另第四題我的作法如下
連結已失效h ttp://tw.knowledge.yahoo.com/question/question?qid=1509061802221
以下令t= tanθ
cos(nθ)+isin(nθ)=(cosθ+isinθ)^n
=C(n,0)*(cosθ)^n+C(n,1)*(cosθ)^(n-1)*(isinθ)+…+C(n,n)*(isinθ)^n
=[C(n,0)*(cosθ)^n-C(n,2)* (cosθ)^(n-2)*(sinθ)^2+-…]+
i[C(n,1)*(cosθ)^(n-1)*(sinθ)-C(n,3)* (cosθ)^(n-3)*(sinθ)^3+-…]
=(cosθ)^n*[C(n,0)-c(n,2)t^2+-…]+i(cosθ)^n*[C(n,1)t-c(n,3)t^3+-…]
故
cos(nθ)= (cosθ)^n*[C(n,0)-c(n,2)t^2+-…]
sin(nθ)= (cosθ)^n*[C(n,1)t-c(n,3)t^3+-…]
得tan(nθ)= sin(nθ)/ cos(nθ)
=[C(n,1)t-C(n,3)t^3+-...]/[C(n,0)-C(n,2)t^2+-...]
知識+菩提兄的作法
tan(nx)
=(-i)[(cosx+i sinx)^n-(cosx-i sinx)^n]/[(cosx+isinx)^n+(cosx-isinx)^n]
=(-i) Im[(1+i t)^n-(1-i t)^n]/ Re[(1+i t)^n+(1- i t)^n] ( t= tanx)
= -ΣC(n, 2k-1)(-1)^k t^(2k-1) / ΣC(n, 2k)(-1)^k t^(2k)
is a rational function of t=tanx
Note: the 1st Σ takes sum for k=1,2,..., [n+1]/2
the 2nd Σ takes sum for k=0,1,...,[n]/2