[armpakorn]

ลองต่อยอดหารูปทั่วไป
ให้ $z = cos(1) + isin(1)$
ดังนั้น $z^x = (cos(1) + isin(1))^x = cos(x) + isin(x)$
$(cos(1) + isin(1))^x = \binom{x}{0}cos(1)^x + \binom{x}{1}cos(1)^{x-1}(isin(1))^{1} + \binom{x}{2}cos(1)^{x-2}(isin(1))^{2} + \binom{x}{3}cos(1)^{x-3}(isin(1))^{3} + \binom{x}{4}cos(1)^{x-4}(isin(1))^{4} + ... + \binom{x}{x}(isin(1))^{x}$
$= cos(1)^x + i\binom{x}{1}cos(1)^{x-1}sin(1) - \binom{x}{2}cos(1)^{x-2}sin(1)^{2} - i\binom{x}{1}cos(1)^{x-3}sin(1)^3 + \binom{x}{4}cos(1)^{x-4}sin(1)^{4} + ... + i^xsin(1)^x$
จะได้
$cos(x) = cos(1)^x - \binom{x}{2}cos(1)^{x-2}sin(1)^2 + \binom{x}{4}cos(1)^{x-4}sin(1)^4 + ... + \cases{sin(1)^x &, x mod 4 = 0 \cr -cos(1)^{x-1} &, x mod 4 = 1 \cr -sin(1)^x &, x mod 4 = 2 \cr cos(1)^{x-1} &, x mod 4 = 3}$
และ
$sin(x) = \binom{x}{1}cos(1)^{x-1}sin(1) - \binom{x}{3}cos(1)^{x-3}sin(1)^3 + \binom{x}{5}cos(1)^{x-5}sin(1)^5 + ... + \cases{-cos(1)^{x-1} &, x mod 4 = 0 \cr sin(1)^{x} &, x mod 4 = 1 \cr cos(1)^{x-1} &, x mod 4 = 2 \cr -sin(1)^x &, x mod 4 = 3}$