Question about sines of angles in an acute triangle

by Noppawee Apichonpongpan   Last Updated April 25, 2018 14:20 PM

Let ABC be a triangle such that each angle is less than 90 degrees. I want to prove that sinA + sinB + sinC > 2.

Here is what I have done:

Since A+B+C=180 and 0 < A,B,C < 90, at least two of A,B,C are in the range 45 < x < 90, without loss of generality, let these angles be A and B.

sinA + sinB + sinC = sinA + sinB + sin(180-A-B) = sinA + sinB + sin(A+B)

Since 45 < A,B < 90 it follows that 2^0.5 < sinA + sinB < 2. Am I near the answer?

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Multiplying both sides by $2R$ and exploiting the sine theorem, we have to prove that:

The perimeter of acute triangle is always greater than four times the circumradius.

Assume that $A,B$ and the circumradius $R$ are fixed. Since $ABC$ is an acute triangle, the circumcenter $O$ of $ABC$ lies inside $ABC$, so $C$ lies between the antipode $A'$ of $A$ and the antipode $B'$ of $B$ in the circumcircle $\Gamma$:

Let $M$ be the midpoint of the arc $A'B'$ and $\Gamma_C$ be the ellipse through $C$ with foci in $A,B$, i.e. the locus of points $P$ for which $PA+PB=CA+CB$. Since $A'$ and $B'$ lie inside $\Gamma_C$, we have $A'A+A'B<CA+CB$, so the perimeter of $ABC$ is minimized when $C\equiv A'$ or $C\equiv B'$, i.e. when $\widehat{C}=\frac{\pi}{2}$. This gives that the minimum perimeter is achieved in the limit case when $A$ and $B\equiv C$ are endpoints of a diameter of $\Gamma$. In such a case, obviously, the perimeter is $4R$.

As an alternative, we can use concavity. Since $\sin x$ is a concave function over $[0,\pi/2]$, $$f(A,B,C)=\sin A+\sin B+\sin C$$ is a concave function over the set $E=\{(A,B,C)\in[0,\pi/2]^3:A+B+C=\pi\}$, so its minima lie on $\partial E$.

Jack D'Aurizio
December 13, 2014 21:01 PM

i am able to simplify $$\sin A + \sin B + \sin (A + B) = \sin A + \sin B + \sin A \cos B + \sin B \cos A \\ = (1+\cos B)\sin A + (1 + \cos A)\sin B \\ = 4\cos^2 B/2\sin A/2 \cos A/2 + 4\cos^2 A/2 \sin B/2 \cos B/2 \\ = 4\cos B/2 \cos A/2(\sin A/2 \cos B/2 + \sin B/2 \cos A/2) \\ = 4\cos B/2 \cos A/2\sin (A/2 + B/2) = 4\cos A/2 \cos B/2 \cos C/2$$

4now, using the fact that $A/2 < 45^\circ, B/2 < 45^\circ$ and $C/2 < 45^\circ,$ i can only conclude $$\sin A + \sin B + \sin C > \sqrt 2.$$

i am going to try to improve the bound. introduce $0 <\alpha, \beta < 45^\circ$ so that $A/2 = 45^\circ - \alpha, B/2 = 45^\circ - \beta, C/2 = \alpha + \beta.$ in terms of these new variables,

$$4\cos A/2 \cos B/2 \cos C/2 = 2(\cos \alpha + \sin \alpha)(\cos \beta + \sin \beta)\cos(\alpha + \beta) \\ = 2(\cos(\alpha + \beta) + \sin(\alpha + \beta))\cos(\alpha + \beta) \\ = 1 + \cos(2\alpha + 2\beta) + \sin(2\alpha + 2\beta) = 1 + \sqrt 2 \sin(2\alpha + 2\beta + 45^\circ)$$

since $0 < 2\alpha + 2 \beta < 90^\circ,$ we get the desired bound

$$2 < 1 + \sqrt 2\sin(2\alpha + 2 \beta) < 1 + \sqrt 2$$

abel
December 13, 2014 21:04 PM

I have found a simpler solution.
Observing the graph of y = $sinx$ and y = $\frac{2}{\pi}x,x\in[0,\frac{\pi}{2}]$.
we can see that: $sinx\ge\frac{\pi}{2}x$.
Let $\bigtriangleup ABC$ be a triangle such that each angle is less than 90 degrees,so we have: $A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),C\in(0,\frac{\pi}{2})$.
Because of $sinx>\frac{\pi}{2}x$,when $x\in(0,\frac{\pi}{2})$,so
$sinA+sinB+sinC > \frac{\pi}{2}(A+B+C)=2$

wzj7531
April 25, 2018 14:18 PM

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