Find the principal value of
`sin^(-1) (1/2)`

Options

A. `10 - 3pi`
B. `10 -2 pi`
C. `10 -(5pi)/(2)`
D. `(7pi)/(2) - 10`

Answer

B

Explanation

We have `f(x) = sin^(-1)x`, having range `[(pi)/(2), (3pi)/(2)]`
Consider the function `h(x) = (pi - sin^(-1) x)`, where
`sin^(-1) x in [-(pi)/(2), (pi)/(2)]`
Clearly, function `f(x) and h(x)` are identical.
Advantage with function `y = h(x)` is that we are well versed with `sin^(-1) x` if `(sin^(-1) x) in [-(pi)/(2), (pi)/(2)]`
`h(sin 10) = pi - sin^(-1) (sin 10)`
`= pi - sin^(-1) (sin (3pi - 10)))`
`= pi - (3pi - 10)`
`= 10 - 2pi`
Now, `f(x) lt (3pi)/(4)`
`:. h(x) lt (3pi)/(4)`
`rArr pi - sin^(-1) x lt (3pi)/(4)`
`rArr sin^(-1) x gt (pi)/(4)`
`rArr (pi)/(4) lt sin^(-1) x le (pi)/(2)`
`rArr (1)/(sqrt2) lt x le 1`