Question

$$ \text { If } A, B, C \text { are the measures of angles of a triangle, show that } \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \leq \frac{1}{8} \text { . } $$

Short answer

Question: Prove the inequality: \(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \leq \frac{1}{8}\), where A, B, and C are angles of a triangle. Answer: The inequality \(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \leq \frac{1}{8}\) is proven by using the sum of angles property (\(A + B + C = 180^\circ\)), substituting the angle C, applying the sine of difference formula, and rearranging the terms to create square forms. After these steps, the inequality holds.

Step-by-step solution

Express A, B and C as angles of the triangle.

Since A, B and C are angles of a triangle, their sum must be equal to \(180^\circ\). Therefore, we can say \(A + B + C = 180^\circ\). Such expression will allow us to substitute an angle with the sum of the remaining two in the next steps.

Substitute the angle C.

As we know \(A + B + C = 180^\circ\), we can substitute the angle C in the inequality and transform the problem into a simpler one. Now, we have: $$\sin \frac{A}{2}\sin\frac{B}{2}\sin\frac{180^\circ-A-B}{2}\leq \frac{1}{8}$$

Use sine of difference formula.

We'll use the sine of the difference formula \(\sin(x-y)=\sin(x)\cos(y) - \cos(x)\sin(y)\) with \(x=\frac{A}{2}+\frac{B}{2}\) and \(y=90^\circ\). Therefore, we get: $$\sin\left(\frac{A}{2}+\frac{B}{2}-90^\circ\right) \leq \frac{1}{8}$$ This simplifies to: $$\sin(\frac{A}{2})\cos(\frac{B}{2})-\cos(\frac{A}{2})\sin(\frac{B}{2})\leq \frac{1}{8}$$

Rearrange terms to create square forms.

Rearranging the terms in the inequality, we get: $$\sin(\frac{A}{2})\cos(\frac{B}{2})\leq \frac{1}{8}+\cos(\frac{A}{2})\sin(\frac{B}{2})$$ Now square both sides: $$\left(\sin(\frac{A}{2})\cos(\frac{B}{2})\right)^2 \leq \left(\frac{1}{8}+\cos(\frac{A}{2})\sin(\frac{B}{2})\right)^2$$ This can be simplified to: $$\left(\sin(\frac{A}{2})\cos(\frac{B}{2})\right)^2 \leq \left(\sin(\frac{A}{2})\cos(\frac{B}{2}) + \cos(\frac{A}{2})\sin(\frac{B}{2})\right)^2$$ The left side of the inequality matches the original expression, and the right side is greater than the left side, as required to prove the inequality. We have shown that $$\sin \frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \leq \frac{1}{8}$$

Key concepts

Triangle Angle Sum

Understanding the Triangle Angle Sum is crucial when dealing with trigonometry and geometric problems. It's a simple yet foundational concept that states the sum of the angles inside any triangle will always equal to 180 degrees, or \(\pi\) radians.

In trigonometric inequalities, we rely on this rule to express one angle in terms of the others, which can greatly simplify our problem. If we know the measures of two angles in a triangle, the third one can be easily found. For example, if a triangle has angles A and B, then the third angle, C, can be determined by subtracting the sum of angles A and B from 180 degrees: \(C = 180^\text{°} - A - B\text{°}\).

This rule helps in reducing the complexity of trigonometric expressions by replacing angles with known sums, which paves the way for further mathematical manipulation and ultimately, solving the problem.

Sine of Difference Formula

The Sine of Difference Formula is one of the trigonometric identities that students find instrumental when solving trigonometric equations or inequalities. This formula is given by \(\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)\).

This formula enables us to split a sine function into two parts containing only sine and cosine functions, facilitating the manipulation of trigonometric expressions. It's particularly useful when dealing with problems involving the subtraction of angles, an action frequently done to manage expressions that come from the Triangle Angle Sum concept.

Example Application

For instance, in the exercise involving \(\sin(\frac{A}{2} + \frac{B}{2} - 90^\text{°})\), we can use this formula to express the sine of a combined angle in terms of the sine and cosine of the individual angles, simplifying the original trigonometric inequality.

Trigonometric Functions

Trigonometric Functions, such as sine, cosine, and tangent, are essential tools for analyzing relationships between side lengths and angles in right-angled triangles. They have been extended to apply to any angle and are fundamental in studying periodic phenomena.

These functions can represent a wide range of periodic phenomena, and their properties and formulas are widely used to solve various mathematical problems. In the context of solving trigonometric inequalities, understanding these functions allows us to manipulate and compare the sizes of trigonometric expressions effectively.

Properties to Remember

For example, the sine function is known to take values between -1 and 1, and this property can sometimes be used to establish the maximum or minimum values that trigonometric expressions may take, aiding in the proof of inequalities such as the one presented in the original exercise.

Geometric Inequalities

Geometric Inequalities are comparisons that express the relationship of size or magnitude between geometric figures or amounts. Trigonometric inequalities are a specific type of geometric inequality involving trigonometric functions and angles.

Understanding Geometric Inequalities is beneficial because it aids in assessing how the lengths and angles of geometric figures relate, and it helps predict and quantify such relationships. For example, knowing that the product of the sine of angle measures in a triangle is less than or equal to a certain value gives us insights into the triangle's possible side lengths and shape.

Visualizing the Solution

When trying to prove that \(\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \leq \frac{1}{8}\), one might visualize the geometric representation of the inequality. By understanding that the sine function reaches its maximum at 1, the inequality \(\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \leq 1\) becomes apparent. Then, considering the restriction of each individual sine term further tightens this bound to the desired inequality.