# Trig minimum

#### jacks

##### Well-known member
If $A+B+C=\pi$. Then Minimum value of $\cot^2(A)+\cot^2(B)+\cos^2(C)$ is

#### CaptainBlack

##### Well-known member
If $A+B+C=\pi$. Then Minimum value of $\cot^2(A)+\cot^2(B)+\cos^2(C)$ is
What have you tried?

If you put $$C=\pi-(A+B)$$ and substitute into the objective you get an unconstrained optimisation problem in two variables.

CB

#### jacks

##### Well-known member
Thanks Caption Black But I did not Understand What steps i do after that means after converting into 2 variable.

would you like to explain it to me

We Know that If $$A+B+C = \pi$$, Then $$\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)$$

Which we can prove easily

$$A+B=\pi-C\Leftrightarrow \tan(A+B) = \tan (\pi-C) = -\tan (C)$$

So $$\frac{\tan(A)+\tan(B)}{1-\tan(A).\tan(B)} = -\tan (C)$$

So $$\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)$$

Now Using $$\mathbb{A.M}\geq \mathbb{G.M}$$

$$\frac{\tan (A)+\tan(B)+\tan(C) }{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}$$

$$\frac{\tan(A).\tan(B).\tan(C)}{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}$$

$$\left(\tan(A).\tan(B).\tan(C)\right)^3\geq 27 \left(\tan(A).\tan(B).\tan(C)\right)$$

So $$\left(\tan(A).\tan(B).\tan(C)\right)\geq 3\sqrt{3}$$

bcz $$\tan(A).\tan(B).\tan(C)> 0$$

So $$\cot(A).\cot(B).\cot(C)\leq \frac{1}{3\sqrt{3}}$$

Now again Using $$\mathbb{A.M}\geq \mathbb{G.M}$$

$$\frac{\cot^2(A)+\cot^2(B)+\cot^2(C)}{3}\geq \left(\cot(A).\cot(B).\cot(C)\right)^{\frac{2}{3}}$$

So $$\cot^2(A)+\cot^2(B)+\cot^2(C)\geq 1$$

and equality hold when $$A=B=C=\frac{\pi}{3}$$

#### CaptainBlack

##### Well-known member
Thanks Caption Black But I did not Understand What steps i do after that means after converting into 2 variable.

would you like to explain it to me

We Know that If $$A+B+C = \pi$$, Then $$\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)$$

Which we can prove easily

$$A+B=\pi-C\Leftrightarrow \tan(A+B) = \tan (\pi-C) = -\tan (C)$$

So $$\frac{\tan(A)+\tan(B)}{1-\tan(A).\tan(B)} = -\tan (C)$$

So $$\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)$$

Now Using $$\mathbb{A.M}\geq \mathbb{G.M}$$

$$\frac{\tan (A)+\tan(B)+\tan(C) }{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}$$

$$\frac{\tan(A).\tan(B).\tan(C)}{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}$$

$$\left(\tan(A).\tan(B).\tan(C)\right)^3\geq 27 \left(\tan(A).\tan(B).\tan(C)\right)$$

So $$\left(\tan(A).\tan(B).\tan(C)\right)\geq 3\sqrt{3}$$

bcz $$\tan(A).\tan(B).\tan(C)> 0$$

So $$\cot(A).\cot(B).\cot(C)\leq \frac{1}{3\sqrt{3}}$$

Now again Using $$\mathbb{A.M}\geq \mathbb{G.M}$$

$$\frac{\cot^2(A)+\cot^2(B)+\cot^2(C)}{3}\geq \left(\cot(A).\cot(B).\cot(C)\right)^{\frac{2}{3}}$$

So $$\cot^2(A)+\cot^2(B)+\cot^2(C)\geq 1$$

and equality hold when $$A=B=C=\frac{\pi}{3}$$
This is a different question from the one you asked in the original post. Please clarify what the question really is.

Also as this is posted in the calculus area of MHB I would expect to see some calculus.

CB

#### CaptainBlack

##### Well-known member
So $$\cot^2(A)+\cot^2(B)+\cot^2(C)\geq 1$$

and equality hold when $$A=B=C=\frac{\pi}{3}$$
By the way equality holds when $$\cot^2(A)=\cot^2(B)=\cot^2(C)$$ which does not imply that $$A=B=C$$ unless you place some restriction on the allowed values of $$A,B$$ and $$C$$, which you have not done.

CB

#### CaptainBlack

##### Well-known member
Thanks Caption Black But I did not Understand What steps i do after that means after converting into 2 variable.

would you like to explain it to me
Given a function $$f(x,y,z)$$ for which you seek a minimum subject to a constraint $$z=g(x,y)$$, we look for the unconstrained minimum of $$h(x,y)=f(x,y,g(x,y))$$.

Such a minimum if it exists is a solution of:

\begin{aligned} \frac{\partial}{\partial x}h(x,y)=0 \\ \frac{\partial}{\partial y}h(x,y)=0 \end{aligned}

Applying this to the given problem of minimising $$f(A,B,C)= \cot^2(A)+\cot^2(B)+\cot^2(C)$$ subject to $$A+B+C=\pi$$ , gives us a condition:

$\frac{\cos(A)}{\sin^3(A)}=\frac{\cos(B)}{\sin^3(B)}=\frac{\cos(C)}{\sin^3(C)}$

and as $$\cos(x)/\sin^3(x)$$ strictly decreasing over $$(0,\pi)$$ and periodic with period $$\pi$$ this implies that $$B=A+n\pi,\ C=A+m\pi ,\ n,m \in \mathbb{Z}$$. So:

$A=\frac{\pi(1-n-m)}{3}$

(Not all of these critical point correspond to minima, in particular the cases where $$1-n-m$$ is a multiple of $$3$$ are not)

Now since $$\cot^2(x)$$ is periodic with period $$\pi$$ at a critical point we have:

$f(A,B,C)= \cot^2(A)+\cot^2(B)+\cot^2(C)=3\;\cot^2(A)=3\;\cot^2\left( \frac{\pi(1-n-m)}{3} \right)$

and a bit more checking shows that the mimimum is $$1$$ and occurs whenever $$1-n-m$$ is not a multiple of $$3$$.

Note: if we had specified that $$0\le A,B,C \le \pi$$ at the start we would have found that the critical point is $$A=B=C=\pi/3$$ without much trouble.

CB

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#### jacks

##### Well-known member
By the way equality holds when $$\cot^2(A)=\cot^2(B)=\cot^2(C)$$ which does not imply that $$A=B=C$$ unless you place some restriction on the allowed values of $$A,B$$ and $$C$$, which you have not done.

CB
Yes CaptionBlack