Proving 2b^2 = A^2 + C^2 Given Trigonometric Ratios In A Triangle

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Introduction

In the realm of trigonometry and geometry, exploring relationships between angles and sides of triangles often leads to fascinating results. This article delves into a specific trigonometric problem: given the condition that $\sin A : \sin C = \sin(A - B) : \sin(B - C)$, we aim to prove that $2b^2 = a^2 + c^2$. This intriguing equation suggests a particular relationship between the sides of the triangle, where a, b, and c are the side lengths opposite to angles A, B, and C, respectively. Let's embark on a detailed exploration of this problem, breaking down the steps and trigonometric identities involved, and arrive at the final proof. Understanding the intricate connections between angles and sides enhances our comprehension of triangular geometry and its applications in various fields.

Trigonometry, as a branch of mathematics, provides us with powerful tools to analyze triangles. The sine rule, cosine rule, and other trigonometric identities form the bedrock of our analysis. By leveraging these fundamental principles, we can establish the desired relationship and gain deeper insights into the properties of triangles that satisfy the given condition.

The journey through this problem involves transforming the initial trigonometric ratio into a form that allows us to apply the sine rule effectively. The sine rule, which states that the ratio of a side length to the sine of its opposite angle is constant for all sides and angles in a triangle, is a cornerstone in solving many trigonometric problems. By carefully manipulating the given ratios and applying trigonometric identities, we can unveil the hidden relationship between the side lengths and demonstrate that $2b^2 = a^2 + c^2$. This result highlights a specific condition under which the sides of a triangle satisfy a particular arithmetic progression, adding another layer of understanding to the rich tapestry of triangle geometry.

Problem Statement

The core of our discussion revolves around the following statement: If $\sin A : \sin C = \sin(A - B) : \sin(B - C)$, then prove that $2b^2 = a^2 + c^2$. This statement presents a conditional relationship within a triangle, linking the sines of the angles to the differences between them, and ultimately connecting it to a relationship between the squares of the side lengths. To unravel this problem, we will employ the sine rule, which is a fundamental concept in trigonometry that relates the sides of a triangle to the sines of their opposite angles. The sine rule states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. Mathematically, this can be expressed as:

asin⁡A=bsin⁡B=csin⁡C=2R\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

where a, b, and c are the side lengths, A, B, and C are the angles opposite these sides, and R is the circumradius of the triangle. The circumradius is the radius of the circle that passes through all three vertices of the triangle. Understanding this fundamental relationship is crucial for navigating through the complexities of the given problem and arriving at the desired conclusion.

To effectively tackle the problem, we need to manipulate the given ratio, $\sin A : \sin C = \sin(A - B) : \sin(B - C)$, and strategically apply trigonometric identities to express the sines of angle differences in terms of individual angles. This transformation will pave the way for the application of the sine rule, allowing us to bridge the gap between the trigonometric ratios and the side lengths of the triangle. By carefully orchestrating these steps, we can reveal the underlying mathematical structure and demonstrate the relationship $2b^2 = a^2 + c^2$. This exploration not only enhances our problem-solving skills in trigonometry but also deepens our appreciation for the interconnectedness of various geometric concepts.

Proof

Let's begin by expressing the given condition as a proportion:

sin⁡Asin⁡C=sin⁡(A−B)sin⁡(B−C)\frac{\sin A}{\sin C} = \frac{\sin(A - B)}{\sin(B - C)}

Cross-multiplying gives us:

sin⁡Asin⁡(B−C)=sin⁡Csin⁡(A−B)\sin A \sin(B - C) = \sin C \sin(A - B)

Now, we will apply the sine subtraction formula, which states that $\sin(x - y) = \sin x \cos y - \cos x \sin y$, to expand both sides of the equation:

sin⁡A(sin⁡Bcos⁡C−cos⁡Bsin⁡C)=sin⁡C(sin⁡Acos⁡B−cos⁡Asin⁡B)\sin A (\sin B \cos C - \cos B \sin C) = \sin C (\sin A \cos B - \cos A \sin B)

Expanding both sides, we get:

sin⁡Asin⁡Bcos⁡C−sin⁡Acos⁡Bsin⁡C=sin⁡Acos⁡Bsin⁡C−cos⁡Asin⁡Bsin⁡C\sin A \sin B \cos C - \sin A \cos B \sin C = \sin A \cos B \sin C - \cos A \sin B \sin C

Notice that the term $\sin A \cos B \sin C$ appears on both sides, but with opposite signs. We made an error in the initial expansion. The correct expansion should be:

sin⁡Asin⁡Bcos⁡C−sin⁡Acos⁡Bsin⁡C=sin⁡Csin⁡Acos⁡B−sin⁡Ccos⁡Asin⁡B\sin A \sin B \cos C - \sin A \cos B \sin C = \sin C \sin A \cos B - \sin C \cos A \sin B

Rearranging the terms, we obtain:

sin⁡Asin⁡Bcos⁡C+sin⁡Csin⁡Bcos⁡A=2sin⁡Acos⁡Bsin⁡C\sin A \sin B \cos C + \sin C \sin B \cos A = 2 \sin A \cos B \sin C

Factoring out $\sin B$ from the left side gives:

sin⁡B(sin⁡Acos⁡C+cos⁡Asin⁡C)=sin⁡Asin⁡Ccos⁡B+sin⁡Csin⁡Acos⁡B\sin B (\sin A \cos C + \cos A \sin C) = \sin A \sin C \cos B + \sin C \sin A \cos B

sin⁡B(sin⁡Acos⁡C+cos⁡Asin⁡C)=sin⁡C(sin⁡Acos⁡B)−sin⁡C(sin⁡Bcos⁡A)\sin B (\sin A \cos C + \cos A \sin C) = \sin C (\sin A \cos B) - \sin C (\sin B \cos A)

sin⁡Asin⁡Bcos⁡C−sin⁡Asin⁡Ccos⁡B=sin⁡Asin⁡Ccos⁡B−sin⁡Bsin⁡Ccos⁡A\sin A \sin B \cos C - \sin A \sin C \cos B = \sin A \sin C \cos B - \sin B \sin C \cos A

Now, move all the terms to one side:

sin⁡Asin⁡Bcos⁡C+sin⁡Bsin⁡Ccos⁡A=sin⁡Csin⁡(A−B)\sin A \sin B \cos C + \sin B \sin C \cos A = \sin C \sin(A-B)

Now we consider sine addition formula: $\sin(A+C) = \sin A \cos C + \cos A \sin C $. Using this, the equation becomes:

sin⁡Bsin⁡(A+C)=2sin⁡Asin⁡Ccos⁡B\sin B \sin(A + C) = 2 \sin A \sin C \cos B

Since $A + B + C = \pi$, we have $A + C = \pi - B$. Thus, $\sin(A + C) = \sin(\pi - B) = \sin B$. Substituting this into the equation, we get:

sin⁡Bsin⁡B=sin⁡2B=2sin⁡Asin⁡Ccos⁡B\sin B \sin B = \sin^2 B = 2 \sin A \sin C \cos B

Now, let's apply the sine rule. According to the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where R is the circumradius of the triangle. Therefore, we have:

\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, and $\sin C = \frac{c}{2R}

Substituting these expressions into our equation, we get:

(b2R)2=2(a2R)(c2R)cos⁡B\left(\frac{b}{2R}\right)^2 = 2 \left(\frac{a}{2R}\right) \left(\frac{c}{2R}\right) \cos B

Simplifying, we get:

b24R2=2accos⁡B4R2\frac{b^2}{4R^2} = \frac{2ac \cos B}{4R^2}

Multiplying both sides by $4R^2$ yields:

b2=2accos⁡Bb^2 = 2ac \cos B

Next, we apply the cosine rule, which states that $b^2 = a^2 + c^2 - 2ac \cos B$. Rearranging the cosine rule, we have:

2accos⁡B=a2+c2−b22ac \cos B = a^2 + c^2 - b^2

Substituting this into our equation $b^2 = 2ac \cos B$, we get:

b2=a2+c2−b2b^2 = a^2 + c^2 - b^2

Adding $b^2$ to both sides gives:

2b2=a2+c22b^2 = a^2 + c^2

Thus, we have proven that if $\sin A : \sin C = \sin(A - B) : \sin(B - C)$, then $2b^2 = a^2 + c^2$.

Conclusion

In conclusion, we have successfully demonstrated that if the ratio of the sines of angles A and C is equal to the ratio of the sines of the differences (A - B) and (B - C), then the relationship $2b^2 = a^2 + c^2$ holds true. This elegant result showcases the intricate interplay between the angles and sides of a triangle. By strategically applying trigonometric identities, such as the sine subtraction formula and the sine rule, and then incorporating the cosine rule, we were able to bridge the gap between the initial condition and the final equation. This journey highlights the power of trigonometric principles in unraveling geometric relationships. The final equation, $2b^2 = a^2 + c^2$, implies that the squares of the side lengths a, b, and c form an arithmetic progression, where $b^2$ is the arithmetic mean of $a^2$ and $c^2$. This adds a layer of geometric interpretation to the algebraic result.

The methods used to solve this problem are widely applicable in various areas of mathematics and physics. The strategic application of trigonometric identities and the sine and cosine rules are essential tools for solving a plethora of problems involving triangles and other geometric figures. The ability to manipulate trigonometric expressions and relate them to side lengths and angles is a cornerstone of mathematical problem-solving. Moreover, the logical progression of steps and the clear articulation of each stage of the proof underscore the importance of rigorous mathematical reasoning.

By exploring such problems, we not only enhance our problem-solving capabilities but also gain a deeper appreciation for the beauty and interconnectedness of mathematical concepts. The relationship we have proven serves as a testament to the elegance of trigonometric principles and their ability to reveal hidden connections within geometric structures. The process of unraveling this problem encourages a systematic approach to mathematical exploration, emphasizing the importance of careful manipulation, strategic application of identities, and logical deduction. Through such exercises, we refine our analytical skills and develop a more nuanced understanding of the world of mathematics.