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

Question

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

Infinity Learn · Accepted Answer

$$S=$$∑$$r=0n$$ 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$$.

Suppose ABC is a triangle and n is a natural number, then sum of the series $S = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r} \cos [n A - (n - r) B]$ is

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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

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Suppose ABC is a triangle and n is a natural number, then sum of the series $S = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r} \cos [n A - (n - r) B]$ is