證明sin x=x-(x^3)/3!+(x^5)/5!-(x^7)/7! + (x^9)/9!-(x^11)/11!+...?最好有解釋。?

設： cos(x)= a(0) + a(1)*(x^1) + a(2)*(x^2) + a(3)*(x^3)...... + a(n)*(x^n)
，其中 a(0)、a(1)、a(2)......a(n)為常數，x^n= x*x*x......*x (n個x相乘)

則對任意實數x，兩邊同時微分亦相等:
-sin(x)= 0 + 1*a(1)*(x^0) + 2*a(2)*(x^1) + 3*a(3)*(x^2)...... + n*a(n)*(x^(n-1))

對任意實數x，再次微分兩邊亦相等:
-cos(x)= 1*2*a(2) + 2*3*a(3)*(x^1) + 3*4*a(4)*(x^3)*(x^2)...... + n(n-1)*a(n)*(x^(n-2))
= - a(0) - a(1)*(x^1) - a(2)*(x^2) ...... - a(n-2)*(x^(n-2))
因對任意實數x，兩函數皆相等，所以x不同次方的各項係數必相等:
1*2*a(2) = - a(0)
2*3*a(3)= - a(1)
......
n(n-1)*a(n)= - a(n-2) -> 推論1
又：cos(0)=1，且cos(x)為偶函數
因此：a(0)＝1，a(1)=a(3)=a(5)=a(7)=......=a(2n+1)=0 -> 推論2

之後用數學歸納法試證：對任意正偶數k，a(k)=( (-1)^(k/2) )/(k!)
1:由推論2得 a(0)=1=( (-1)^(0/2) )/(1!)=(-1)^0=1 -> 在k=0成立

2:假設k = n時公式成立，則得a(n)=( (-1)^(n/2) )/(n!)
又由推論1知：
a(n+2)= a(n+2-2) /(-(n+2)(n+1))
->a(n+2)
=a(n)/(-(n+2)(n+1))
=( ((-1)^(n/2) )/(n!) ) / (-(n+2)(n+1))
=(-1)^(n/2) * (-1) / ( n! *(n+2)(n+1) )
=( (-1)^(n/2+1) ) / (n+2)!
=( (-1)^((n+2)/2) ) / (n+2)!
故此假設若於k=n時成立，則於k=n+2時亦成立。
3:
由1,2及數學歸納法知：對任意正偶數k，a(k)=( (-1)^(k/2) )/(k!)

故cos(x)= 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! ......+ (x^k)( (-1)^(k/2) )/(k!) (k為任意正偶數)
故sin(x)= - cos'(x) = x-(x^3)/3!+(x^5)/5!-(x^7)/7! + (x^9)/9!-(x^11)/11!+...