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Question
Find the value of the given expression.
`sin^(-1) (sin (2pi)/3)`
Solution
`sin^(-1) (sin 2pi/3)`
We know that sin−1 (sin x) = x if x in `[-pi/2, pi/2]`, which is the principal value branch of sin−1x.
Here, `2pi/3 in [(-pi)/2, pi/2]`
Now `sin^(-1) (sin 2pi/3)` can be written as
`sin^(-1) (sin (2pi)/3) `
`= sin^(-1) [sin (pi - (2pi)/3)] `
`= sin^(-1) (sin pi/3) "where" pi/3 in [(-pi)/2, pi/ 2]`
`:. sin^(-1) (sin (2pi)/2) `
`= sin^(-1) (sin pi/3) `
` = pi/3`
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