求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2
来源:学生作业帮 编辑:大师作文网作业帮 分类:综合作业 时间:2024/09/23 05:32:38
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]
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)]
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2
COSX+COS2X+COS3X+COS4X+COS5X+COS6X+...+COSNX=1/2|{SIN(N+1/2)
求证 sinx(cos^2 2x-sin^2 2x) + 2cosx cos2x sin2x= sin 5x
求证 cos^8(x)-sin^8(x)=cos2x【1-1/2sin^2(2x)】
请问怎么证明cosnx*sinx+sinnx*cosx=sin(n+1)*x?
求证:sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)=2/cos2x
1-2sinx cosx /COS^2X-SIN^2X =1-tanx/1+tanx 求证
sin(x+1/2π)=cosx?还是cos-x
求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(ta
已知4sin^2x-cos^2x+3cosx=0求(cos2x-cos^2x)/(1-cot^2x)
y=2cos2x+sin平方x-4cosx.化简
lim[(1-cosx*cos2x****cosnx)/x^2]在x趋于0时