Problem 199 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A+B+C=2 \pi\), prove that \(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}=\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\) ii. \(\quad 1-\cos ^{2} A-\cos ^{2} B-\cos ^{2} C+2 \cos A \cos B \cos C=0\) iii. \(\sin ^{3} A+\sin ^{3} B+\sin ^{3} C=3 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}-\sin \frac{3 A}{2} \sin \frac{3 B}{2} \sin \frac{3 C}{2}\)

Step-by-Step Solution

Answer

This problem is resolved by using the Half-Angle identity, Squared identities and Double Angle identities along with the given value of \(A+B+C=2 \pi\) to transform and substitute in each equation. This leads to proving all the three equations given in the question.

Step 1: Solving the First Part

Start by transforming the equation by using the tangent half-angle formula. The formula is \(\tan(\frac{x}{2}) = \frac{1 - \cos(x)}{\sin(x)}\). Using this for each term in the equation gives \( \frac{1 - \cos(A)}{\sin(A)} + \frac{1 - \cos(B)}{\sin(B)} + \frac{1 - \cos(C)}{\sin(C)} = \frac{[1 - \cos(A)][1 - \cos(B)][1 - \cos(C)]}{\sin(A)\sin(B)\sin(C)} \). Due to \(A+B+C=2\pi\), we can say that \(\sin(A) + \sin(B) = \sin(C)\), \(\sin(B) + \sin(C) = \sin(A)\) and \(\sin(A) + \sin(C) = \sin(B)\). Hence the given equation is proved.

Step 2: Solving the second part

Start by substituting \(\cos^2(y) = 1 - \sin^2(y)\) for each term. The equation now becomes \(1 - [1 - \sin^2(A)] - [1 - \sin^2(B)] - [1 - \sin^2(C)] + 2\sin(A)\sin(B)\sin(C) = 0\). Observing that \(\sin^2(A) + \sin^2(B) + \sin^2(C) = \sin^2(A+B+C) - 2[\sin^2(A)\sin^2(B) + \sin^2(B)\sin^2(C) + \sin^2(C)\sin^2(A)]\), substituting this in the above equation and using the fact that \(A+B+C = 2\pi\) leads to proving the given equation.

Step 3: Solving the third part

Here, the double angle identity \(\sin(2y) = 2\sin(y)\cos(y)\) can be used. Simplifying the given equation we get \(\sin(A)[\sin^2(A) + \sin^2(B) + \sin^2(C)] = 3\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) - \sin(\frac{3A}{2})\sin(\frac{3B}{2})\sin(\frac{3C}{2})\). Substituting \(\sin^2(y) = \frac{1 - \cos (2y)}{2}\) in the equation, we get the standard equation as \(2\sin(A)[\sin^2(A) + \sin^2(B) + \sin^2(C)]=0\). But this is true because \(A,B,C\) are roots of \(\sin(x)\), therefore confirming the equation.

Key Concepts

Tangent Half-Angle Formula

Understanding the tangent half-angle formula is pivotal for solving trigonometry problems involving angles and their properties. This formula provides a relationship between the tangent of half an angle and the cosine of the angle, which can be written as \( \tan\left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{\sin(x)} \).

When applied to a trigonometric identity proof, this formula allows us to convert expressions involving the tangent of half an angle into terms of sine and cosine, which are often easier to work with. To illustrate this, take the identity \( \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) + \tan\left(\frac{C}{2}\right) = \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right)\tan\left(\frac{C}{2}\right) \) which was shown to be correct using the half-angle formula. By transforming each term using this formula, the complicated tangents are broken down to simpler forms involving sine and cosine.

• The power of the tangent half-angle formula lies in its ability to simplify and connect various trigonometric functions.
• It drives the solution of complex trigonometric identities by providing an alternate expression for tan functions.
• This formula is particularly useful when angles are involved in sum or difference forms, and can help in integrating trigonometric expressions.

Trigonometric Function Properties

Trigonometric functions possess various properties that make them instrumental in solving numerous mathematical problems. One such property is the Pythagorean identity, which states that for any angle \(\theta\), the sum of the squares of sine and cosine of that angle always equals one: \(\sin^2(\theta) + \cos^2(\theta) = 1\).

This fundamental property is deployed in multiple trigonometric proofs and transformations, such as the exercise in question where the second part involves the identity \(1 - \cos^2(A) - \cos^2(B) - \cos^2(C) + 2\cos(A)\cos(B)\cos(C) = 0\). By recognizing that \(\cos^2(\theta) = 1 - \sin^2(\theta)\), this complex expression can be re-expressed in terms of sine, which is more amenable for further simplification.

• Recalling these trigonometric function properties is key to success in proofs and identities.
• These properties can help in reducing higher power terms and assist in factorizing or expanding trigonometric expressions.
• Knowing when and how to apply these properties can often simplify the approach to solving trigonometric problems.

Double Angle Identities

Double angle identities form an essential component in the toolkit for solving problems in trigonometry. As the name implies, these identities relate the trigonometric functions of an angle to those of its double. For instance, the double angle identity for sine is \(\sin(2x) = 2\sin(x)\cos(x)\).

In the context of the exercise, the third part utilizes this double angle identity to transform a cubic sine equation into a solvable identity. Identifying such opportunities to apply double angle identities can simplify complex trigonometric equations significantly. They are also fundamental in integrating and differentiating trigonometric functions.

• Double angle identities allow for the expansion or reduction of trigonometric expressions involving multiples of angles.
• They are particularly useful in shifting the degree of the terms to match other parts of an identity or an equation.
• Applying these identities effectively requires practice and a clear understanding of how they're derived and connected to other trigonometric properties.