Taylor polynomials of $\sin(x)$ are, for $x > 0,$ alternatively upper resp. lower bounds?

Question

For $x > 0$, I observed:

• $\sin(x) < x$
• $\sin(x) > x - x^3 \mathbin{/} 3!$
• $\sin(x) < x - x^3 \mathbin{/} 3! + x^5 \mathbin{/} 5!$
• etc.

Is it true that the Taylor polynomials of $\sin(x)$ are, for $x > 0$, alternatively upper resp. lower bounds? If yes, how to prove it?