Is |sin y| <= |y| True for Every Real y?

[island-boy]

|sin y| <= |y| for every real y.

I suspect that it is.

It is easy if |y| > 1 since |sin y| <= 1.

but what about for |y| <= 1?

how do I prove this without resorting to the use of graphs?

 

[Galileo]

There are probably many, many ways to prove this.
Try it with the mean value theorem and a little imagination.

 

[island-boy]

using the mean value theorem:
Let f(y) = sin y

thenby MVT

f(y) - f(0) = f'(c) (y - 0) where c is between y and 0

getting absolute values of both sides
|f(y) - f(0)| = |y f'(c)|

this is equal to
|sin y - sin0| = |y cos(c)|

equal to
|sin y| = |y cos(c)| <= |y||cos(c)| <= |y| since |cos(c)| <= 1

would this proof suffice?

thanks by the way for mentioning MVT.