prove-that-9pi-8-9-4-sin-1-1-3-9-4-sin-1-2-2-3-

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

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 \dots

Commented by peter frank last updated on 29/Sep/18

v e r y n i c e