Question

Show that for all x and y, absolute value of sin(x)-sin(y) is less than or equal to the absolute value of x-y
\[\left| \sin(x)-\sin(y) \right|\le \left| x-y \right|\]

Answer 1

The mean value theorem is a good way to go.

Answer 2

i got it

Answer 3

the Mean Value Theorem
\[\frac{f(x) - f(y)} {x - y} = f '(c)\] for some c
==> \[\frac{\sin x - \sin y}{x - y} = \cos c\] for some c in (y, x).

Taking absolute values
\[\frac{|\sin x - \sin y|}{ |x - y| }= |\cos c| \le 1.\]

Therefore,
\[|\sin x - \sin y| \le |x - y|.\]

Answer 4

for the record kazuo36, it's bad form to delete your questions after they're answered. They should remain here as a resource for others. So leave this one here please.

Answer 5

oh sorry will do