# Thread: [SOLVED] Prove the range of value of cos A + cos B + cos C?

1. ## [SOLVED] Prove the range of value of cos A + cos B + cos C?

Prove that the value of cos A + cos B + cos C lies between 1 & 3/2 where A + B + C = π?

2. Originally Posted by fardeen_gen
Prove that the value of cos A + cos B + cos C lies between 1 & 3/2 where A + B + C = π?
Is thist

$\displaystyle 1\leq\cos\left(A\right)+\cos\left(B\right)+\cos\le ft(C\right)\leq\frac{3}{2}$

Where $\displaystyle A+B+C=\pi$

3. This is easiest using partial derivatives.

Since $\displaystyle C=\pi-A-B$ just plug in to get:

$\displaystyle \cos(A)+\cos(B)+\cos(\pi-A-B)=\cos(A)+\cos(B)-\cos(A+B)$

So define

$\displaystyle f(A,B)=\cos(A)+\cos(B)-\cos(A+B)$

Note that this function is $\displaystyle 2\pi$ periodic in both variables so we can restrict ourselves to finding the maxima and minima over that square. If you know what partial derivatives are:

just calculate $\displaystyle \frac{\partial}{\partial A}f$ and $\displaystyle \frac{\partial}{\partial B}f$

Set them both to 0, and solve (make sure you get ALL the solutions in $\displaystyle [0,2\pi)$. Once you have all the critical points (there are no boundaries (well we have imposed some, however, they don't matter since the function is periodic at our boundaries) or places where the derivative does not exist) just evaluate $\displaystyle f$ at the critical points. Since the function in continuous, the largest value(s) of $\displaystyle f$ is the global maximum, and the smallest is the global minimum. Therefore you just found upper and lower bounds for your expression.

If you don't know what partial derivatives are, let me know, there is another way of explaining that does the same thing.

Make sense?

4. Originally Posted by meymathis

Set them both to 0, and solve (make sure you get ALL the solutions in $\displaystyle [0,2\pi)$.
Make sense?
For clarification, by this do you mean

Find all $\displaystyle (a,b)\backepsilon~f_A=f_B=0$

Where $\displaystyle a\in[0,2\pi)\quad\text{and}\quad{b\in[0,2\pi)}$?

5. Originally Posted by Mathstud28
For clarification, by this do you mean

Find all $\displaystyle (a,b)\backepsilon~f_A=f_B=0$

Where $\displaystyle a\in[0,2\pi)\quad\text{and}\quad{b\in[0,2\pi)}$?
Thank you, yes. Although just to avoid confusion I would say:
Find all $\displaystyle (a,b)\backepsilon~f_A=0 \text{ and }f_B=0$

Where $\displaystyle a\in[0,2\pi)\text{ and }b\in[0,2\pi)$.

I say that just so that no one thinks that we should be setting f_A=f_B.

By the way, as a check that you've got it, the critical points are:
$\displaystyle A=0$ ($\displaystyle B$ could be anything in the interval $\displaystyle [0,2\pi)$)

$\displaystyle B=0$ ($\displaystyle A$ A could be anything in the interval $\displaystyle [0,2\pi)$)

and $\displaystyle A=B=\pi/3$

6. It looks like $\displaystyle A, B, C$ are supposed to be the interior angles of a triangle.

For the record, I attempted what Meymathis suggested here:

http://www.mathhelpforum.com/math-he...nequality.html

However, ThePerfectHacker said the method wasn’t very good – so I suppose there’s some better, non-calculus way of doing it.

7. Originally Posted by meymathis
Thank you, yes. Although just to avoid confusion I would say:
Find all $\displaystyle (a,b)\backepsilon~f_A=0 \text{ and }f_B=0$

Where $\displaystyle a\in[0,2\pi)\text{ and }b\in[0,2\pi)$.

I say that just so that no one thinks that we should be setting f_A=f_B.

