cosnx*sinx+sinnx*cosx=sin(n+1)x.是怎么算的?
请问怎么证明cosnx*sinx+sinnx*cosx=sin(n+1)*x?
设n∈N*,且sinx+cosx=-1,则sinnx+cosnx=______.
三角恒等变换证明cosx+cos2x+…+cosnx=[cos(n+1/2)·sinnx/2]/sinx/2怎么证明?
证明:cosnx+i sinnx 等于(cosx+i sinx)的n次方.
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2
已知函数f(x)=cosnx/(sinnx-1),且f'(x)不在x=π/4上连续,则n的最小正整数值为
用数学归纳法证明:sinx+sin2x+sin3x+……+sinnx=[sin(nx/2)sin((n+1)x/2)]/
COSX+COS2X+COS3X+COS4X+COS5X+COS6X+...+COSNX=1/2|{SIN(N+1/2)
cosX+sinX=根号2sin(X+π/4)是怎么推出来的,
(1+2sinxcosx+cos^2x-sin^2x)+1这个是怎么到这步的[(cosx+sinx)(cosx+sinx
y=sinx的n次方乘以cosnx求导?
为什么cosX+cos²X-sinX-sin²X=(cosX-sinX)(1+sinX+cosX)