答案] sinxsin2xsin3x =1/2(cos2x-cos4x)sin2x =1/4sin4x-1/4(sin6x-sin2x) =1/4(sin4x-sin6x+sin2x) 所以∫sinxsin2xsin3xdx =1/4∫(sin4x-sin6x+sin2x)dx =1/4(-1/4*cos4x+1/6*cos6x-1/2*cos2x)+C =1/24cos6x-1/16cos4x-1/8cos2x+C

单钧汤4217如何求y=sinx*sin2x*sin3x的二阶导数sinx,sin2x ,sin3x 都分别为一个整体 * 表示乘 -
殷姚寿18452911528 ______[答案] y=sinxsin2xsin3x =-1/2(cos(x+2x)-cos(x-2x))sin3x =-1/2sin3xcos3x+1/2sin3xcosx =-1/4sin6x+1/4(sin(3x+x)+sin(3x-x) =-1/4sin6x+1/4sin4x+1/4sin2x y'=-1/4cos6x*6 +1/4cos4x*4+1/4cos2x*2 =-3/2cos6x+cos4x+1/2cos2x y''=3/2sin6x*6-sin4x*4-1/2sin2x*2 ...