Problem 25 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A value of \(\theta \in(0, \pi / 3)\), for which \(\left|\begin{array}{ccc}1+\cos ^{2} \theta & \sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & 1+\sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & \sin ^{2} \theta & 1+4 \cos 6 \theta\end{array}\right|=0\), is [April 12, 2019 (II)] (a) \(\frac{\pi}{9}\) (b) \(\frac{\pi}{18}\) (c) \(\frac{7 \pi}{24}\) (d) \(\frac{7 \pi}{36}\)

Step-by-Step Solution

Answer

The value of \( \theta \) is \( \frac{\pi}{9} \) (option a).

Step 1: Understand the Matrix and Determinant

Identify the matrix given in the problem: \[ \begin{bmatrix} 1+\cos^2 \theta & \sin^2 \theta & 4 \cos 6\theta \ \cos^2 \theta & 1+\sin^2 \theta & 4 \cos 6\theta \ \cos^2 \theta & \sin^2 \theta & 1+4 \cos 6 \theta \end{bmatrix} \]. The problem asks us to determine for which value of \( \theta \) the determinant of this matrix is equal to zero.

Step 2: Simplify Matrix Elements

Notice that for trigonometric identities, we have \( \sin^2 \theta + \cos^2 \theta = 1 \). This can be used to express the first two elements on the diagonal as complete squares: \( 1 + \cos^2 \theta = 1 + (1 - \sin^2 \theta) = 2 - \sin^2 \theta \) and similarly, \( 1 + \sin^2 \theta = 2 - \cos^2 \theta \).

Step 3: Determinant Calculation

Write the determinant and expand it: \[ \det(A) = (2 - \sin^2 \theta)((2 - \cos^2 \theta)(1+4 \cos 6 \theta) - (\sin^2 \theta)(4\cos 6 \theta)) - (\sin^2 \theta)(\cos^2 \theta(1+4\cos 6 \theta) - \sin^2 \theta(4\cos 6 \theta)) \], and \[ +4\cos 6 \theta \left((\cos^2 \theta)(\sin^2 \theta) - ((2-\cos^2 \theta)(\cos^2 \theta))\right) \].

Step 4: Substitute and Equate to Zero

Simplifying the determinant equation involves substituting values if needed. However, an easier approach might be comparing the angular periodic component. Substitute options for \( \theta \) and expand any simplifications needed to set the equation to zero and evaluate each option.

Step 5: Solve for Zero Determinant

It's often easier to substitute each given angle option into the determinant equation to check which results in zero. Start with angle \( \theta = \frac{\pi}{9} \) and compute the determinants, followed by the others, until you find when \( \det(A) = 0 \). Evaluate trigonometric values such as \( \cos 6\theta \), especially because they recur in the determinant elements.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in solving and simplifying expressions involving trigonometric functions like sine, cosine, and tangent. One of the most fundamental identities is

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity helps simplify expressions such as the diagonal elements in the matrix by rewriting them as complete squares, facilitating easier determinant calculations. Another useful identity in problems concerning angles multiple of \( \theta \) is the multiple angle formulas for cosine and sine:

• \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta \)
• \( \sin(2\theta) = 2\sin \theta \cos \theta \)

These identities allow transforming complex trigonometric expressions into simpler ones that are easier to substitute and evaluate when solving equations to find values of \( \theta \). In this exercise, particularly, knowing that \( \cos(6\theta) \) can be linked to known values helps simplify the determinant to zero at the correct angle value.

Matrix Algebra

Matrix Algebra involves operations with matrices, such as addition, multiplication, and finding determinants. A determinant is a scalar value that can be calculated from the elements of a square matrix and provides insights into matrix properties like invertibility.
In this exercise, we are tasked with finding a value of \( \theta \) for which the determinant of the given matrix is zero. This involves simplifying the determinant formula to find a meaningful solution:

• \( \det(A) = (2 - \sin^2 \theta)((2 - \cos^2 \theta)(1+4 \cos 6 \theta) - (\sin^2 \theta)(4\cos 6 \theta)) \)
• \( - (\sin^2 \theta)(\cos^2 \theta(1+4\cos 6 \theta) - \sin^2 \theta(4\cos 6 \theta)) \)
• \( +4\cos 6 \theta \left((\cos^2 \theta)(\sin^2 \theta) - ((2-\cos^2 \theta)(\cos^2 \theta))\right) \)

Finding where the determinant equals zero requires substituting the given \( \theta \) values and simplifying the expression until the result is zero. This simplification process involves careful algebraic manipulation and evaluation of the trigonometric terms.

JEE Main Mathematics

In the context of competitive exams like JEE Main, Mathematics problems emphasize conceptual understanding and application skills, particularly in topics like trigonometry and algebra. Such problems often present

• Multistep processes, requiring a strong grip on formulas and identities.
• Application of concepts like matrix determinants into practical calculations.

Mastering these concepts involves both understanding formula derivations and practice in applying them, as evidenced by this determinant problem. This exercise blends trigonometric identities with matrix algebra, testing students' abilities to simplify and accurately compute determinant expressions.
Engaging actively with these kinds of problems helps build the critical thinking skills necessary for success in exams. Remember, practice and familiarity with such exercises equip students with the intuition to identify correct strategies quickly, essential during high-pressure exam environments.