if ∅ lies between −(π/4) and (π/4) then prove that ∅^2 =tan^2 ∅ −(1+(1/3))((tan^4 ∅)/2) +(1+(1/3)+(1/5))((tan^6 ∅)/3) +−−−−− −−−to ∞ terms


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 8224 by trapti rathaur@ gmail.com last updated on 03/Oct/16

$i f \emptyset l i e s b e t w e e n - \frac{π}{4} a n d \frac{π}{4} t h e n p r o v e t h a t$ $\emptyset^{2} = \tan^{2} \emptyset - (1 + \frac{1}{3}) \frac{\tan^{4} \emptyset}{2} + (1 + \frac{1}{3} + \frac{1}{5}) \frac{\tan^{6} \emptyset}{3} + - - - - -$ $- - - t o \infty t e r m s$
Commented by Yozzias last updated on 03/Oct/16
$RTP: If b∈(((−π)/4),(π/4))⇒b^2 =Σ_(r=1) ^∞ (−1)^(r−1) (1/r){Σ_(k=1) ^r (1/(2k−1))}tan^(2r) b. or b^2 =lim_(n→∞) [Σ_(r=1) ^n (−1)^(r−1) (1/r){Σ_(k=1) ^r (1/(2k−1))}tan^(2r) b]$
$RTP : If b \in (\frac{- π}{4}, \frac{π}{4}) \Rightarrow b^{2} = \underset{r = 1}{\sum^{\infty}} {(- 1)}^{r - 1} \frac{1}{r} {\underset{k = 1}{\sum^{r}} \frac{1}{2 k - 1}} \tan^{2 r} b .$ $or b^{2} = \lim_{n \to \infty} [\underset{r = 1}{\sum^{n}} {(- 1)}^{r - 1} \frac{1}{r} {\underset{k = 1}{\sum^{r}} \frac{1}{2 k - 1}} \tan^{2 r} b]$
Commented by trapti rathaur@ gmail.com last updated on 03/Oct/16

$H i n t - \tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = \tan^{- 1} \frac{x + y + z - x y z}{1 - x y - y z - z x}$
Commented by trapti rathaur@ gmail.com last updated on 03/Oct/16

$i n t h e q u e s t i o n g i v e n a b o v e h i n t$ $s o h o w t o s o l v e t h i s q u e s t i o n u s i n g h i n t .$
	Answered by prakash jain last updated on 03/Oct/16
	$Taylor series expansion for tan^(−1) x tan^(−1) x=x−(x^3 /3)+(x^5 /5)+−... (for −1<x<1) (tan^(−1) x)^2 =(x−(x^3 /3)+(x^5 /5)−+...)(x−(x^3 /3)+(x^5 /5)−+...) when n is even coefficient of x^n is obtained by multiplying coeeficients for example coefficient of x^6 =a_1 a_5 +a_3 a_3 +a_5 a_1 where a_n is coeffiient of x^n in tan^(−1) x c_n =coefficient of x^n = { ((Σ_(k=1) ^(n/2) (1/((2k−1)(2(n−k)+1)))),(n even)),(0,(n odd)) :} Σ_(k=1) ^(n/2) (1/((2k−1)(2(n−k)+1))) =Σ_(k=1) ^(n/2) (1/n)((1/(2k−1))+(1/(2((n/2)−k)+1))) =(1/n)(Σ_(k=1) ^(n/2) (1/(2k−1))+Σ_(k=1) ^(n/2) (1/(2((n/2)−k)+1))) =(1/n)(Σ_(k=1) ^(n/2) (1/(2k−1))+Σ_(k=1) ^(n/2) (1/(2k−1))) c_2 =(2/n)Σ_(k=1) ^(n/2) (1/(2k−1)) c_n = { (((2/n)Σ_(k=1) ^(n/2) (1/(2k−1))),(n even)),(0,(n odd)) :} put x=tan ∅ (−(π/4)<∅<(π/4)) to get the required result$
	$Taylor series expansion for \tan^{- 1} x$ $\tan^{- 1} x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + - . . . (for - 1 < x < 1)$ ${(\tan^{- 1} x)}^{2} = (x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - + . . .) (x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - + . . .)$ $w h e n n i s e v e n c o e f f i c i e n t o f x^{n} i s o b t a i n e d$ $b y m u l t i p l y i n g c o e e f i c i e n t s$ $f o r e x a m p l e c o e f f i c i e n t o f x^{6} = a_{1} a_{5} + a_{3} a_{3} + a_{5} a_{1}$ $w h e r e a_{n} i s c o e f f i i e n t o f x^{n} i n \tan^{- 1} x$ $c_{n} = coefficient of x^{n} = {\begin{cases} \underset{k = 1}{\sum^{n / 2}} \frac{1}{(2 k - 1) (2 (n - k) + 1)} & n e v e n \\ 0 & n o d d \end{cases}$ $\underset{k = 1}{\sum^{n / 2}} \frac{1}{(2 k - 1) (2 (n - k) + 1)}$ $= \underset{k = 1}{\sum^{n / 2}} \frac{1}{n} (\frac{1}{2 k - 1} + \frac{1}{2 (\frac{n}{2} - k) + 1})$ $= \frac{1}{n} (\underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 (\frac{n}{2} - k) + 1})$ $= \frac{1}{n} (\underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1})$ $c_{2} = \frac{2}{n} \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1}$ $c_{n} = {\begin{cases} \frac{2}{n} \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} & n e v e n \\ 0 & n o d d \end{cases}$ $put$ $x = \tan \emptyset (- \frac{π}{4} < \emptyset < \frac{π}{4})$ $to get the required result$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum