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How do I find the range of :

$$ \dfrac{\sin(\alpha +\beta +\gamma )}{\sin\alpha + \sin\beta + \sin\gamma} $$ Where, $$ \alpha , \beta\; and \;\gamma \in \left(0, \frac{\pi}{2}\right) $$

I tried using jensen's inequality on $\alpha, \beta \;and\; \gamma $, and got : $$ \frac{\sin\alpha+\sin\beta+\sin\gamma}{3} \;\le\; \sin\left(\frac{\alpha + \beta+\gamma}{3}\right) $$ $$ \implies \sin\alpha+\sin\beta+\sin\gamma \;\le\; 3\sin\left(\frac{\alpha + \beta+\gamma}{3}\right) $$ Now if I assume that $\alpha+\beta+\gamma \; \le\; \pi $ then I can say: $$ \frac{\sin\left(\alpha + \beta + \gamma\right)}{\sin\alpha+\sin\beta+\sin\gamma}\; \ge\; \frac{\sin\left(\alpha + \beta + \gamma\right)}{3\sin\left(\frac{\alpha + \beta+\gamma}{3}\right)}$$ Solving this I got: $$ \frac{\sin\left(\alpha + \beta + \gamma\right)}{\sin\alpha+\sin\beta+\sin\gamma}\; \ge\; 1\; - \; \frac{4}{3}\sin^2\left(\frac{\alpha + \beta+\gamma}{3}\right) $$ $$\implies \frac{\sin\left(\alpha + \beta + \gamma\right)}{\sin\alpha+\sin\beta+\sin\gamma}\; > \; -\frac{1}{3}$$

But this is not useful as it is based on the assumption that $\alpha+\beta+\gamma \le \pi $.

So , I would like someone to provide me with a solution to my problem (I am not familiar with methods of higher mathematics so it would be helpful if these are not used, however use of complex numbers or any other basic inequalities in the solution is much appreciated ).

It would also be helpful if I can get more than one solution.

Thank you.

Bipul Kumar
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3 Answers3

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Using

$\displaystyle \sin(\alpha+\beta+\gamma)=\sin\alpha\cos\beta\cos\gamma+\cos\alpha\sin\beta\cos\gamma+\cos\alpha\cos\beta\sin\gamma-\sin\alpha\sin\beta\sin\gamma$

$\displaystyle \Longrightarrow \sin(\alpha+\beta+\gamma)-\sin\alpha-\sin\beta-\sin\gamma=\sin\alpha(\underbrace{\cos\beta\cos\gamma-1}_{<0})+\sin\beta(\underbrace{\cos\alpha\cos\gamma-1}_{<0})+\sin\gamma(\underbrace{\cos\alpha\cos\beta-1}_{<0})-\sin\alpha\sin\beta\sin\gamma<0$

$\displaystyle \sin(\alpha+\beta+\gamma)<\sin\alpha+\sin\beta+\sin\gamma$

$\displaystyle \frac{\sin(\alpha+\beta+\gamma)}{\sin\alpha+\sin\beta+\sin\gamma}<1$

jacky
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Well, idk if this answer is rigorous enough, but I will give it a shot

$\alpha, \beta, \gamma \in (0, \frac{\pi}{2}) \implies \alpha+\beta+\gamma \in (0, \frac{3\pi}{2}) \implies sin(\alpha+\beta+\gamma) \in (-1, 1] $

and also

$\alpha, \beta, \gamma \in (0, \frac{\pi}{2}) \implies sin(\alpha)+sin(\beta)+sin(\gamma) \in (0, 3) $

We can see that $sin(\alpha+\beta+\gamma) > -1$ and we can conclude that it equals -1 when $\alpha = \beta = \gamma = \frac{\pi}{2}$

These values for $\alpha,\beta,\gamma$ also happen to be the value where $sin(\alpha)+sin(\beta)+sin(\gamma)$ attain their maximum value of 3. In other words, the least value of $\frac{sin(\alpha+\beta+\gamma)}{sin(\alpha)+sin(\beta)+sin(\gamma)} = \frac{-1}{3}$, but it never quite reaches this value and is always greater than it, but never equal to it.

Thus we can conclude,

$$\frac{sin(\alpha+\beta+\gamma)}{sin(\alpha)+sin(\beta)+sin(\gamma)} > \frac{-1}{3}$$

Hope it helps!

Anirudh Panguluri
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  • Yes i guess this works well but i would like to know if there is an upper bound of range for the given expression because the original question was actually to find whether the given expression is less than 1. – Bipul Kumar Oct 02 '23 at 07:55
  • Oh, I don't think there will be an upper bound, you can easily verify it by taking numerical examples but also by identifying the fact that, if sin(a) + sin(b) + sin(c) = 0, then sin(a+b+c) isn't necessarily equal to 0, so you can take a case sin(a) + sin(b) + sin(c) becomes smaller and smaller approaching 0, and sin(a+b+c) is some positive, and thus the expression grows to infinity without bounds. – Anirudh Panguluri Oct 02 '23 at 08:05
  • But if sin(a) + sin(b) + sin(c) approaches zero then that must mean that a, b and c approach 0 given the condition, so in that case the given expression would actually tend towards 1 right ? – Bipul Kumar Oct 02 '23 at 08:09
  • @BipulKumar Oh my bad, I hadn’t considered the condition when I wrote that comment, let me try to think about it and get back to you! – Anirudh Panguluri Oct 02 '23 at 09:10
1

More elementary way :

$\displaystyle \sin(\alpha+\beta+\gamma)=\sin\alpha\cos(\beta+\gamma)+\cos\alpha\sin(\beta+\gamma)$

$\displaystyle =\sin\alpha(\cos\beta\cos\gamma)+\sin\beta(~cos\gamma\cos\alpha)+\sin\gamma(\cos\alpha\cos\beta)-\sin\alpha\sin\beta\sin\gamma$

$\displaystyle <\sin\alpha+\sin\beta+\sin\gamma\;\forall\; \alpha,\beta,\gamma\in\bigg(0,\frac{\pi}{2}\bigg)$

$\displaystyle \Longrightarrow \frac{\sin(\alpha+\beta+\gamma)}{\sin\alpha+\sin\beta+\sin\gamma}<1$

jacky
  • 5,194