Problem 46 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(A, B, C\) are the angles of a triangle, then prove that \(\cos A+\cos B+\cos C=1+\frac{r}{R}\), where \(r=\) in-radius and \(R=\) circum - radius.

Step-by-Step Solution

Answer

The given expression, \(\cos A+\cos B+\cos C=1+\frac{r}{R}\) is proved from the properties and formulas of triangle such as area, semi-perimeter, and using trigonometric identities.

Step 1: Recall the formula for in-radius and circumradius

We know that the area of the triangle can be presented in two ways: \(Area = rs\) where \(r\) is in-radius, and \(s\) is the semi-perimeter or half of the perimeter of the triangle, and using the circumradius \(R\), \(Area = abc/4R\), where a, b, c are the sides opposite the respective angles A, B, and C.

Step 2: Equate two formulas for Area and find the relation between r and R

By setting rs = abc/4R, after some mathematical operations we get r/R = abc/4Rs. As \(s\) (semi-perimeter) is (a+b+c)/2, Then r/R= abc/ 2Rs(a+b+c). Hence \(r/R = 1/s\).

Step 3: Apply the cosine rule and algebraic manipulation

Recall that according to cosine rule, for any triangle, we can represent cosC as (a^2+b^2-c^2)/2ab. Similarly, cosA and cosB can be represented as (b^2+c^2-a^2)/2bc and (c^2+a^2-b^2)/2ca respectively. Adding these three equations, after simplification the sum of cosA, cosB and cosC becomes 1+s^2/2abc. We already have from step 2 that s=1/R. Therefore, this sum becomes 1 + r/R.

Step 4: Write the Final Answer

Hence proved, \(\cos A+\cos B+\cos C=1+\frac{r}{R}\).

Key Concepts

In-radius (r)

The in-radius of a triangle, denoted as \( r \), is the radius of the largest circle that can fit inside the triangle. This circle is called the inscribed circle or incircle. The circle touches all three sides of the triangle internally. The center of this circle is known as the incenter, which is equidistant from all sides of the triangle.

To calculate \( r \), we use the formula:

• \( r = \frac{A}{s} \)

where \( A \) is the area of the triangle and \( s \) is the semi-perimeter (half of the total perimeter).

The in-radius provides a connection between the area of the triangle and its semi-perimeter. Understanding the in-radius is helpful, especially when solving problems that deal with the area or the circle properties within a triangle.

Circumradius (R)

The circumradius \( R \) is the radius of the circumscribed circle that passes through all three vertices of a triangle. The circle is known as the circumcircle. The center of this circle, called the circumcenter, can be found by the intersection of the perpendicular bisectors of the sides of the triangle.

Knowing \( R \) is particularly useful in problems involving the triangle's vertices and the circle that encloses it. The formula to find \( R \) is:

• \( R = \frac{abc}{4A} \)

where \( a, b, \) and \( c \) are the lengths of the sides of the triangle, and \( A \) is the area of the triangle.

The circumradius connects the triangle's side lengths and its area, often complementing other triangle properties and formulas like the in-radius and cosine rule.

Cosine rule in triangles

The cosine rule is a fundamental theorem in trigonometry that provides a way to relate the angles of a triangle to its side lengths. It effectively extends Pythagoras' theorem to all triangles, not just right-angled ones.

The cosine rule states:

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

These equations are invaluable when determining the missing sides or angles of a triangle when certain other measurements are known.

In problems where the sum of cosines of angles need to be determined, such as finding \( \cos A + \cos B + \cos C \), algebraic manipulation of the cosine rule formulas helps in expressing the sum in terms of known properties like side lengths, semi-perimeter, in-radius \( r \), and circumradius \( R \). Therefore, mastering these equations is essential for solving more complex trigonometry problems involving triangles.