cosx=[e^ix+e^(-ix)]/2
e^x cosx=[e^(x+ix)+e^(x-ix)]/2
=1/2*∑[(x+ix)^n+(x-ix)^n]/n!
=1/2* ∑[x^n/n!*( (1+i)^n+(1-i)^n]
因
1+i=√2(cosπ/4+isinπ/4)
1-i=√2[cos(-π/4)+isin(-π/4)]
(1+i)^n+(1-i)^n=(√2)^n* 2cosnπ/4
故e^xcosx=∑[x^n/n! *(√2)^n cosnπ/4]
類似地：
sinx=[e^ix-e^(-ix)]/2i
e^x sinx=[e^(x+ix)-e^(x-ix)]/2i
=1/2*∑[(x+ix)^n-(x-ix)^n]/n!
=1/2* ∑[x^n/n!*( (1+i)^n+(1-i)^n]
因
1+i=√2(cosπ/4+isinπ/4)
1-i=√2[cos(-π/4)+isin(-π/4)]
(1+i)^n-(1-i)^n=(√2)^n* 2isin(nπ/4)
故e^xsinx=∑[x^n/n! *(√2)^n sinnπ/4]