Using mean value theorem show that: $x< sin^-1x$ , for $x>0$ ?

Question

Using mean value theorem show that:
$x< sin^-1x$ , for $x>0$

15473 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-21T15:26:33

See below.

According to the mean value theorem, there exists $0 < zeta < x < 1$

such that

$(d/(dx)sin^-1(x))(zeta) = (sin^-1(x)-sin^-1(0))/(x-0) = (sin^-1(x))/x$

but

$(d/(dx)sin^-1(x))(zeta)=1/sqrt(1-zeta^2) > 1$

finally

$(sin^-1(x))/x = 1/sqrt(1-zeta^2) > 1$ so

$sin^-1(x) > x$ or $x < sin^-1(x)$