`sin{2cos^(-1)((-3)/(5))}=`

Explanation

We have `A = 2 tan^(-1) (2 sqrt2 -1) = 2 tan^(-1) (1.828)`
`rArr A gt 2 tan^(-1) sqrt3`
`rArr A gt (2pi)/(3)`
Now, `sin^(-1) ((1)/(3)) lt sin^(-1) ((1)/(2))`
`rArr sin^(-1) ((1)/(3)) lt (pi)/(6)`
`rArr 3 sin^(-1).(1)/(3) lt (pi)/(2)`
Further, `sin^(-1) ((3)/(5)) = sin^(-1) (0.6) lt sin^(-1) ((sqrt3)/(2))`
`rArr sin^(-1) ((3)/(5)) lt (pi)/(3)`
`rArr B = 3 sin^(-1) ((1)/(3)) + sin^(-1) ((3)/(5)) lt (pi)/(2) + (pi)/(3)`
`rArr B lt (5pi)/(6)`
From this, we really cannot relate A and B.
Now, `3 sin^(-1) ((1)/(3)) = sin^(-1) [3.(1)/(3) - 4((1)/(3))^(3)]`
`= sin^(-1) ((23)/(27))`
`= sin^(-1) (0.852)`
`rArr 3 sin^(-1) ((1)/(3)) lt sin^(-1) ((sqrt3)/(2)) = (pi)/(3)`
Hence, `B = 3 sin^(-1) ((1)/(3)) + sin^(-1) ((3)/(5)) lt (pi)/(3) + (pi)/(3) = (2pi)/(3)`
`:. A gt B`