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Trig minimum

jacks

Well-known member
Apr 5, 2012
226
If $A+B+C=\pi$. Then Minimum value of $\cot^2(A)+\cot^2(B)+\cos^2(C)$ is
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
Apr 5, 2012
226
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 [tex]A+B+C = \pi[/tex], Then [tex]\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)[/tex]


Which we can prove easily


[tex]A+B=\pi-C\Leftrightarrow \tan(A+B) = \tan (\pi-C) = -\tan (C)[/tex]


So [tex]\frac{\tan(A)+\tan(B)}{1-\tan(A).\tan(B)} = -\tan (C)[/tex]


So [tex]\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)[/tex]


Now Using [tex]\mathbb{A.M}\geq \mathbb{G.M}[/tex]


[tex]\frac{\tan (A)+\tan(B)+\tan(C) }{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}[/tex]


[tex]\frac{\tan(A).\tan(B).\tan(C)}{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}[/tex]


[tex]\left(\tan(A).\tan(B).\tan(C)\right)^3\geq 27 \left(\tan(A).\tan(B).\tan(C)\right)[/tex]


So [tex]\left(\tan(A).\tan(B).\tan(C)\right)\geq 3\sqrt{3}[/tex]


bcz [tex]\tan(A).\tan(B).\tan(C)> 0[/tex]


So [tex]\cot(A).\cot(B).\cot(C)\leq \frac{1}{3\sqrt{3}}[/tex]


Now again Using [tex]\mathbb{A.M}\geq \mathbb{G.M}[/tex]


[tex]\frac{\cot^2(A)+\cot^2(B)+\cot^2(C)}{3}\geq \left(\cot(A).\cot(B).\cot(C)\right)^{\frac{2}{3}}[/tex]


So [tex]\cot^2(A)+\cot^2(B)+\cot^2(C)\geq 1[/tex]


and equality hold when [tex]A=B=C=\frac{\pi}{3}[/tex]
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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 [tex]A+B+C = \pi[/tex], Then [tex]\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)[/tex]


Which we can prove easily


[tex]A+B=\pi-C\Leftrightarrow \tan(A+B) = \tan (\pi-C) = -\tan (C)[/tex]


So [tex]\frac{\tan(A)+\tan(B)}{1-\tan(A).\tan(B)} = -\tan (C)[/tex]


So [tex]\tan (A)+\tan(B)+\tan(C) = \tan(A).\tan(B).\tan(C)[/tex]


Now Using [tex]\mathbb{A.M}\geq \mathbb{G.M}[/tex]


[tex]\frac{\tan (A)+\tan(B)+\tan(C) }{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}[/tex]


[tex]\frac{\tan(A).\tan(B).\tan(C)}{3}\geq \left(\tan(A).\tan(B).\tan(C)\right)^{\frac{1}{3}}[/tex]


[tex]\left(\tan(A).\tan(B).\tan(C)\right)^3\geq 27 \left(\tan(A).\tan(B).\tan(C)\right)[/tex]


So [tex]\left(\tan(A).\tan(B).\tan(C)\right)\geq 3\sqrt{3}[/tex]


bcz [tex]\tan(A).\tan(B).\tan(C)> 0[/tex]


So [tex]\cot(A).\cot(B).\cot(C)\leq \frac{1}{3\sqrt{3}}[/tex]


Now again Using [tex]\mathbb{A.M}\geq \mathbb{G.M}[/tex]


[tex]\frac{\cot^2(A)+\cot^2(B)+\cot^2(C)}{3}\geq \left(\cot(A).\cot(B).\cot(C)\right)^{\frac{2}{3}}[/tex]


So [tex]\cot^2(A)+\cot^2(B)+\cot^2(C)\geq 1[/tex]


and equality hold when [tex]A=B=C=\frac{\pi}{3}[/tex]
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
Jan 26, 2012
890
So [tex]\cot^2(A)+\cot^2(B)+\cot^2(C)\geq 1[/tex]


and equality hold when [tex]A=B=C=\frac{\pi}{3}[/tex]
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
Jan 26, 2012
890
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
Apr 5, 2012
226
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