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#

15085 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)#