Range of a trigonometric function Question:
Prove that:
$$0 \leq \frac{1 + \cos\theta}{2 + \sin\theta}\leq \frac{4}{3}$$
I have absolutely no idea how to proceed in this question. Please help me!
 A: Since $\sin(\theta) , \cos(\theta) \in [-1,1]$ LHS part is obviously true.
for RHS part we have to prove 
$$ \frac{1+\cos(\theta}{2+\sin(\theta)} - \frac{4}{3} \le 0 $$ 
which is equivalent to, 
$$ \frac{3\cos(\theta)-4\sin(\theta)-5}{2+\sin(\theta)} \le 0$$ 
So it suffice to show that $$ 3\cos(\theta)-4\sin(\theta)-5 \le 0$$ since $2+\sin(\theta)$ will always be positive.
This is well known result that, 
$$ -\sqrt{a^2+b^2} \le a\cos(\theta)+b\sin(\theta) \le \sqrt{a^2+b^2}$$ 
Apply here, and we are done :)
A: $$\frac{1-(-\cos\theta)}{2-(-\sin\theta)}$$ represents the slope of a line on which $(2,1)$ and $(-\sin\theta,-\cos\theta)$ exist.
Since $x=-\sin\theta,y=-\cos\theta$ satisfies $x^2+y^2=1$, we know that a point $(-\sin\theta,-\cos\theta)$ is on the circle whose center is the origin with the radius $1$. 
Now, you can draw both the circle and a point $(2,1)$, and can see the followings :
The maximum or the minimum of the slope is given when a line on which a point $(2,1)$ exists is tangent to the circle. I hope this helps!
This means that the distance between the origin and a line $$y-1=a(x-2)\iff ax-y-2a+1=0$$ on which a point $(2,1)$ exists equals $1$ (here, $a$ is the slope we are talking about). Hence, we have
$$1=\frac{|a\cdot 0-0-2a+1|}{\sqrt{a^2+1^2}}\iff \sqrt{a^2+1}=|-2a+1|$$$$\iff a^2+1=(-2a+1)^2\iff a=0,\frac 43.$$
A: Another way of tackling this would be to make use of the half angle formulas for cos and sin:$$\sin(\theta)=\frac{2t}{1+t^2}$$$$\cos(\theta)=\frac{1-t^2}{1+t^2}$$this leads to:$$\frac{1+\cos(\theta)}{2+\sin(\theta)}=\frac{1}{t^2+t+1}=\frac{1}{(t+\frac{1}{2})^2+\frac{3}{4}}$$and from this one can observe that the maximum occurs when $(t+\frac{1}{2})$ is zero giving a maximum of:$$\frac{1}{0+\frac{3}{4}}=\frac{4}{3}$$
A: Let $\displaystyle\dfrac{1+\cos\theta}{2+\sin\theta}=R$
As $\displaystyle0\le\sin\theta,\cos\theta\le1, R\ge0$
$\displaystyle\iff1+\cos\theta=2R+R\sin\theta\iff\sqrt{1+R^2}\cos\left(\theta+\arctan R\right)=2R-1$
Now, $\displaystyle-1\le\cos\left(\theta+\arctan R\right)\le1\iff -1\le\frac{2R-1}{\sqrt{1+R^2}}\le1$
$\displaystyle\implies0\le\frac{(2R-1)^2}{1+R^2}\le1$
$\displaystyle\iff4R^2-4R+1\le1+R^2\iff R\left(R-\frac34\right)\le0\iff0\le R\le\frac43$ as $R\ge0$
A: Let $\displaystyle\dfrac{1+\cos\theta}{2+\sin\theta}=R$
As $\displaystyle0\le\sin\theta,\cos\theta\le1, R\ge0$
Using Weierstrass Substitution, setting $\displaystyle c=\cot\frac\theta2$
We have $$R=\frac{1+\dfrac{c^2-1}{c^2+1}}{2+\dfrac{2c}{c^2+1}}$$
$$\iff c^2=R(c^2+c+1)\iff c^2(R-1)+Rc +R=0$$ which is a Quadratic Equation in $c$
As $\theta$ is real, so is $\displaystyle c=\cot\frac\theta2\implies$ the discriminant must be $\ge0$
i.e., $\displaystyle R^2\ge4(R-1)R\iff R\left(R-\frac43\right)\ge0$ which conforms to my other answer
A: Let 
$$
f(\theta) = \frac{1+\cos \theta}{2+\cos \theta}.
$$
You are asked to prove that $\min f=0$ and $\max f=4/3$. You can find maximum and minimum values by studying the sign of the derivative.
