In triangle $ABC$, find maximum value of $$\sin A \cos B + \sin B \cos C + \sin C \cos A$$

We could make $\cos C = - \cos(A+B)$ and $\sin C = \sin(A+B)$.

But then we have a rather awkward expression that doesn't share the same power

$$ \sin A \cos B + \sin A \sin^{2}B + \sin B \cos^2A $$

The answer btw is not hard to be guessed, $\frac{3}{4} \sqrt{3}$, but not sure how to prove it.

Actually looks like we can solve like below:

if $A < B < C$, then $cosA > cosB > cosC$, $sinC > sinB > sinA$

Then $sinA cosB + sinB cosC + sinC cosA \leq cosAsinC + sinBcosB + sinAcosC = sinBcosB + sinB = sinB \sqrt{1-sin^2 B} + sinB$

did I do this correctly?


2 Answers 2


Let $\{\alpha,\beta,\gamma\}=\{x,y,z\},$ where $x\geq y\geq z$.

Thus, since $\cos$ decreases on$(0,\pi)$, by Rearrangement we obtain: $$\cos\alpha\sin\gamma+\cos\beta\sin\alpha+\cos\gamma\sin\beta\leq\cos{x}\sin{z}+\cos{y}\sin{y}+\cos{z}\sin{x}=$$ $$=\sin{(x+z)}+\frac{1}{2}\sin2y=\sin{y}+\frac{1}{2}\sin2y.$$ Can you end it now?

  • $\begingroup$ Yes, I just posted my own solution, 2 seconds bofore you, lol $\endgroup$
    – Mark Klein
    Dec 16, 2019 at 18:27
  • $\begingroup$ @HY.fantasy I think your solution is not so good because you can not assume $A\geq B\geq C$. The inequality is cyclic and not symmetric. See please better my solution. $\endgroup$ Dec 16, 2019 at 18:29

You can use Lagrange multipliers. You are maximizing $$ \sin A\cos B+\sin B\cos C+\sin C\cos A $$ under $$ A+B+C=\pi. $$ Using Lagrange multipliers and the identity $\cos (x+y)=\cos x\cos y-\sin x\sin y$, you easily get $$ \cos(A+B)-\lambda=\cos(B+C)-\lambda=\cos(A+C)-\lambda=0, $$ from where it follows that $\cos C=\cos A=\cos B=-\lambda$ and then necessarily (given that at least two of them are acute) $A=B=C$ and the maximum is achieved for the equilateral triangle.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .