Proving Triangle Properties Using the Cosine Rule and Trigonometric Identities

Often, geometric problems involve proving specific properties of triangles using fundamental theorems and identities. One such problem involves proving that if in triangle ABC, a cos A b cos B, then triangle ABC is either isosceles or right-angled. This article outlines a detailed step-by-step solution, leveraging the Cosine Rule and Trigonometric Identities to arrive at the desired conclusion.

Using the Cosine Rule to Prove Triangle Properties

Given the condition:

a cos A b cos B

where a, b, c are the sides of triangle ABC opposite to angles A, B, C, respectively. Our goal is to use the Cosine Rule to derive the property of the triangle.

Step 1: Recall the Cosine Rule

The Cosine Rule states for any triangle:

c^2 a^2 b^2 - 2ab cos C

In terms of cosines:

cos A frac{b^2 c^2 - a^2}{2bc}

cos B frac{a^2 c^2 - b^2}{2ac}

Step 2: Substitute into the Given Equation

Substitute the expressions for cos A and cos B into the given condition:

a cos A b cos B

This gives:

a left(frac{b^2 c^2 - a^2}{2bc}right) b left(frac{a^2 c^2 - b^2}{2ac}right)

Step 3: Simplify the Equation

Cross-multiplying to clear the denominators:

a^2(b^2 c^2 - a^2) b^2(a^2 c^2 - b^2)

Expanding both sides:

a^2b^2 a^2c^2 - a^4 b^2a^2 b^2c^2 - b^4

Rearranging terms:

a^2b^2 - b^2a^2 b^4 - a^4 a^2c^2 - b^2c^2

This simplifies to:

b^4 - a^4 a^2c^2 - b^2c^2

Step 4: Factor the Equation

The left-hand side can be factored using the difference of squares:

b^4 - a^4 (b^2 - a^2)(b^2 a^2)

The right-hand side can be factored as:

a^2c^2 - b^2c^2 c^2(a^2 - b^2)

So the equation becomes:

c^2(a^2 - b^2) (b^2 - a^2)(b^2 a^2)

Rearranging terms:

c^2(a^2 - b^2) -(a^2 - b^2)(b^2 a^2)

Step 5: Analyze the Factors

From the factored equation we have two cases:

Case 1: a^2 - b^2 0

implies

a b

which means triangle ABC is isosceles with A B.

Case 2: c^2 -(b^2 a^2)

Rearranging this gives:

c^2 a^2 b^2

Using the Pythagorean Theorem, this implies that triangle ABC is a right-angled triangle specifically with angle C being 90^circ.

Using Trigonometric Identities to Simplify the Proof

Alternatively, consider a more streamlined approach using trigonometric identities:

Step 1: Recall the Trigonometric Identity

We know that:

frac{a}{sin A} frac{b}{sin B}

which implies:

a sin B b sin A

Step 2: Divide to Form the Equation

Dividing the given condition by this identity gives:

frac{cos A}{sin B} frac{cos B}{sin A}

This simplifies to:

cos A sin A cos B sin B

Step 3: Further Simplification

Subtracting terms:

cos A sin B - cos B sin A 0

This is a form of the sine difference identity:

sin (A - B) 0

Step 4: Analyze the Result

From the sine difference identity, we get:

A - B 0