Proving $1 - \frac{2 \vartheta}{\pi} \sin \vartheta \leq \cos \vartheta$, for $\vartheta \in [0, \frac{\pi}{2}]$ For my thesis I need the inequality $1 - \frac{2 \vartheta}{\pi} \sin \vartheta \leq 2 \cos \vartheta$ for $\vartheta \in [0, \frac{\pi}{2}]$ which can be proved by exploiting the fact that $\cos \vartheta$ is concave on $[0, \frac{\pi}{2}]$.
When I plotted the graph of $1 - \frac{2\vartheta}{\pi} \sin \vartheta$ and the graph of $\cos \vartheta$ I noticed that indeed the stronger inequality
$$1 - \frac{2 \vartheta}{\pi} \sin \vartheta \leq \cos \vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway. 
What I have tried:


*

*Trying to find the zeros of $h(\vartheta)=\cos \vartheta - 1 + \frac{2 \vartheta}{\pi} \sin \vartheta$.

*Writing down the Taylor-expansion of $h(\vartheta)$ and comparing the positive and the negative terms.


Both approaches ended up in a mess. Does anyone  have an idea on how to prove this?

The inequality obviously holds for $θ = 0$. For $θ \in \left(0, \dfrac{π}{2}\right]$, note that\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}}{} 1 - \frac{2}{π} θ \sin θ \leqslant \cos θ\\
&\Longleftrightarrow 2\sin^2 \frac{θ}{2} = 1 - \cos θ \leqslant \frac{2}{π} θ \sin θ = \frac{4}{π} θ \sin\frac{θ}{2} \cos\frac{θ}{2}\\
&\Longleftrightarrow \frac{\tan \dfrac{θ}{2}}{\dfrac{θ}{2}} \leqslant \frac{4}{π}.
\end{align*}
Define $f(t) = \dfrac{\tan t}{t}$ for $t \in \left(0, \dfrac{π}{4}\right]$, then $f'(t) = \dfrac{2t - \sin 2t}{2t^2 \cos^2 t} \geqslant 0$, which implies that$$
\frac{\tan \dfrac{θ}{2}}{\dfrac{θ}{2}} \leqslant f\left( \frac{π}{4} \right) = \frac{4}{π}.
$$
