So I came across a question in my textbook:

In triangle ABC, if $A$,$B$,$C$ represent angles, then find the maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?

So I have already tried and best and put my blood,sweat and tears into this question..But I'm not able to go solve further!

So here's my approach: By Using $\sin(C)+\sin(D)$ and $A+B+C= \pi$;

  1. $2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+\cos(\frac{A+B}{2})$ Now, Using $\cos(2A)$ formula i.e, $1-2\sin^2(A) $
  2. $2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1-2\sin^2(\frac{A+B}{4})$
  3. So I got quadratic in variable $\sin(\frac{A+B}{4})$
  4. $-2\sin^2(\frac{A+B}{4})+2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1$

But I dunno what to do After that

Can I solve this question using this method or I have to use a different approach!

BTW, the answer is 3/2

Edit:I have just finished my high school and preparing for entrance exam IIT-JEE,So please don't use hard terms to solve this question.

This solution is sent by my teacher, atleast make me understand this one [https://i.stack.imgur.com/51pCB.png]

  • 1
    $\begingroup$ Welcome to MSE. Please use MathJax to format your posts. To begin with, surround math expressions (including numbers) with $ signs and use _ for subscripts. $x_1$ comes out as $x_1$. $\endgroup$
    – saulspatz
    Dec 11, 2020 at 17:02
  • $\begingroup$ Since the maximum value of $\sin$ is $1$, it is impossible that the sum of three sines be any greater than $3$. $\endgroup$
    – saulspatz
    Dec 11, 2020 at 17:03
  • $\begingroup$ But the situation is in a triangle, so there are more restrictions. We can find a sharper bound. $\endgroup$
    – user459879
    Dec 11, 2020 at 17:05
  • $\begingroup$ I think this question is very similar to math.stackexchange.com/questions/990418/… $\endgroup$
    – user459879
    Dec 11, 2020 at 17:24
  • $\begingroup$ I updated my answer to give a differentiation-only option. $\endgroup$
    – J.G.
    Dec 11, 2020 at 17:42

4 Answers 4


Since $\sin x$ is concave on acute $x$, by Jensen's inequality the maximum is found at $A/2=B/2=C/2=\pi/6$, as $3\sin\pi/6=3/2$.

Edit: since the OP mentioned in a comment on @B.Goddard's answer that they know differentiation, here's another proof the equilateral case achieves a maximum:

Keep using $\frac{C}{2}=\frac{\pi}{2}-\frac{A+B}{2}$. To extemize $\sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{A+B}{2}$ simultaneously solve$$\tfrac12\cos\tfrac{A}{2}-\tfrac12\cos\tfrac{C}{2}=0,\,\tfrac12\cos\tfrac{B}{2}-\tfrac12\cos\tfrac{C}{2}=0$$viz. $A=B=C$. I'll leave the reader to check it's a maximum by considering second derivatives.

  • $\begingroup$ I dunno what is Jensen's inequality! $\endgroup$ Dec 11, 2020 at 17:16
  • $\begingroup$ Essentially it is maximal when the triangle is equilateral. $\endgroup$
    – user459879
    Dec 11, 2020 at 17:16
  • $\begingroup$ @KumarShuvam I've added some links. For concave $f$, the mean of $f(x)$ is at most $f$ evaluated at the mean of $x$.(If you draw a diagram, you can probably see why this is quite intuitive; for a full proof of the convex case, see here.) In this case, we average over $A/2,\,B/2,\,C/2$. $\endgroup$
    – J.G.
    Dec 11, 2020 at 17:18
  • 1
    $\begingroup$ So what's the flaw in my method?? $\endgroup$ Dec 11, 2020 at 17:18
  • $\begingroup$ Yes I think this solution goes above the OP's head for now. Nice application of Jensen 's though, kudos! $\endgroup$
    – user459879
    Dec 11, 2020 at 17:19

Where you have stopped, let $$z=-2\sin^2x+1+2\sin x\cos y$$

$$\iff2\sin^2x-2\sin x\cos y+z-1=0$$

As $\sin x$ is real, the discriminant must be $\ge0$

$\implies8(z-1)\le(-2\cos y)^2\le2^2$


The equality occurs if $\cos^2y=1\iff\sin y=0$

and consequently $\sin x=\mp\dfrac{\cos y}2=\mp\dfrac12$

  • $\begingroup$ Sir from where 4or(2^2) came in discriminant ? $\endgroup$ Dec 11, 2020 at 18:22
  • $\begingroup$ @Shuvam, What is the maximum value of $\cos^2y$ $\endgroup$ Dec 11, 2020 at 18:25
  • $\begingroup$ Sir it's 1 ..I mean it's 1 $\endgroup$ Dec 11, 2020 at 18:26
  • $\begingroup$ @Kumar,So, $$(-2\cos y)^2\le?$$ $\endgroup$ Dec 11, 2020 at 18:29
  • $\begingroup$ It maximum value i.e 4..yeah! $\endgroup$ Dec 11, 2020 at 18:31

You can do it with Lagrange multipliers. Maximize $f=\sin x/2 + \sin y/2+\sin z/2$ under the constraint $g=x+y+z = \pi$.


$$\nabla f = \langle \cos(x/2)/2, \cos(y/2)/2, \cos(z/2)/2 \rangle =\lambda\langle 1,1,1 \rangle = \nabla g.$$

This shows that $x=y=z$ and the maximal triangle is equilateral.

  • $\begingroup$ Lagrange multipliers is a Multivariable Calculus technique. It's nice if you have the background, but if you don't understand it, don't worry, it's about 3 or 4 courses for you in the future. $\endgroup$ Dec 11, 2020 at 17:27
  • $\begingroup$ I have just read precalculus..I just know how to do simple differnation and integration! $\endgroup$ Dec 11, 2020 at 17:30

In a triangle ABC, $A+B+C=\pi$ $$f(x)=\sin(x/2) \implies f''(x)=-\frac{1}{4}\sin(x/2)<0, x\in[0,2\pi].$$ So by Jemsen's inequality $$\frac{f(A/2)+f(B/2)+f(C/2)}{3} \le f(\frac{A+B+C}{6}).$$ $$\implies \frac{\sin (A/2)+\sin(B/2)+\sin{C/2}}{3} \le \sin\frac{\pi}{6}.$$ $$\implies \sin(A/2)+\sin(B/2)+\sin(C/2) \le \frac{3}{2}$$

  • $\begingroup$ We already tried that. Apparently, that method is too advanced. $\endgroup$
    – J.G.
    Dec 11, 2020 at 17:37

You must log in to answer this question.

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