Use the Mean Value Theorem and the fact that for f(x) = cos(x), f′(x) = −sin(x), to prove that, for x, y ∈ R,
| cos x − cos y| ≤ |x − y|.
f(x) = cos(x) is continuous on R & differentiable on R, hence, it is continuous and differentiable on any interval in R.
By mean value theorem on the interval [x,y] for the function f, we have,
{f(x) - f(y)}/(x-y) = f'(z) for some z in (x,y)
Now, f'(z) = -sin(z) and, |f'(z)| = |-sin(z)| = |sin(z)| 1
So, |{f(x) - f(y)}/(x-y)| 1
So, |f(x) - f(y)| |x-y|
So, |cos(x) - cos(y)| |x - y| for all x, y in R
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