prove that ((9π)/(8 ))−(9/4)sin^(−1) (1/3)=(9/4)sin^(−1) ((2(√2))/3)


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Question Number 44480 by peter frank last updated on 29/Sep/18

$p r o v e t h a t \frac{9 π}{8} - \frac{9}{4} \sin^{- 1} \frac{1}{3} = \frac{9}{4} \sin^{- 1} \frac{2 \sqrt{2}}{3}$
	Answered by math1967 last updated on 30/Sep/18

	$L . H . S = \frac{9}{4} (\frac{π}{2} - \sin^{- 1} \frac{1}{3})$ $= \frac{9}{4} (\cos^{- 1} \frac{1}{3}) [∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}]$ $= \frac{9}{4} \sin^{- 1} (\sqrt{1 - {(\frac{1}{3})}^{2}})$ $= \frac{9}{4} \sin^{- 1} \frac{2 \sqrt{2}}{3} = R . H . S p r o v e d$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18

	$\frac{9}{4} ({s i n}^{- 1} \frac{2 \sqrt{2}}{3} + {s i n}^{- 1} \frac{1}{3})$ $\frac{9}{4} ({t a n}^{- 1} \frac{2 \sqrt{2}}{1} + {t a n}^{- 1} \frac{1}{2 \sqrt{2}})$ $\frac{9}{4} ({t a n}^{- 1} \frac{2 \sqrt{2} + \frac{1}{2 \sqrt{2}}}{1 - 2 \sqrt{2} \times \frac{1}{2 \sqrt{2}}})$ $\frac{9}{4} {t a n}^{- 1} (\infty)$ $\frac{9}{4} \times \frac{π}{2} = \frac{9 π}{8}$ $s o \frac{9 π}{8} - \frac{9}{4} {s i n}^{- 1} (\frac{2 \sqrt{2}}{3}) = \frac{9}{4} {s i n}^{- 1} (\frac{1}{3}) p r o v e d$
	Commented by peter frank last updated on 29/Sep/18

	$s i r \frac{9 π}{8} \neq \sin^{- 1} \frac{2 \sqrt{2}}{3}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18

	$l o o k t h e m e t h o d o f s o l v e$ $g i v e n t o p r o v e a - b = c$ $i h a v e p r o v e a = b + c$ $i f a = b + c t h a t m e a n s a - b = c$ $s o b e f o r e c o m m e n t p l s s e e t h e m e t h o d o f s o l v e$
	Commented by peter frank last updated on 29/Sep/18

	$o k a y s i r t h a n k y o u n o w i u n d e r s t o o d$
	Commented by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18
	$or method to prove ((9π)/8)−(9/4)sin^(−1) (((2(√2))/3))=(9/4)sin^(−1) ((1/3)) let k=(9/4)((π/2)−sin^(−1) (((2(√2))/3)) ((4k)/9)=(π/2)−sin^(−1) (((2(√2))/3)) sin(((4k)/9))=sin{(π/2)−sin^(−1) (((2(√2))/3))} sin(((4k)/9))=cos({sin^(−1) (((2(√2))/3))} sin(((4k)/9))=(1/3) [reason sinα=((2(√2))/3) so cosα=(1/3)] ((4k)/9)=sin^(−1) ((1/3)) k=(9/4)sin^(−1) ((1/3)) hence prlved...$
	$o r m e t h o d$ $t o p r o v e$ $\frac{9 π}{8} - \frac{9}{4} {s i n}^{- 1} (\frac{2 \sqrt{2}}{3}) = \frac{9}{4} {s i n}^{- 1} (\frac{1}{3})$ $l e t k = \frac{9}{4} (\frac{π}{2} - {s i n}^{- 1} (\frac{2 \sqrt{2}}{3})$ $\frac{4 k}{9} = \frac{π}{2} - {s i n}^{- 1} (\frac{2 \sqrt{2}}{3})$ $s i n (\frac{4 k}{9}) = s i n {\frac{π}{2} - {s i n}^{- 1} (\frac{2 \sqrt{2}}{3})}$ $s i n (\frac{4 k}{9}) = c o s ({{s i n}^{- 1} (\frac{2 \sqrt{2}}{3})}$ $s i n (\frac{4 k}{9}) = \frac{1}{3} [r e a s o n s i n α = \frac{2 \sqrt{2}}{3} s o c o s α = \frac{1}{3}]$ $\frac{4 k}{9} = {s i n}^{- 1} (\frac{1}{3})$ $k = \frac{9}{4} {s i n}^{- 1} (\frac{1}{3}) h e n c e p r l v e d . . .$
	Commented by peter frank last updated on 29/Sep/18

	$v e r y n i c e$
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