Q.390. Application of Mean Value Theorem

Question 390.
Using Lagrange's mean value theorem proove that :
x < sin ^-1 x < x / [square root of (1-x^2) ] for 0 < x < 1.

Answer 390.
Let f (x) = sin^-1 x
f (x) is continuous in [0, 1] and differentiable in (0, 1)
Hence, MVT is applicable and
f (a + h) = f (a) + h f '(a + θh), 0 < θ < 1
Taking a = 0, h = x and f '(x) = 1/√(1 - x^2)
=> sin^-1 x = x / √(1 - θ^2x^2)
=> √(1 - θ^2x^2) = x/sin^-1 x ... ( 1 )

0 < θ < 1
=> 0 < θ^2 x^2 < x^2
=> 0 > - θ^2 x^2 > - x^2
=> - x^2 < - θ^2 x^2 < 0
=> 1 - x^2 < 1 - θ^2 x^2 < 1
=> √(1 - x^2) < √(1 - θ^2 x^2) < 1

Plugging √(1 - θ^2x^2) = x/sin^-1 x from ( 1 ),
=> √(1 - x^2) < x/sin^-1 x < 1
=> 1/√(1 - x^2) > sin^-1x / x > 1
=> 1 < sin^-1 x / x < 1/√(1 - x^2)
=> x < sin^-1 x < x/√(1 - x^2).

Link to YA!