please prove the question

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! +...................
Put x = ix, where i is the imaginary number sqrt(-1). We get
e^(ix) = 1 + (ix) + (ix)^2/2! + (ix)^3/3! + (ix)^4/4! + (ix)^5/5! + ..................
= 1 + ix - x^2/2! -ix^3/3! + x^4/4! + ix^5/5! + .................
= (1 - x^2/2! + x^4/4! + ............) + i(x - x^3/3! + x^5/5! + .................).
But e^(ix) = cosx + isinx. Therefore,
cosx = Re[e^(ix)] = 1 - x^2/2! +x^4/4! - x^6/6! +................. And
sinx = Im[e^(ix)] = x - x^3/3! + x^5/5! -x^7/7! + ...................

2008-06-27 17:27:59 補充：
You may also use the Maclaurin theorem to prove it.

首先將sin x和cos x 用通項表示

http://en.wikipedia.org/wiki/Taylor_series

對sin x

|R|=x^(2n+3)/(2n+3)!

Since lim n-> infinity |x|^n/n! tends to 0

So sinx=x-(x^3/3!)+(x^5/5!)-(x^7/7!)+...
the series is convergent for all values for x and angle x must be measured in radians

對cos x

|R|=x^(2n+2)/(2n+2)!

Since lim n-> infinity |x|^n/n! tends to 0

So cosx=(1-x^2/2!+x^4/4!-x^6/6!+...)
the series is convergent for all values for x and angle x must be measured in radians