Prove that $0 \le \frac{1+\cos\theta}{2+\sin\theta} \le \frac{4}{3}$ for all real $\theta$. I've tried substituting $1+\cos\theta=2\cos^2(\frac{\theta}{2})$ but it didn't give me the answer. 
 A: Using half angle formula we convert it to tan function. (Hoping you know the formulae for sin,cos to convert it to $\tan (x/2) $). So we get it as $$\frac {1}{1+\tan^2 (\frac {x}{2})+\tan (\frac {x}{2})}$$. 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 $2\tan (\frac {x}{2})=-1$ knowing that we can cancel out $\sec^2 (\frac {x}{2}) $ as it is never $0$. Thus minimum is achieved at $\tan (\frac {x}{2})=-1/2$ . 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.
A: Let $\dfrac{1+\cos\theta}{2+\sin\theta}=y$
$$\iff2y-1=\cos\theta-y\sin\theta=\sqrt{y^2+1}\cos\left(\theta+\arccos\dfrac1{\sqrt{y^2+1}}\right)$$
$$-\sqrt{y^2+1}\le2y-1\le\sqrt{y^2+1}$$
$$\implies(2y-1)^2\le y^2+1\iff0\ge3y^2-4y=3y\left(y-\dfrac43\right)$$
$$\implies0\le y\le\dfrac43$$
