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Question Number 167158 by mnjuly1970 last updated on 08/Mar/22

	Answered by mindispower last updated on 10/Mar/22
	$cos(3n)=4cos^3 (n)−3cos(n)⇒cos^3 (n)=((cos(3n)+3cos(n))/4) Θ=(1/4)Re{Σ_(n≥1) (((−1)^(n−1) )/n^2 ).(e^(3in) +3e^(in) )} =−(1/4)ReΣ_(n≥1) (((−e^(3i) )^n )/n^2 )−(3/4)ReΣ_(n≥1) (((−e^i )^n )/n^2 ) =−(1/4)(Re(Li_2 (−e^(3i) )+3Li_2 (−e^i )) =−(1/8)(Li_2 (−e^(3i) )+Li_2 (−e^(−3i) )+3Li_2 (−e^i )+3Li_2 (−e^(−i) )) li_2 (a)+li_2 ((1/a))=−ζ(2)−((ln^2 (−a))/2) =−(1/8)(−ζ(2)−(1/2)ln^2 (e^(3i) )+3(−ζ(2)−(1/2)ln^2 (e^i )) =((ζ(2))/2)−(1/8)(6)=(π^2 /(12))−(6/8)$
	$c o s (3 n) = 4 {c o s}^{3} (n) - 3 c o s (n) \Rightarrow {c o s}^{3} (n) = \frac{c o s (3 n) + 3 c o s (n)}{4}$ $Θ = \frac{1}{4} R e {\sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{n^{2}} . (e^{3 i n} + 3 e^{i n})}$ $= - \frac{1}{4} R e \sum_{n ⩾ 1} \frac{{(- e^{3 i})}^{n}}{n^{2}} - \frac{3}{4} R e \sum_{n ⩾ 1} \frac{{(- e^{i})}^{n}}{n^{2}}$ $= - \frac{1}{4} (R e ({L i}_{2} (- e^{3 i}) + 3 {L i}_{2} (- e^{i}))$ $= - \frac{1}{8} ({L i}_{2} (- e^{3 i}) + {L i}_{2} (- e^{- 3 i}) + 3 {L i}_{2} (- e^{i}) + 3 {L i}_{2} (- e^{- i}))$ ${l i}_{2} (a) + {l i}_{2} (\frac{1}{a}) = - ζ (2) - \frac{{l n}^{2} (- a)}{2}$ $= - \frac{1}{8} (- ζ (2) - \frac{1}{2} {l n}^{2} (e^{3 i}) + 3 (- ζ (2) - \frac{1}{2} {l n}^{2} (e^{i}))$ $= \frac{ζ (2)}{2} - \frac{1}{8} (6) = \frac{π^{2}}{12} - \frac{6}{8}$
	Commented by mnjuly1970 last updated on 12/Mar/22

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