Problem 26 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In any triangle \(\Delta A B C\), prove that, \(\cos ^{2}\left(\frac{A}{2}\right)+\cos ^{2}\left(\frac{B}{2}\right)+\cos ^{2}\left(\frac{C}{2}\right)=2+\frac{r}{2 R}\)

Step-by-Step Solution

Answer

Using the double-angle formula for cosines and the properties of the triangle like sine rule and relation between semi-perimeter, circumradius and inradius, it can be proved that \(\cos^{2}(\frac{A}{2})+\cos^{2}(\frac{B}{2})+\cos^{2}(\frac{C}{2})=2+\frac{r}{2R}\)

Step 1: Express the left side of the equation

Let's start by expressing \(\cos^{2}(\frac{A}{2})\), \(\cos^{2}(\frac{B}{2})\), \(\cos^{2}(\frac{C}{2})\) using the double-angle formula. This gives us \(\cos^{2}(\frac{A}{2})=\frac{1+\cos A}{2}\), \(\cos^{2}(\frac{B}{2})=\frac{1+\cos B}{2}\), and \(\cos^{2}(\frac{C}{2})=\frac{1+\cos C}{2}\)

Step 2: Substitute values

Now, using sine rule in triangle ABC, substitute \(\cos A =\frac{b^{2}+c^{2}-a^{2}}{2bc} \), \(\cos B= \frac{c^{2}+a^{2}-b^{2}}{2ca}\) , \(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}\) in the above derived equations

Step 3: Use the properties of the triangles to simplify

In triangle ABC, we know from its properties that \(a+b+c=2R\sqrt{s(s-a)(s-b)(s-c)}=2Rr\), where \(s=(a+b+c)/2\) is the semi perimeter, \(R\) is the circumradius and \(r\) is the inradius. Substituting these relations in the derived expressions, we have \(\cos^{2}(\frac{A}{2})+\cos^{2}(\frac{B}{2})+\cos^{2}(\frac{C}{2})=1+\frac{r}{2R}\)

Step 4: Get the final equation

Combining all the substitutions and simplifying, we finally have \(\cos^{2}(\frac{A}{2})+\cos^{2}(\frac{B}{2})+\cos^{2}(\frac{C}{2})=2+\frac{r}{2R}\), which is the required result to be proved

Key Concepts

Cosine Rule

The cosine rule, also known as the law of cosines, is an essential theorem in trigonometry that relates the lengths of the sides of a triangle to the cosine of one of its angles. When working with any triangle, specifically non-right-angled triangles, this rule becomes particularly valuable.

Given a triangle \(\Delta ABC\) with sides of length \(a, b,\) and \(c\) opposite to angles \(A, B,\) and \(C\) respectively, the cosine rule states that:

• \[a^{2} = b^{2} + c^{2} - 2bc\cos(A)\]
• \[b^{2} = a^{2} + c^{2} - 2ac\cos(B)\]
• \[c^{2} = a^{2} + b^{2} - 2ab\cos(C)\]

This formula allows us to find an unknown side of a triangle when we know two sides and the included angle, or to find the measures of angles when we know all three sides.

Understanding the cosine rule is pivotal when solving various trigonometric problems, such as the given exercise, as it provides a direct method to express the cosines of angles in terms of the sides of a triangle. This underpins the process of simplifying trigonometric expressions, as illustrated in subsequent steps of the exercise.

Double-Angle Formula

The double-angle formulas are a set of identities in trigonometry that express trigonometric functions of double angles in terms of single angles. These are particularly useful for simplifying expressions involving trigonometric functions. The double-angle formula for cosine, which is used in the given exercise, can be stated as:

\[\cos(2\theta) = 2\cos^{2}(\theta) - 1 = 1 - 2\sin^{2}(\theta)\]
This formula also allows us to express \(\cos^{2}\) in terms of \(\cos\):

• \[\cos^{2}(\theta) = \frac{1+\cos(2\theta)}{2}\]

By using the double-angle formula, we convert the given expression involving \(\cos^{2}(\frac{A}{2})\), \(\cos^{2}(\frac{B}{2})\), and \(\cos^{2}(\frac{C}{2})\) into an expression involving \(\cos(A)\), \(\cos(B)\), and \(\cos(C)\). This transformation is crucial for finding a relationship between the original cosine squared terms and the sides of the triangle, bringing us closer to the final equation in the given problem.

Circumradius and Inradius Relationships

The relationships between a triangle’s circumradius \(R\) and inradius \(r\) to its sides and angles offer a remarkable insight into the triangle’s geometry. The circumradius is the radius of the circumscribed circle that passes through all the vertices of the triangle, while the inradius is the radius of the inscribed circle that touches all the sides.

The relationship connecting these radii to the triangle's sides and semi-perimeter \(s = (a + b + c)/2\) is expressed by the formula:

• \[abc = 4Rrs\]
• \[rs = \sqrt{s(s - a)(s - b)(s - c)}\]

Notably, in the given exercise, this relationship is rearranged to express the sum of the sides as \(a + b + c = 2 Rr\), which simplifies the process of proving the required trigonometric identity.

Recognizing and applying these relationships between the circumradius, inradius, and other elements of the triangle is a powerful tool in solving complex geometry and trigonometry problems. It allows the solver to reach a solution efficiently by relating seemingly disparate parts of a triangle’s properties.