题目

求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]

提问时间：2020-12-19

答案

cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)]=[cos(n+1)x/2][cos((n-1)x/2)+cos(((n-3)x/2)+...+cos((n-(2n-1))x/2)=[cos(n+1)x/2/sin(x/2)]*[sin(x/2)*cos((n-1)x/2)+sin(x/2)*cos(((n-3)x/2)+...+sin(x/2)*cos((n-(2n-1))x/2)=1/2[cos(n+1)x/2/sin(x/2)][sin(nx/2)+sin((2-n)x/2)+sin((n-2)x/2)+sin((4-n)x/2)+...+sin((2-n)x/2)+sin(nx/2)]={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]