 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, givingThe cosine function is decreasing on $[0,\pi ]$, so $\cos x\le \cos \xi \le 1$. Multiplying by the nonnegative factor $x^5/120$ and adding $x-x^3/6$ proves both required bounds. At $x=0$ both are equalities.