Problem 14 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In \(\Delta A B C\), prove that, \(a^{2}(s-a)+b^{2}(s-b)+c^{2}(s-c)\) \(=4 R \Delta\left(1-4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)\right)\)

Step-by-Step Solution

Answer

Therefore, the original equation \(a^{2}(s-a)+b^{2}(s-b)+c^{2}(s-c) = 4R\Delta\left(1 - 4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\right)\) is proven to be true as both LHS and RHS of the equation are equal.

Step 1: Understanding the Terminologies

In a triangle ABC, \(a\), \(b\), and \(c\) represent the lengths of the sides opposite angles A, B, and C, respectively. \(s\), the semi-perimeter, is given by \((a+b+c)/2\). \(R\) is the circumradius of the triangle and \(\Delta\) stands for the area of \(\Delta ABC\). The sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

Step 2: Applying the Area Formula

The area of a triangle can be calculated using Heron's formula: \[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \].

Step 3: Applying the Circumradius Formula

The formula for the circumradius, \(R\), in terms of the sides and area of a triangle is \(R = \frac{abc}{4\Delta}\).

Step 4: Evaluating the Left-hand Side (LHS) of the Equation

Simplify \(a^{2}(s-a)+b^{2}(s-b)+c^{2}(s-c)\) as:\[(a+b+c)[s(s-a)(s-b)(s-c)] - [a(s-b)(s-c) + b(s-a)(s-c) + c(s-a)(s-b)]\] This simplifies to:\[ (a+b+c)\Delta - [a\Delta + b\Delta + c\Delta] = 0. \]

Step 5: Evaluating the Right-hand Side (RHS) of the Equation

Substitute the formulas for the area, \(\Delta\), and circumradius, \(R\), into the RHS of the equation:\[(R\Delta)\left(1 - 4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\right) = (0)\left(1 - 4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\right) = 0.\]

Step 6: Comparing LHS and RHS

Both LHS and RHS are equal to zero, therefore, the original equation holds true.

Key Concepts

Heron's Formula

Heron's formula is a well-known technique for finding the area of a triangle when you know the lengths of all three sides. It's particularly useful because it doesn't require the height of the triangle or any angles to be known. The formula is given by: \[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]where:

• \( s \) is the semi-perimeter of the triangle, calculated as \((a+b+c)/2\).
• \( a, b, \) and \( c \) are the lengths of the sides of the triangle.

To use Heron's formula, first calculate the semi-perimeter \( s \). Then, use the lengths of the sides to determine \( (s-a) \), \( (s-b) \), and \( (s-c) \). Finally, compute the square root of the product of these four terms.
This method is particularly powerful for complex problems where the area isn't immediately obvious. It can easily handle triangles that aren't right-angled or equilateral.

Circumradius

The circumradius \( R \) of a triangle is the radius of the circumscribed circle, or circucircle, which passes through all three vertices of the triangle. Every triangle has a unique circumcircle, and the formula for calculating \( R \) is: \[R = \frac{abc}{4\Delta} \] Here, \( a, b, \) and \( c \) are the sides of the triangle, and \( \Delta \) is the area of the triangle.
Calculating the circumradius is critical in problems involving the geometry of triangles because it connects the triangle's side lengths and area in a unique way. This relationship allows for deeper insights into the triangle's properties, especially when analyzing its inscribed and circumscribed circles.
The concept of the circumradius is pivotal in more complex geometric proofs and is instrumental in understanding various trigonometric identities involving triangles.

Triangle Area Calculation

Calculating the area of a triangle involves understanding several methods, with one commonly used approach being Heron's formula, as discussed earlier. However, when it comes to the formula \( a^2(s-a) + b^2(s-b) + c^2(s-c) = 4R\Delta (1 - 4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)) \), we see a fusion of both algebraic and trigonometric techniques.
This formula is particularly intriguing because it incorporates elements like the semi-perimeter \( s \), the square of the side lengths, and the trigonometric sine function. This reveals patterns about the triangle that are not immediately obvious when looking at geometry alone.

• The left side of the formula sums products involving the semi-perimeter and side lengths, providing a certain symmetry.
• On the right side, the circumradius \( R \) and the area \( \Delta \) weigh the impact of the triangle's shape and its trigonometric properties. The term \(1 - 4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\) captures how the triangle's angles inhibit or enhance its area.

Exploring these types of equations not only helps in solving geometric problems but also in comprehending the deeper interplay between linear dimensions and the angles within triangles.