cos(-19/6n)=
cos(-19/6n)=
求证:(cosπ/6+isinπ/6)^n=cos(nπ/6)+i *sin( nπ/6)谢谢
设f(x)=cos^(nπ+x).sin^(nπ-x)/cos^[(2n+1)π-x](n∈z)求f(π/6)的值
cos(-19/6π)=?
n=0:127; x=cos(0.04*pi*n)+cos(0.08*pi*n)+cos(0.4*pi*n); w=ra
求当n趋向于无穷时,lim[cos(θ/n)])^n)^n=?
帮忙解一道极限的题n=>无穷lim[cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n)]
.若f(x)={sin(n派-x)cos(n派+x)/cos[(n+1)派-x]}*tan(x-n派)cot[(n派/2
cos(-19/6∏)=
cos(-19/6π)
数列an=n^2((cos(nπ/3))^2-(sin(nπ/3))^2)
cos=(-19兀/3)答案cos=19兀/3=cos(兀/3+6兀)=cos兀/3=1/2