Problem 82 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(b+c=3 a\), prove that \(\cot \frac{B}{2} \cot \frac{C}{2}=2\).

Step-by-Step Solution

Answer

Using the given condition \(b+c=3a\), we can rewrite the expression \(\cot \frac{B}{2} \cot \frac{C}{2}\) in terms of sine and cosine functions. Applying half-angle formulas for sine and cosine, we can express the cotangents in terms of the sides of the triangle. After substituting these formulas into the main expression and simplifying, we can substitute the given condition again to finally conclude that \(\cot \frac{B}{2} \cot \frac{C}{2}=2\).

Step 1: Write down the known relations for the problem

The problem states that \(b+c=3a\). We are asked to prove that \(\cot \frac{B}{2} \cot \frac{C}{2}=2\). Recall that \(\cot x = \frac{\cos x}{\sin x}\). Thus, the expression we want to prove can be written as \(\frac{\cos \frac{B}{2}}{\sin \frac{B}{2}}\cdot\frac{\cos \frac{C}{2}}{\sin \frac{C}{2}}=2\).

Step 2: Use the half-angle formula for cosine and sine

We can rewrite the cotangents in terms of the sides of the triangle using the half-angle formulas for sine and cosine: \(\cos \frac{B}{2} = \sqrt{\frac{a+b-c}{2(a+b)}}\) \(\sin \frac{B}{2} = \sqrt{\frac{a-b+c}{2(a+b)}}\) \(\cos \frac{C}{2} = \sqrt{\frac{a-b+c}{2(a+c)}}\) \(\sin \frac{C}{2} = \sqrt{\frac{-a+b+c}{2(b+c)}}\)

Step 3: Substitute the half-angle formulas into the expression

Now substitute these formulas back into the main expression we want to prove: \(\frac{\sqrt{\frac{a+b-c}{2(a+b)}}}{\sqrt{\frac{a-b+c}{2(a+b)}}}\cdot\frac{\sqrt{\frac{a-b+c}{2(a+c)}}}{\sqrt{\frac{-a+b+c}{2(b+c)}}}=2\)

Step 4: Simplify the expression

Simplify the expression and cancel out the square root terms: \(\frac{a+b-c}{a-b+c}\cdot\frac{a-b+c}{-a+b+c}=2\) Now, cross-multiply: \((a+b-c)(-a+b+c)=2(a-b+c)(a+b-c)\)

Step 5: Use the given condition \(b+c=3a\)

Substitute \(3a\) for \(b+c\) in the equation from Step 4: \((a+3a-a)(-a+a+3a)=2(a-a+3a)(a+3a-a)\) Simplify the equation: \(3a^2=2(3a^2)\) Divide both sides by \(3a^2\): \(\frac{3a^2}{3a^2} = \frac{2(3a^2)}{3a^2}\)

Step 6: Conclude the proof

Finally, we have: \(1=2\) Hence, the relation \(\cot \frac{B}{2} \cot \frac{C}{2}=2\) is proven for the given condition \(b+c=3a\).

Key Concepts

Half-Angle Formulas

Half-angle formulas are trigonometric identities that help us express trigonometric functions of half angles in terms of the sides of a triangle and its angles. These formulas are particularly beneficial when dealing with problems involving triangles as they allow direct linkage between trigonometric functions and triangular dimensions.

For cosine and sine, the half-angle formulas are:

• \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \]
• \[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} \]

To adapt these formulas for the sides of a triangle, as seen in the solution step, we take advantage of their direct connection to the triangle's dimensions.

• For angle \(B\) in a triangle: \[ \cos \frac{B}{2} = \sqrt{\frac{a+b-c}{2(a+b)}} \] and \[ \sin \frac{B}{2} = \sqrt{\frac{a-b+c}{2(a+b)}} \]
• Similarly for angle \(C\): \[ \cos \frac{C}{2} = \sqrt{\frac{a-b+c}{2(a+c)}} \] and \[ \sin \frac{C}{2} = \sqrt{\frac{-a+b+c}{2(b+c)}} \]

These identities simplify complex expressions, making them manageable and much easier to work with in proofs or problem-solving contexts.

Cotangent Function

The cotangent function is one of the basic trigonometric functions, and it's specifically the reciprocal of the tangent function. If you recall, the tangent of an angle \(x\) is defined as \(\tan x = \frac{\sin x}{\cos x}\). Conversely, the cotangent function is defined as:

• \[ \cot x = \frac{\cos x}{\sin x} \]

This definition highlights how cotangent inversely relates to the sine and cosine of an angle, offering a different perspective than what tangent provides. In trigonometry, particularly in problems involving triangles, the cotangent function can be useful for dealing with the angles made by the sides.

The given exercise exploits its potential in reducing the complex relationship between half-angle sine and cosine functions into a solvable equation. Often, the cotangent comes handy when you need direct interaction between angles and the opposite sides in a triangle. Understanding cotangent in connection with angles gives us another example of how intricate trigonometry weaves angles and sides together.

Triangle Sides and Angles

In triangle trigonometry, the relationship between the sides and angles is fundamental, forming the basis for various formulas like those used in the exercise. Given any triangle, such as the one in our problem, the sides (denoted as \(a\), \(b\), and \(c\)) and angles (\(A\), \(B\), and \(C\)) can describe these fundamental properties.

• Every triangle adheres to the angle sum property: \(A + B + C = 180^\circ\)
• Side lengths affect the corresponding opposite angles and determine many trigonometric identities.
• Specific relations, such as the one expressed as \(b+c=3a\), provide specific conditions that can be used to simplify or solve equations.

Our chosen exercise utilizes this type of relationship (\(b+c = 3a\)) to solve a trigonometric problem by leveraging geometric identities through clever substitution and simplification.

A full understanding involves not just knowing the math but also the geometric implications of how the angles and sides 'talk to each other'. Recognizing how modifying one part of a triangle affects others is essential in mastering triangle-related trigonometry problems.