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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!

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6 Answers 6

up vote 6 down vote accepted

$$\frac{1-(-\cos\theta)}{2-(-\sin\theta)}$$ 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 $$1=\frac{|a\cdot 0-0-2a+1|}{\sqrt{a^2+1^2}}\iff \sqrt{a^2+1}=|-2a+1|$$$$\iff a^2+1=(-2a+1)^2\iff a=0,\frac 43.$$

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....... I'm afraid I don't understand? –  Gummy bears Aug 26 at 15:06
    
@Gummybears: $x=-\sin\theta,y=-\cos\theta$ satisfies $x^2+y^2=1$, doesn't it? –  mathlove Aug 26 at 15:09
    
Yep it does. And that would be the equation of a circle? –  Gummy bears Aug 26 at 15:11
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@Gummybears: No worry. I added a bit. Take a look. –  mathlove Aug 26 at 15:18
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@Gummybears: Again, I added a bit. I hope this helps. –  mathlove Aug 26 at 15:26

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 :)

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The range of $\sin\theta$ and $\cos\theta$ is from -1 to 1. –  Gummy bears Aug 26 at 15:05
    
Thanks for telling. It was typo –  Shivang jindal Aug 26 at 15:06
    
I still am not able to get the answer? Using your final formula, doesn't the range come from -10 to 0?? –  Gummy bears Aug 26 at 15:25
    
@Gummybears Precisely, and these values are all $\leq0$. –  Jean-Claude Arbaut Aug 26 at 15:27
    
And what about 4/3?? –  Gummy bears Aug 26 at 15:30

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:$$\frac{1}{0+\frac{3}{4}}=\frac{4}{3}$$

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Ummmm..... I lost you. What exactly are those formulas for? –  Gummy bears Aug 26 at 15:34
    
They are the half agle formulas where $t=\tan(\theta/2)$ –  Mufasa Aug 26 at 15:35
    
Ohhh... I see. Slightly more complicated as I see it, as it requires the use of memorized formulas. However, also a good method :D –  Gummy bears Aug 26 at 15:36

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.

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Differentiating this isn't the best option, or the easiest one at that. –  Gummy bears Aug 26 at 15:07

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 $R\ge0$

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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

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