1)prove that arctan(a) +arctanb =arctan(((a+b)/(1−ab))) with ab≠1 2)find the value of S_N = Σ_(n=1) ^N (−1)^n arctan(((2n+1)/(n^2 +n−1)))


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Question Number 57848 by Abdo msup. last updated on 13/Apr/19

$1) p r o v e t h a t a r c t a n (a) + a r c t a n b = a r c t a n (\frac{a + b}{1 - a b})$ $w i t h a b \neq 1$ $2) f i n d t h e v a l u e o f S_{N} = \sum_{n = 1}^{N} {(- 1)}^{n} a r c t a n (\frac{2 n + 1}{n^{2} + n - 1})$
Commented by maxmathsup by imad last updated on 13/Apr/19
$1) let put α =arctan(a) and β =arctan(b) ⇒ tan(α+β) =((tan(α)+tan(β))/(1−tan(α)tan(β))) =((a+b)/(1−ab)) (we suppose ab ≠1) ⇒ α+β =arctan(((a+b)/(1−ab))) 2) we have S_n =−Σ_(n=1) ^N (−1)^n arctan(((2n+1)/(1−n(n+1)))) =−Σ_(n=1) ^N (−1)^n arctan(((n +n+1)/(1−n(n+1)))) =−Σ_(n=1) ^N (−1)^n {arctan(n) +arctan(n+1)} ⇒−S_n =−arctan(1) −arctan(2) +arctan(2) +arctan(3)−... +(−1)^n arctan(n) +(−1)^n arctan(n+1) =(−1)^n arctan(n+1) −(π/4) ⇒ S_N =(π/4) −(−1)^n arctan(n+1) .$
$1) l e t p u t α = a r c t a n (a) a n d β = a r c t a n (b) \Rightarrow$ $t a n (α + β) = \frac{t a n (α) + t a n (β)}{1 - t a n (α) t a n (β)} = \frac{a + b}{1 - a b} (w e s u p p o s e a b \neq 1) \Rightarrow$ $α + β = a r c t a n (\frac{a + b}{1 - a b})$ $2) w e h a v e S_{n} = - \sum_{n = 1}^{N} {(- 1)}^{n} a r c t a n (\frac{2 n + 1}{1 - n (n + 1)})$ $= - \sum_{n = 1}^{N} {(- 1)}^{n} a r c t a n (\frac{n + n + 1}{1 - n (n + 1)}) = - \sum_{n = 1}^{N} {(- 1)}^{n} {a r c t a n (n) + a r c t a n (n + 1)}$ $\Rightarrow - S_{n} = - a r c t a n (1) - a r c t a n (2) + a r c t a n (2) + a r c t a n (3) - . . .$ $+ {(- 1)}^{n} a r c t a n (n) + {(- 1)}^{n} a r c t a n (n + 1)$ $= {(- 1)}^{n} a r c t a n (n + 1) - \frac{π}{4} \Rightarrow S_{N} = \frac{π}{4} - {(- 1)}^{n} a r c t a n (n + 1) .$
Commented by maxmathsup by imad last updated on 13/Apr/19

$S_{N} = \frac{π}{4} - {(- 1)}^{N} a r c t a n (N + 1) .$
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