例如，当 n = 2 时，(cosx)2=1+cos2x2，这里虽然出现了 cos 2x，但它是作为 cos x ...
通过积化和差、套缩求和、和差化积：∑k=1ncoskx=cscx2∑k=1n(sinx2coskx)=12...
(cosx+cos2x+…+cosnx)=( sin 3x 2−sin x 2)+ (sin 5x 2− 3x 2)+…+ (sin 1+2n 2x−sin 1−2n 2x)= sin 1+2n 2x−sin x 2= 2cos n+1 2xsin n 2x.∴cos+cos2x+…+cosnx= cos n+1 2x•sin n 2x sin x 2. 利用 2sin x 2cosnx= sin( x 2+nx)+ sin( x 2...
cosx+cos2x+.+cosnx=1/2sin(x/2)*(cosx*2sin(x/2)+cos2x*2sin(x/2)+.+cosnx*2sin(x/2))=1/2sin(x/2)*(sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+.+sin(n+1/2)x-sin(n-1/2)x)=1/2sin(x/2)*(sin(n+1/2)x-sin(x/2))=1/2sin(x/2)*(2*sinnx*cos(n+1)x...
sinx+2sin2x+⋯+nsinnx=nsin(n+1)x−(n+1)sinnx2(cosx−1). 解:记前式为A,后式为B,设 ω=A+Bi,z=cosx+isinx, 则ω=(cosx+isinx)+2(cos2x+isin2x)+⋯+n(cosnx+isinnx) =z+2z2+⋯+nzn, ω(1−z)=ω−ωz =(z+2z2+⋯+nzn)−(z2+2z3+⋯+nzn+1) =z+z2...
lim x→0(1-cosxcos2x...cosnx)/(x^2)=lim(sinx/cosx+2sin2x/cos2x+...+nsinnx/cosnx)cosxcos2x...cosnx/(2x)=1/2lim(tanx/x+2tan2x/x+...+ntannx/x)cosxcos2x..cosnx =1/2(1+2^2+3^2+...+n^2)*1 =n(n+1)(2n+1)/12 ...
因此最终得到∑k=1ncoskx=12(sin[(n+12)x]sin(x/2)−1)PS：呃……你说和其他人得到的答案不一样？主要是因为三角恒等变换比较多，其实我们再算一算就能得到相同的结果。例如，12(sin[(n+12)x]sin(x/2)−1)=sin[(n+12)x]−sin(x/2)2sin(x/2)=2cos...
cosnx=sin( x 2 +nx)+sin( x 2 -nx).及和差化积即可得出. 解答:证明:∵2sin x 2 cosnx=sin( x 2 +nx)+sin( x 2 -nx). ∴2sin x 2 (cosx+cos2x+…+cosnx)=(sin 3x 2 -sin x 2 )+(sin 5x 2 - 3x 2 )+…+(sin
解析 1.令 2sinx 则△f(n-1)=f(n)-f(n-1)=cosn- sin(n+号)x-sin(n-)x 2sinx = 0 = Af(n).又 f(1)=1/2+cosx-(3sinx/2-tsinx/2)/(2sin1/2x)=-1+ cosx+2sin^2x/2=0 ,故 f(n) = f(1) = 0. 反馈 收藏
(cosx+cos2x+…+cosnx)=( sin 3x 2−sin x 2)+ (sin 5x 2− 3x 2)+…+ (sin 1+2n 2x−sin 1−2n 2x)= sin 1+2n 2x−sin x 2= 2cos n+1 2xsin n 2x.∴cos+cos2x+…+cosnx= cos n+1 2x•sin n 2x sin x 2. 利用 2sin x 2cosnx= sin( x 2+nx)+ sin( x 2...