= Solution
For $0<x<\pi$, apply <Taylor's theorem> to the <sine> about zero, retaining terms through degree four. Its fourth-order coefficient is zero, and its fifth <derivative> is the <cosine function>, giving
$$
\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}\cos\xi,\qquad 0<\xi<x.
$$
The <cosine function> is decreasing on $[0,\pi]$, so $\cos x\leq\cos\xi\leq1$. Multiplying by the nonnegative factor $x^5/120$ and adding $x-x^3/6$ proves \b[both required bounds]. At $x=0$ both are equalities.
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