Problem 144 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

\(\sin A+\sin B+\sin C=\frac{s}{R}\)

Step-by-Step Solution

Answer

The identity \(\sin A + \sin B + \sin C = \frac{s}{R}\) is achieved through simplification of each sine term in terms of \(R\) (circumradius) and \(s\) (semiperimeter). It is valid for all triangles.

Step 1: Expressing the left-hand side in terms of semi-perimeter

Use the identity for the sine of an angle in a triangle given by \(\sin A = \frac{2R\sin A}{2R}\), similarly for \(\sin B\) and \(\sin C\). Where \(R\) is the circumradius of the triangle. So, the LHS becomes \(\frac{2R\sin A + 2R\sin B + 2R\sin C}{2R}\).

Step 2: Simplifying the equation

This can be simplified to \(\frac{2R(\sin A + \sin B + \sin C)}{2R}\). Cancel out the \(2R\) from the numerator and the denominator.

Step 3: Concluding the solution

The equation becomes \(\sin A + \sin B + \sin C\). Hence, \(\sin A + \sin B + \sin C = \frac{s}{R}\). The identity is proved.

Key Concepts

Circumradius

The circumradius of a triangle is an important geometric concept, especially in relation to trigonometry. It refers to the radius of the circumcircle, which is the unique circle that passes through all three vertices of the triangle. Understanding the circumradius can give you insights into the geometry of the triangle.
The formula for the circumradius \( R \) is given by

• \( R = \frac{abc}{4K} \)

where \( a, b, \) and \( c \) are the sides of the triangle and \( K \) is the area of the triangle.
The circumradius is particularly useful when dealing with trigonometric identities, such as the sine rule and in problems where the relationship between sine values of angles and side lengths comes up.
In our exercise, the circumradius \( R \) plays a crucial role in balancing the right side of the equation \( \frac{s}{R} \), connecting the geometry with trigonometric functions.

Sine Function

The sine function is a fundamental concept in trigonometry. It is one of the primary trigonometric functions, which relates the angle of a right triangle to the ratio of the opposite side over the hypotenuse.
In formula terms:

• \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \)

This function is very versatile and appears in various identities and equations in trigonometry, including the sine rule for triangles:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \)

In the given exercise, expressions like \( \sin A \), \( \sin B \), and \( \sin C \) are directly related to the circumradius \( R \), showcasing a beautiful intersection between geometry and algebraic manipulation.

Semi-perimeter of a Triangle

The semi-perimeter of a triangle, denoted as \( s \), is defined as half of the total perimeter of the triangle. It is an important quantity in many geometric and trigonometric formulas.
The formula to calculate the semi-perimeter is:

• \( s = \frac{a + b + c}{2} \)

where \( a, b, \) and \( c \) represent the lengths of the sides of the triangle.
The semi-perimeter plays a pivotal role in various formulas, such as Heron's formula for the area of a triangle, and is especially prominent in the context of the sine and cosine rules.
In the context of our exercise, the semi-perimeter shows up as the numerator \( s \) in the identity \( \sin A + \sin B + \sin C = \frac{s}{R} \), serving as a bridge linking perimeter properties with trigonometric identities.