sin2x除以sin3x

夏任侮4071cos3x除以cosx等于什么? -
别亚修15523739272 ______[答案] cos3x=cos(2x+x)=cos2x.cosx--sin2x.sinx=cosx--4cosx.sinx.sinx

夏任侮4071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 -
别亚修15523739272 ______[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 ...

夏任侮4071求解sinx+sin2x+sin3x+……sinnx -
别亚修15523739272 ______ sinx+sin2x+sin3x+...+sin(nx)等于1/2*(cot(x/2)-cot(x/2)*cos(nx)+sin(nx)). 解:因为cos(A-B)-cos(A+B)=2sinAsinB, 那么2sinx*sin(x/2)=cos(x-x/2)-cos(x+x/2)=cos(x/2)-cos(3x/2) 2sin(2x)*sin(x/2)=cos(2x-x/2)-cos(2x+x/2)=cos(3x/2)-cos(5x/2) 2sin(3x)*...

夏任侮4071数学数列求和sinx+sin2x+sin3x+……+sinnx的和详细解法乘以2sinx/2后怎么整理 -
别亚修15523739272 ______[答案] 乘以2sinx/2后,用积化和差公式,相当于拆项,然后前后相消: 2sinx/2sinkx=cos(k-1/2)x-cos(k+1/2)x 这样就为: cos0.5x-cos1.5x+cos1.5x-cos2.5x+cos2.5x-cos3.5x+.....+cos(n-1/2)x-cos(n+1/2)x =cos0.5x-cos(n+1/2)x 最后再除以2sinx/2就得到结果了...

夏任侮4071如何求sinx+sin2xsin3x+...+sinnx -
别亚修15523739272 ______ 用复数比较方便 e^(xi)=cosx+isinx e^(2xi)=cos2x+isin2x ... e^(nxi)=cosnx+isinnx 相加得 e^(xi)+e^(2xi)+...+e^(nxi)=(cosx+cos2x+...+cosnx)+i(sinx+sin2x+...+sinnx) e^(xi)+e^(2xi)+...+e^(nxi), 用等比数列求和 =e^(xi) [1-e^(nxi)]/(1-e^(xi)) =Cos[(1 + n) x/...

夏任侮4071sinx+sin2x+sin3x+........+sinnx=?求和..
别亚修15523739272 ______ 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)

夏任侮4071三角函数的求和我记得有个公式好像是sinx+sin2x+sin3
别亚修15523739272 ______ A =sin x+sin 2x+sin 3x+……+sin nx A*2*sin(x/2) = [2*sin(x/2)]*sin x +[2*sin(x/2)]*sin 2x +....+[2*sin(x/2)]*sin nx = [-cos(3x/2)+cos(x/2)] +[-cos(5x/2)+cos(3x/2)] +. .....

夏任侮4071y=sin3x 乘 ln sin2x 的导数 -
别亚修15523739272 ______ y=sin3x ln(sin2x) y'=3cos3x ln(sin2x)+sin3x /sin2x *2cos2x=3cos3xln(sin2x)+2sin3xcot2x

夏任侮4071sinx+sin2x+sin3x+.+sinnx=?求和.数列求和里的一道题. -
别亚修15523739272 ______[答案] 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)

夏任侮4071已知tanx=3,计算:(2cosx - 3sinx)/(2sinx+3cosx) -
别亚修15523739272 ______[答案] (2cosx-3sinx)/(2sinx+3cosx) 【分子分母同除以cosx】 =(2-3tanx)/(2tanx+3) =-7/9