Question

Prove: \(\frac {9\pi}{4} - \frac 94 sin^{-1}\frac 13 = \frac 94sin^{-1}\frac {2\sqrt 2}{3}\)

Answer 1

L.H.S = \(\frac {9\pi}8{}\) - \(\frac 94\)sin^-1\(\frac 13\)
=\(\frac 94\)(\(\frac {\pi}{2}\) - sin^-1\(\frac 13\)) 
=\(\frac 94\)(cos^-1\(\frac 13\)) ……....(1)  [sin^-1x + cos^-1x = \(\frac {\pi}{2}\)] 
Now, let cos^-1\(\frac 13\) = x. Then, cos x = \(\frac 13\) \(\implies\)sin x =  \(\sqrt {1-(\frac 13)^2}\)  = \(\frac {2\sqrt 2}{3}\). 
Therefore x = sin^-1\(\frac {2\sqrt 2}{3}\) 
Therefore L.H.S = \(\frac 94\)sin^-1\(\frac {2\sqrt 2}{3}\)
                           =R.H.S