Suppose ABC is a triangle and n is a natural number, then sum of the series S=∑r=0n nCran−rbrcos⁡[nA−(n−r)B] is

S=∑r=0n nCran−rbrRe⁡ei(nB−(n−r)A   =Re⁡∑r=0n nCrae−iBn−rbeiAr   =Re⁡ae−iB+beiAn    =Re⁡[a(cos⁡B−isin⁡B)+b(cos⁡A+isin⁡A)]n   =Re⁡[(acos⁡B+bcos⁡A)      −i(asin⁡B−bsin⁡A)]nBut acosB+bcosA=c                    [projection formula]and asinB=bsinA                            [Law of sines]∴ S=Re⁡cn=cn.