By the way, as a check that you've got it, the critical points are:
$\displaystyle A=0$ ($\displaystyle B$ could be anything in the interval $\displaystyle [0,2\pi)$)

$\displaystyle B=0$ ($\displaystyle A$ A could be anything in the interval $\displaystyle [0,2\pi)$)

and $\displaystyle A=B=\pi/3$
Yes, I suppose one could misconstrue my original for solving $\displaystyle f_A=f_B$ instead of solving the simultaneous equation $\displaystyle f_A=0$ and $\displaystyle f_B=0$

8. Originally Posted by JaneBennet
It looks like $\displaystyle A, B, C$ are supposed to be the interior angles of a triangle.

For the record, I attempted what Meymathis suggested here:

http://www.mathhelpforum.com/math-he...nequality.html

However, ThePerfectHacker said the method wasn’t very good – so I suppose there’s some better, non-calculus way of doing it.
Looking over that post I have 3 comments.

1) I think what you did was just fine. You changed the constrained (on a plane) min/max of a surface of 3 variables into a min/max of a surface of 2 variables on a square (except I think your domain is too small - at least without justification). I am making use of the Extreme Value Theorem without specifically naming it. I argued above why you don't need to worry about the boundaries. It's a bit hand-wavy so to really firm up the argument:

Close the domain to $\displaystyle A,B\in[0,2\pi]$. Now you have a continuous function on a compact set. By EVT must be global max/min must be local max/min or else on the boundary. The boundaries are easy to check. You get 1. The critical points you get now happen to be the boundaries (which you already checked) and $\displaystyle A=B=\pi/3$. Check that one. Since the extreme global min/max must occur on the boundaries and/or critical points, and you just found that the smallest value the function takes on that set is 1 and the largest is 3/2, you have your inequality.

2) Because of the above, there really was no need to take second derivatives.

3) I'm sure you can certainly prove it without calculus. This was just the way that occured to me first. It's a pretty nice general way of showing these things. For example, it is a nice way to show that the geometric mean of positive numbers is less that the arithmetic mean.

9. Originally Posted by fardeen_gen
Prove that the value of cos A + cos B + cos C lies between 1 & 3/2 where A + B + C = π?
As JaneBennet said, A, B and C are assumed to be the interior angles of a triangle.

since $\displaystyle \cos C = -\cos(A+B)=1-2 \cos^2 \left(\frac{A+B}{2} \right),$ we'll have:

$\displaystyle \cos A + \cos B + \cos C = 2\cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A-B}{2} \right) -2\cos^2 \left(\frac{A+B}{2} \right) + 1 \ \ \ \ \ (*)$

$\displaystyle =2 \cos \left(\frac{A+B}{2}\right) \left(\cos \left(\frac{A-B}{2}\right) - \cos \left(\frac{A+B}{2}\right) \right) + 1$

$\displaystyle =4 \cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A}{2} \right) \sin \left(\frac{B}{2} \right) + 1$
$\displaystyle =4 \sin \left(\frac{A}{2} \right) \sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2}\right) + 1 \geq 1.$

for the upper bound: since $\displaystyle \cos \left(\frac{A+B}{2}\right)=\sin \left(\frac{C}{2}\right),$ by $\displaystyle (*)$ we have:

$\displaystyle \cos A + \cos B + \cos C \leq 2 \sin \left(\frac{C}{2} \right) - 2 \sin^2 \left(\frac{C}{2} \right) + 1=\frac{3}{2}-2 \left(\sin \left(\frac{C}{2} \right) - \frac{1}{2} \right)^2 \leq \frac{3}{2}. \ \ \ \square$

Edit: After few times editing and deleting once, it's finally perfect! That was a "typing" nightmare!

10. Actually this problem is pretty popular and has a number of solutions.

Read page number 13, Theorem 2.1.2 for a couple of methods(that includes TPH's suggestion in JaneBennet's thread) in Hojoo Lee's Book.

I have attached the pdf.