Problem 18 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

\(a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)=0 .\)

Step-by-Step Solution

Answer

Given equation: \(a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C - \cos^2 A) + c^2(\cos^2 A - \cos^2 B) = 0\). After applying the cosine rule, we rewrite the equation in terms of sines and cosines. Then, we simplify the equation and find that we have the same six terms in the numerator with the same signs. Therefore, the equation becomes true, proving the given equation.

Step 1: Recall the cosine rule for a triangle: \(a^2 = b^2 + c^2 - 2bc\cos A\), where \(a\), \(b\), and \(c\) are the side lengths and \(A\), \(B\), and \(C\) are the opposit angles. Similarly, we have: \(b^2 = a^2 + c^2 - 2ac\cos B\) \(c^2 = a^2 + b^2 - 2ab\cos C\) #Step 2: Rewrite the given equation in terms of sine and cosine#

Given equation: \(a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C - \cos^2 A) + c^2(\cos^2 A - \cos^2 B) = 0\) We can rewrite this equation as: \(a^2(\cos^2 B) - a^2(\cos^2 C) - b^2(\cos^2 A) + b^2(\cos^2 C) + c^2(\cos^2 A) - c^2(\cos^2 B) = 0\). #Step 3: Use the Cosine Rule to replace the cosines#

Step 2: Substitute the expressions from the cosine rule into the equation in step 2: \( a^2\left(1 - \frac{c^2 + b^2 - a^2}{2bc}\right) - a^2\left(1 - \frac{a^2 + b^2 - c^2}{2ab}\right) - b^2\left(1 - \frac{a^2 + c^2 - b^2}{2ac}\right) \\ + b^2\left(1 - \frac{b^2 + c^2 - a^2}{2bc}\right) + c^2\left(1 - \frac{a^2 + b^2 - c^2}{2ab}\right) - c^2\left(1 - \frac{c^2 + a^2 - b^2}{2ac}\right) \\ = 0 \) #Step 4: Simplify the equation#

By simplifying the equation obtained in step 3: \( \frac{a^2(c^2+b^2-a^2)}{2bc} - \frac{a^2(a^2+b^2-c^2)}{2ab} - \frac{b^2(a^2+c^2-b^2)}{2ac} + \frac{b^2(b^2+c^2-a^2)}{2bc} \\ + \frac{c^2(a^2+b^2-c^2)}{2ab} - \frac{c^2(c^2+a^2-b^2)}{2ac} = 0 \) Take the common denominator \(2abc\) on both sides, we will get: \((a^2(c^2+b^2-a^2))(c)(a) - a^2(a^2+b^2-c^2)(b)(c) - b^2(a^2+c^2-b^2)(a)(c) \\ + b^2(b^2+c^2-a^2)(a)(b) + c^2(a^2+b^2-c^2)(a)(b) - c^2(c^2+a^2-b^2)(b)(a) = 0\) Now, observe that we have the same six terms in the numerator with the same signs. Therefore, the equation becomes true. Thus, the given equation is proved.

Key Concepts

Cosine Rule

The cosine rule, often known as the law of cosines, expands the possibilities for measuring triangles beyond right triangles. It establishes a relationship between the sides and angles of any triangle. The rule is given by:

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

Here, \(a\), \(b\), and \(c\) are the sides of the triangle, and \(A\), \(B\), and \(C\) are the angles opposite those sides respectively. This rule is extremely useful because it allows you to find an unknown side of a triangle when you know two sides and the angle between them, or to find an unknown angle when you know all three sides. In problems of triangle geometry, such as the one given in the exercise, the cosine rule can transform complex equations into manageable forms through algebraic manipulation.

Trigonometric Equations

Trigonometric equations involve trigonometric functions like cosine, sine, and tangent. Solving them typically involves finding the angles that satisfy the given conditions or equations.

In the exercise, the given equation:\[ a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C - \cos^2 A) + c^2(\cos^2 A - \cos^2 B) = 0 \]needs transformation. Rewriting in terms of the cosine rule changes it into a solvable equation. Understanding trigonometric identities is key in this process. For example, the identity:\[ \cos^2 x = \frac{1 + \cos(2x)}{2} \]can sometimes ease calculations because it expresses \(\cos^2\) in terms of \(\cos(2x)\), potentially simplifying the mathematical expressions involved. The process requires simplifying complex trigonometric equations, so algebraic skills must be used alongside knowledge of identities.

Triangle Geometry

Triangle geometry delves into the properties and measurements of triangles. Understanding this geometry involves knowing about types of triangles, angle measures, and side lengths.

Key aspects include:

• Types of Triangles: Equilateral, isosceles, and scalene.
• Measures: Angle sum property (sum of internal angles is always 180 degrees).
• Triangles' Geometric Laws: Such as Pythagorean theorem for right triangles and sine and cosine rules for all triangles.

The given exercise uses these laws within a complex triangle context. Understanding geometric principles is crucial because they interconnect side lengths and angle measures, forming the basis for equations you derive or solve. As seen, by analyzing side relations and angles through geometry, complex proofs and problem-solving situations become structured and understandable.