Another POTW proposal

Prove that \(|\sin(nx)|\le n|\sin(x)|\) for all \(x \in \mathbb{R}\) and \(n \in \mathbb{N}_+\).

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Solution: This is just an excesise in induction.

The base case \(n=1\) holds obviously and trivialy.

Now suppose that for some \(k \in \mathbb{N}_+\):

\[|\sin(kx)| \le k |\sin(x)|\].

Now consider:

\[|\sin((k+1)x)|=| \sin(kx)\cos(x)+\cos(kx)\sin(x)| \].


Then by the triangle inequality we get:

\[|\sin((k+1)x)|\le |\sin(kx)\cos(x)|+|\cos(kx)\sin(x)| \\ \phantom{[ \sin((k+1)x)xxx}\le |\sin(kx)| + |\sin(x)|=(k+1)|\sin(x)|\].

QED