無窮三角級數求和

Define a complex number z = cosA + isinA
It is well known that zk = coskA + isinkA
It is also konwn that e-x = -x + x2/2! - x3/3! + x4/4! - ...
Therefore x - x2/2! + x3/3! - x4/4! + ... = 1 - e-x
Now define the sums
R = cosA - cos2A/2! + cos3A/3! - cos4A/4! + ..., and
M = sinA - sin2A/2! + sin3A/3! - cos4A/4! + ...
Then the complex number R + iM
= z - z2/2! + z3/3! - z4/4! + ...
= 1 - e-z
= 1 - e-(cosA + isinA)
= 1 - (e-cosA)(e-isinA)
= 1 - (e-cosA)[cos(sinA) - isin(sin(A)]
The imaginary part of the complex number is
M = e-cosAsin(sinA)
Note : sin(sinA) mean the sine of sinA

對不起，看錯題目，

螞蟻雄兵 :
可否解釋 求了 P + Qi 後如何求 Q ?還有答案
Q=(SinACosA+SinA+SinACosA)/(2+2CosA)
的分子你是否未化簡,(有 2 個 SinACosA ?)
我不知道答案的!