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I would like to show that:

$$ 1<\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}$$

where $\alpha, \beta, \gamma$ are the angles of a triangle.

I know that the inequality $$ 1<\cos \alpha+\cos \beta+\cos \alpha $$

is a direct consequence of the identity $$ \cos \alpha+\cos \beta+\cos \alpha =1+\frac{r}{R}$$

with circumradius $R$ and inradius $r$.

So is there a similar expression for $$ \sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}?$$

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Have you learned the identity $sin(x) = cos(x+\frac{\pi}{2}) = cos (x - \frac{\pi}{2})$? –  Vilid Sep 1 '12 at 15:37
    
$\sin x=-\cos(x+\pi/2)$ –  Chon Sep 1 '12 at 15:49
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2 Answers

up vote 8 down vote accepted

Rewrite the inequality as $\sin \frac{\alpha + \beta + \gamma}{2} < \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} + \sin \frac{\gamma}{2}$. Notice that $\sin (a+b) < \sin a + \sin b$ for $a, b, (a+b) \in (0, {\pi \over 2})$. Extend the same statement for three variables.

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So it's simply: $\sin \frac{\alpha+\beta+\gamma}{2}=\sin \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}+\sin \frac{\beta}{2}\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}+\sin \frac{\gamma}{2}\cos \frac{\alpha}{2}\cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \sin\frac{\beta}{2} \sin\frac{\gamma}{2}< \sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}$ Thanks! –  Chon Sep 1 '12 at 16:03
    
I didn't check your equations but that's the idea. Also, this is a direct consequence of $(\sin x)'' < 0$ for $x \in (0, {\pi \over 2})$ (the fact that $\sin x$ is below any of it's tangents in that interval). The algebraic derivation is hardly necessary. –  Karolis Juodelė Sep 1 '12 at 16:13
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In order to demonstrate the inequality above we may actually prove a more general inequality, which can come in handy another time:

For all natural $n > 1$ and $x_1, x_2, \ldots, x_n \in (0, \pi)$ we have: \[ |\sin(x_1 + x_2 + ...+x_n)| < \sin x_1 + \sin x_2 + ... + \sin x_n.\]

This can be proved by induction:

  • for $n = 2$ \[ \begin{split} |\sin(x_1 + x_2)| &= |\sin x_1 \cos x_2 + \sin x_2 \cos x_1| \leq |\sin x_1 \cos x_2| + |\sin x_2 \cos x_1| \\ &= |\sin x_1| \cdot |\cos x_2| + |\sin x_2| \cdot |\cos x_1| < \sin x_1 + \cos x_2 \end{split} \] by the properties of absolute value and since $\cos x < 1$ for $x \in (0, \pi)$

  • induction step \[ \begin{split} |\sin (x_1 + \ldots + x_{n+1})| &= |\sin(x_1 + \ldots + x_n) \cos x_{n+1} + \sin x_{n+1} \cos(x_1 + \ldots + x_n)| \\ &\leq |\sin(x_1 + \ldots + x_n)| \cdot |\cos x_{n+1}| + |\sin x_{n+1}| \cdot |\cos(x_1 + \ldots + x_n)| \\ &< |\sin(x_1 + \ldots + x_n)| + |\sin x_{n+1}| \\ &< \sin x_1 + \sin x_2 + \ldots + \sin x_{n+1}. \end{split} \] by the induction hypothesis and the fact $|\sin x| = \sin x$ for $x \in (0, \pi)$

When plugging in halves of the angles of a triangle we obtain the desired

\[ \sin(\dfrac{\alpha + \beta + \gamma}{2}) = 1 < \sin \dfrac{\alpha}{2} + \sin\dfrac{\beta}{2} + \sin\dfrac{\gamma}{2} .\]

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