# Prove that 0 &#x2264;<!-- ≤ --> 1 + cos &#x2061;<!-- ⁡ -->

Prove that $0\le \frac{1+\mathrm{cos}\theta }{2+\mathrm{sin}\theta }\le \frac{4}{3}$ for all real $\theta$
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Elianna Wilkinson
Using half angle formula we convert it to tan function. (Hoping you know the formulae for sin,cos to convert it to $\mathrm{tan}\left(x/2\right)$). So we get it as
$\frac{1}{1+{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)+\mathrm{tan}\left(\frac{x}{2}\right)}$
The maximum of denominator is obvious which is infinity so minimum of the whole function is 0. Now we use calculus to find the minimum of the denominator. Differentiating only the denominator we have $2\mathrm{tan}\left(\frac{x}{2}\right)=-1$ knowing that we can cancel out ${\mathrm{sec}}^{2}\left(\frac{x}{2}\right)$ as it is never 0. Thus minimum is achieved at $\mathrm{tan}\left(\frac{x}{2}\right)=-1/2$ . Verify using the second derivative test. Thus min of denominator is maximum of the function . Substituting −1/2 back we get the max as $\frac{4}{3}$ as desired.