AL Pure - Complex Number

看不到答案!!!

2009-07-02 22:17:35 補充：
Let a complex number z = cosx + i sinx
and it is well known that
z^k = coskx + i sinkx ... (1)
1+x+x^2+x^3+...+x^k = (1 - x^k)/(1-k) ... (2)

Consider Sum(r=0 to n)[(z/cosx)^r]
=[1 - (z/cosx)^(n+1)]/[1 - z/cosx] [refer to (2) above]
=[1 - z^(n+1)/cos^(n+1)x]/[1 - 1 - i sinx/cosx]
=[1 - cos(n+1)x/cos^(n+1)x - i sin(n+1)x/cos^(n+1)x]/(-i sinx/cosx)
=[i - i cos(n+1)x/cos^(n+1)x + sin(n+1)x/cos^(n+1)x]/(sinx/cosx)

Take the real part on both sides,
Sum(r=0 to n)[(cosrx/cosx)^r]= [sin(n+1)x/cos^(n+1)x]/(sinx/cosx)
=sin(n+1)x/[sinxcos^nx] ... proved

I hope you can read my work.

http://i707.photobucket.com/albums/ww74/stevieg90/01-47.gif

2009-07-02 21:47:15 補充：
我補充左...但reload又唔見左個補充...>

2009-07-02 23:07:00 補充：
我冇用到COMPLEX NUMBER去SOLVE=.="...樓下做法好d!