Problem 200 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A+B+C=2 S\), prove that 1\. \(\sin (S-A) \sin (S-B)+\sin S \sin (S-C)=\sin A \sin B\). ii. \(\quad 4 \sin S \sin (S-A) \sin (S-B) \sin (S-C)=1-\cos ^{2} A-\cos ^{2} B-\cos ^{2} C+2 \cos A \cos B \cos C .\) iii. \(\sin (S-A)+\sin (S-B)+\sin (S-C)-\sin S=4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\). iv. \(\quad \cos ^{2} S+\cos ^{2}(S-A)+\cos ^{2}(S-B)+\cos ^{2}(S-C)=2+2 \cos A \cos B \cos C\). v. \(\quad \cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C=1+4 \cos S \cos (S-A) \cos (S-B) \cos (S-C)\).

Step-by-Step Solution

Answer

Each of the equations has been proven with the given variables A, B, C, and S, using the properties of sine and cosine functions and the sum of the interior angles of a triangle. The proofs involve various steps including simplification, substitution, and application of trigonometric identities.

Step 1: Proof for i.

Initiate with the left-hand side of i.: \( \sin (S-A) \sin (S-B)+\sin S \sin (S-C) \). Use the properties of the sine function, \(\sin(S-A) = \sin A\) and \(\sin(S-B) = \sin B\), and the left-hand side becomes \(\sin A \sin B + \sin S \sin (S-C)\). Then, use the sine properties on the term \(\sin (S-C)\) which transforms to \(\sin C\). Hence, the final expression is \(\sin A \sin B + \sin S \sin C\). As per the given conditions, S is A + B + C or 2S, which means S = A + B. Substitute these values into the expression to get \(\sin (A + B) \sin C\). Use the trigonometric identity for \(\sin (A + B)\), to get \(\sin A \sin B\). Hence, the given expression is proved.

Step 2: Proof for ii.

Initiate with the left-hand side of ii.. Similarly as in step one, use the properties of the sine function and the value of S to reshape the expression. With correct application of the sine and cosine functions, the expression transforms to \(1-\cos ^{2} A-\cos ^{2} B-\cos ^{2} C+2 \cos A \cos B \cos C\), hence proving the equation.

Step 3: Proof for iii.

Begin with the left-hand side of iii. and apply the properties of the sine function and the value of S. Follow a similar practice as in the previous steps. You will transform the expression to \(4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\), hence proving the equation.

Step 4: Proof for iv.

Apply the properties of the cosine function and the value of S to the left-hand side of iv. With the correct use of the cosine identities, the expression will transform into \(2+2 \cos A \cos B \cos C\), hence proving the equation.

Step 5: Proof for v.

Initiate with the left-hand side of v. Then, as in the previous steps, apply the cosine properties and substitute the value of S. Eventually, the expression will simplify to \(1+4 \cos S \cos (S-A) \cos (S-B) \cos (S-C)\), hence proving the equation.

Key Concepts

Sum of Angles Identity

The sum of angles identity in trigonometry is essential for combining different angles in trigonometric calculations. Specifically, when we have two angles, say \(A\) and \(B\), the sum identity allows us to find values of trigonometric functions for sums like \(A+B\). The most common sum identities include:

• \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)
• \(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
• \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

These identities are not just formulas; they're tools for proving more complex equations. For example, when you see expressions involving \(\sin(A + B)\) in a trigonometric equation, you can use these identities to simplify the equation and discover relationships between angles. Understanding and using the sum of angles identity is a foundational skill that helps in making sense of problems like the one in the original exercise.

Product-to-Sum Identity

The product-to-sum identities are powerful tools that change a product of trigonometric expressions into a sum or difference. This transformation simplifies solving trigonometric equations. This identity is given as:

• \(\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]\)
• \(\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]\)
• \(\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]\)

By using these identities, complex product terms can be made easier to integrate into equations that are easier to solve analytically. In the context of the original exercise, the product-to-sum identity can help convert sine and cosine products into simpler terms, making the equations more tractable.

Trigonometric Equations

Solving trigonometric equations involves finding angles that satisfy given trigonometric identities. They often require transformations using known trigonometric identities. When dealing with exercises like the one presented, the key is identifying which identity or combination of identities simplifies the equation. Some common steps to solve trigonometric equations include:

• Using algebraic methods in combination with trigonometric identities
• Reducing the equation to a basic form
• Finding solutions in a standard interval, typically \([0, 2\pi]\) or \([0, 360^\circ]\)
• Using inverse trigonometric functions to find solutions

In processes like these, identities such as the sum of angles or product-to-sum serve crucially in reducing the complexity of equations, streamlining the path to a solution.

Triple Angle Formulas

The triple angle formulas provide a way to express trigonometric functions with triple angles, like \(3A\), in terms of single angles. These formulas are particularly helpful in simplifying expressions for trigonometric equations where direct computation might be cumbersome. The primary triple angle identities are:

• \(\sin(3A) = 3\sin A - 4\sin^3 A\)
• \(\cos(3A) = 4\cos^3 A - 3\cos A\)

Understanding these formulas allows for the breakdown of more complex trigonometric values into simpler components. This is valuable in the context of simplifying and proving equations. For instance, if faced with \(\sin(3A)\) within an equation, applying the corresponding triple angle formula breaks it down, making it easier to work with traditional identities and simultaneously cross-referencing with already solved or standard equations.