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Question
Show that `(9pi)/8 - 9/4 sin^-1 (1/3) = 9/4 sin^-1 ((2sqrt2)/3)`.
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Solution
We have to show that
`(9pi)/8 - 9/4 sin^-1 (1/3) = 9/4 sin^-1 ((2sqrt2)/3)`
i.e. to show that,
`9/4 sin^-1 (1/3) + 9/4 sin^-1 ((2sqrt2)/3) = (9pi)/8`
Let `sin^-1 (1/3)` = x
∴ sin x = `1/3, "where" 0 < x < pi/3`
∴ cos x > 0
Now, `cos "x" = sqrt(1 - sin^2"x") = sqrt(1 - 1/9) = sqrt(8/9) = ((2sqrt2)/3)`
∴ x = `cos^-1 ((2sqrt2)/3)`
∴ `sin^-1 (1/3) = cos^-1 ((2sqrt2)/3)` ....(1)
∴ LHS = `9/4 sin^-1 (1/3) + 9/4 sin^-1 ((2sqrt2)/3)`
= `9/4 [sin^-1 (1/3) + sin^-1 ((2sqrt2)/3)]`
`= 9/4 [cos^-1 ((2sqrt2)/3) + sin^-1 ((2sqrt2)/3)]` ...[By (1)]
`= 9/4 (pi/2) ....[because sin^-1"x" + cos^-1"x" = pi/2]`
`= (9pi)/8`
= RHS.
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