4
$\begingroup$

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}?$

  • 0
    $\sin x=-\cos(x+\pi/2)$2012-09-01

4 Answers 4

9

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.

  • 0
    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.2012-09-01
2

(Using $\alpha$, $\beta$, $\gamma$ to name the half-angles themselves ...)

enter image description here

$\alpha + \beta + \gamma = 90^\circ \quad\implies\quad 1 \leq \sin \alpha + \sin\beta + \sin\gamma$ with equality when one of $\alpha$, $\beta$, $\gamma$ is $90^\circ$.

(By the Law of Sines, an angle $\theta$, inscribed in a circle of diameter $1$, subtends a chord of length $\sin\theta$.)

Note. More generally, for $\alpha+\beta+\gamma \leq 180^\circ$,

$\sin\left(\alpha+\beta+\gamma\right) \leq \sin \alpha + \sin\beta + \sin\gamma$

and even more generally, the relation clearly holds for an arbitrary number of angles.

1

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

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

0

Let $\alpha\geq\beta\geq\gamma$.

$\sin$ is a concave function on $\left(0^{\circ},90^{\circ}\right)$ and $\left(90^{\circ},0^{\circ},0^{\circ}\right)\succ\left(\frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2}\right)$. Thus, by Karamata: $1=\sin90^{\circ}+\sin0^{\circ}+\sin0^{\circ}<\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}.$ Done!