If `x in [1, 0)`, then find the value of `cos^(-1) (2x^(2) - 1) - 2 sin^(-1) x`

If `x in [1, 0)`, then find the value of `cos^(-1) (2x^(2) - 1) - 2 si

1.	If `x in [1, 0)`, then find the value of `cos^(-1) (2x^(2) - 1) - 2 sin^(-1) x`
Answer» Let `x = cso theta, " for " x in [-1, 0), theta in (pi//2, pi]` `rArr cos^(-1) (2x^(2) -1) - 2 sin^(-1) x` `= cos^(-1) (2 cos^(2) theta -1) -2 sin^(-1) (cos theta)` `= cos^(-1) (cos 2 theta) - 2 sin^(-1) (sin(pi//2 - theta))` `= cos^(-1) (cos 2 theta) - 2 ((pi)/(2) - theta)` `= 2 pi - 2 theta + 2 theta - pi` `= pi`

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