# 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$$\frac{\tan \dfrac{θ}{2}}{\dfrac{θ}{2}} \leqslant f\left( \frac{π}{4} \right) = \frac{4}{π}.$$