Prove that:`(9pi)/8-9/4sin^(-1)(1/3)=9/4sin^(-1)((2sqrt(2))/3)`

1.	Prove that:`(9pi)/8-9/4sin^(-1)(1/3)=9/4sin^(-1)((2sqrt(2))/3)`
Answer» `L.H.S. = (9pi)/8 -9/4sin^-1(1/3)``=9/4(pi/2-sin^-1(1/3))` As `pi/2-sin^-1x = cos^-1x`.So,our expression now is, `=9/4cos^-1(1/3)` If, we create a triangle with base 1 and hypotenuse 3, perpendicular comes `2sqrt2`. So, `cos^-1(1/3) = sin^-1((2sqrt2)/3)` Now, our expression becomes, `9/4sin^-1((2sqrt2)/3) = R.H.S.`

Discussion

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