Prove Acute Triangle Inequality: $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$

In summary, The Acute Triangle Inequality states that in a triangle, the length of any side is always less than the sum of the lengths of the other two sides. To prove this inequality, we can use the fact that the sine function is an increasing function in the interval $[0, \frac{\pi}{2}]$. This means that as the angle increases, the sine value also increases. The Acute Triangle Inequality is an important concept in geometry and is applicable to all types of triangles. Other important inequalities related to triangles include the Law of Cosines, the Law of Sines, and the Triangle Inequality.
  • #1
anemone
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Let $\alpha,\,\beta$ and $\gamma$ be the interior angles of an acute triangle.

Prove that if $\alpha \lt \beta \lt \gamma$, then $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.
 
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  • #2
anemone said:
Let $\alpha,\,\beta$ and $\gamma$ be the interior angles of an acute triangle.

Prove that if $\alpha \lt \beta \lt \gamma$, then $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

There are 2 cases

no angle is less than $\frac{\pi}{4}$ and from given condition no angle is greater than or equal to $\frac{\pi}{2}$
so we have

$\frac{\pi}{4} \le \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$

hence $\frac{\pi}{2} \le 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$

as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

case 2 :
one angle $\alpha \lt \frac{\pi}{4}$ and hence $\frac{\pi}{2} - \alpha <\beta$ otherwise $\gamma \ge \frac{\pi}{2}$
$\frac{\pi}{4} \le \frac{\pi}{2}- \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$
hence $\frac{\pi}{2} \le \pi- 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$
as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin (\pi-2\alpha) \gt \sin 2\beta \gt \sin 2\gamma$.
or $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.
 
  • #3
kaliprasad said:
There are 2 cases

no angle is less than $\frac{\pi}{4}$ and from given condition no angle is greater than or equal to $\frac{\pi}{2}$
so we have

$\frac{\pi}{4} \le \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$

hence $\frac{\pi}{2} \le 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$

as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

case 2 :
one angle $\alpha \lt \frac{\pi}{4}$ and hence $\frac{\pi}{2} - \alpha <\beta$ otherwise $\gamma \ge \frac{\pi}{2}$
$\frac{\pi}{4} \le \frac{\pi}{2}- \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$
hence $\frac{\pi}{2} \le \pi- 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$
as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin (\pi-2\alpha) \gt \sin 2\beta \gt \sin 2\gamma$.
or $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.
Very well done kaliprasad!(Cool)
 

FAQ: Prove Acute Triangle Inequality: $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$

1. What is the Acute Triangle Inequality?

The Acute Triangle Inequality states that in a triangle, the length of any side is always less than the sum of the lengths of the other two sides. This is true for all types of triangles, including acute triangles.

2. How can we prove the Acute Triangle Inequality for $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$?

To prove this inequality, we can use the fact that the sine function is an increasing function in the interval $[0, \frac{\pi}{2}]$. This means that as the angle increases, the sine value also increases. By using this property, we can show that the side opposite the largest angle in a triangle is always the longest, and therefore, the length of any side is always less than the sum of the lengths of the other two sides.

3. Why is it important to prove the Acute Triangle Inequality?

The Acute Triangle Inequality is an important concept in geometry as it helps us understand and analyze the relationships between the sides and angles of a triangle. It also serves as a useful tool in solving problems involving triangles, and is a fundamental concept in trigonometry.

4. Can the Acute Triangle Inequality be applied to all triangles?

Yes, the Acute Triangle Inequality holds true for all types of triangles, including acute, right, and obtuse triangles. This is because the inequality is based on the properties of the sine function, which applies to all types of triangles.

5. Are there any other inequalities related to triangles?

Yes, there are several other important inequalities related to triangles, such as the Law of Cosines, the Law of Sines, and the Triangle Inequality. These inequalities play a crucial role in geometry and trigonometry, and can be used to solve a variety of problems involving triangles.

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