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Question
`sin^-1(sin (17pi)/8)`
Solution
We know
`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`
We have
`sin^-1(sin (17pi)/8)=sin^-1{sin(2pi+pi/8)}`
`=sin^-1(sin pi/8)`
`=pi/8`
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