# Maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?

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!

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]

• 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$x_1$. Dec 11, 2020 at 17:02 • Since the maximum value of$\sin$is$1$, it is impossible that the sum of three sines be any greater than$3$. Dec 11, 2020 at 17:03 • But the situation is in a triangle, so there are more restrictions. We can find a sharper bound. – user459879 Dec 11, 2020 at 17:05 • I think this question is very similar to math.stackexchange.com/questions/990418/… – user459879 Dec 11, 2020 at 17:24 • I updated my answer to give a differentiation-only option. – J.G. Dec 11, 2020 at 17:42 ## 4 Answers 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. • I dunno what is Jensen's inequality! Dec 11, 2020 at 17:16 • Essentially it is maximal when the triangle is equilateral. – user459879 Dec 11, 2020 at 17:16 • @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$A/2,\,B/2,\,C/2$. – J.G. Dec 11, 2020 at 17:18 • So what's the flaw in my method?? Dec 11, 2020 at 17:18 • Yes I think this solution goes above the OP's head for now. Nice application of Jensen 's though, kudos! – 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 $$\ge0$$ $$\implies8(z-1)\le(-2\cos y)^2\le2^2$$ $$\implies8z\le4+8$$ The equality occurs if $$\cos^2y=1\iff\sin y=0$$ and consequently $$\sin x=\mp\dfrac{\cos y}2=\mp\dfrac12$$ • Sir from where 4or(2^2) came in discriminant ? Dec 11, 2020 at 18:22 • @Shuvam, What is the maximum value of$\cos^2y\$ Dec 11, 2020 at 18:25
• Sir it's 1 ..I mean it's 1 Dec 11, 2020 at 18:26
• @Kumar,So, $$(-2\cos y)^2\le?$$ Dec 11, 2020 at 18:29
• It maximum value i.e 4..yeah! 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 $$g=x+y+z = \pi$$.

Then

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

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

• 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. Dec 11, 2020 at 17:27
• I have just read precalculus..I just know how to do simple differnation and integration! 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 $$\frac{f(A/2)+f(B/2)+f(C/2)}{3} \le f(\frac{A+B+C}{6}).$$ $$\implies \frac{\sin (A/2)+\sin(B/2)+\sin{C/2}}{3} \le \sin\frac{\pi}{6}.$$ $$\implies \sin(A/2)+\sin(B/2)+\sin(C/2) \le \frac{3}{2}$$