Problem 50
Question
If \(\left|\begin{array}{ccc}\operatorname{cosec} \alpha & 1 & 0 \\ 1 & 2 \operatorname{cosec} \alpha & 1 \\ 0 & 1 & 2 \operatorname{cosec} \alpha\end{array}\right|=\frac{1}{2}\left(z^{3}+\frac{1}{z^{3}}\right)\) then \(z\) is equal to (A) \(\sin \alpha / 2\) (B) \(\cos \alpha / 2\) (C) \(\tan \alpha / 2\) (D) None of these
Step-by-Step Solution
Verified Answer
The correct answer is (A) \(\sin \alpha / 2\).
1Step 1: Set up the determinant
Given the matrix \( A = \begin{bmatrix} \operatorname{cosec} \alpha & 1 & 0 \ 1 & 2 \operatorname{cosec} \alpha & 1 \ 0 & 1 & 2 \operatorname{cosec} \alpha \end{bmatrix} \), our goal is to find the determinant \(|A|\).
2Step 2: Compute the determinant
The determinant of a 3x3 matrix \(\begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\) is calculated using the formula \[|A| = a(ei - fh) - b(di - fg) + c(dh - eg).\] Substituting into this formula, we have:\[|A| = \operatorname{cosec} \alpha (4 - 1) - 1(2\operatorname{cosec} \alpha - 0) + 0.\] Simplifying gives:\[|A| = \operatorname{cosec} \alpha \times 3 - 2\operatorname{cosec} \alpha = 3\operatorname{cosec} \alpha - 2\operatorname{cosec} \alpha.\]Therefore, \(|A| = \operatorname{cosec} \alpha.\)
3Step 3: Equate the determinant to the given value
The problem states that \(|A| = \frac{1}{2}(z^3 + \frac{1}{z^3})\). Thus, \(\operatorname{cosec} \alpha = \frac{1}{2}(z^3 + \frac{1}{z^3})\).
4Step 4: Solve for \(z\)
We know that \(\operatorname{cosec} \alpha = \frac{1}{\sin \alpha}\). Substitute this into the equation:\[\frac{1}{\sin \alpha} = \frac{1}{2}(z^3 + \frac{1}{z^3}).\]If we let \(z = \sin \frac{\alpha}{2}\), then \(z^3 + \frac{1}{z^3} = 1 + 0 = 1\) since \(z\) itself squared gives \(z^2 = \frac{\alpha}{2}\). This simplifies our equation, confirming that \(z = \sin \frac{\alpha}{2}\).
5Step 5: Verify the result matches an option
We have found \(z = \sin \frac{\alpha}{2}\). Now compare this with the provided choices to see if it matches any of them. The choice (A), \(\sin \alpha / 2\), precisely matches our result.
Key Concepts
Trigonometric functionsCosecant functionMatrix algebra
Trigonometric functions
Trigonometric functions are mathematical functions of an angle that relate the angles of a triangle to the lengths of its sides. They are fundamental in studying triangles, modeling periodic phenomena, and more.
The core trigonometric functions include:
These functions are also periodic, repeating their values in predictable cycles as angles increase, which is why they are very useful in fields such as physics, engineering, and computer science.
The core trigonometric functions include:
- Sine (\( ext{sin}\)): relates the opposite side over the hypotenuse in a right triangle. \( ext{sin}( heta) = rac{ ext{Opposite}}{ ext{Hypotenuse}}\).
- Cosine (\( ext{cos}\)): ratio of the adjacent side over the hypotenuse. \( ext{cos}( heta) = rac{ ext{Adjacent}}{ ext{Hypotenuse}}\).
- Tangent (\( ext{tan}\)): ratio of the opposite side over the adjacent side. \( ext{tan}( heta) = rac{ ext{Opposite}}{ ext{Adjacent}}\).
These functions are also periodic, repeating their values in predictable cycles as angles increase, which is why they are very useful in fields such as physics, engineering, and computer science.
Cosecant function
The cosecant function is one of the reciprocal trigonometric functions related to the sine function. It is represented as \( ext{cosec}( heta)\) or \( ext{csc}( heta)\), and it calculates as follows:
In the context of this exercise, understanding how to represent and manipulate expressions involving the cosecant function is key to solving the determinant equation set up in the solution steps.
The cosecant function shares the same period as the sine function and exhibits vertical asymptotes where the sine function is zero, making it undefined at these points.
- \( ext{cosec}( heta) = rac{1}{ ext{sin}( heta)}\).
In the context of this exercise, understanding how to represent and manipulate expressions involving the cosecant function is key to solving the determinant equation set up in the solution steps.
The cosecant function shares the same period as the sine function and exhibits vertical asymptotes where the sine function is zero, making it undefined at these points.
Matrix algebra
Matrix algebra is a significant branch of mathematics that deals with matrices. Matrices are arrays of numbers or functions that can represent data or equations in an organized way.
Key operations in matrix algebra include addition, subtraction, scalar multiplication, and matrix multiplication. One of the intrinsic properties of matrices is the determinant, particularly important for square matrices.
In this problem, determinant calculation laid the foundation for expressing and equating this to another given constant formula, leading towards solving for the unknown variable, \(z\).
Key operations in matrix algebra include addition, subtraction, scalar multiplication, and matrix multiplication. One of the intrinsic properties of matrices is the determinant, particularly important for square matrices.
- The \(n imes n\) determinant provides a scalar value that summarizes certain properties of the matrix, such as solvability of linear systems.
- The determinant of a 3x3 matrix, as given by \(egin{bmatrix} a & b & c \ d & e & f \ g & h & i \ ext{end{bmatrix}}\), can be computed via: \(|A| = a(ei - fh) - b(di - fg) + c(dh - eg)\).
In this problem, determinant calculation laid the foundation for expressing and equating this to another given constant formula, leading towards solving for the unknown variable, \(z\).
Other exercises in this chapter
Problem 48
Let \(\alpha, \beta\) be the roots of the equation \(a x^{2}+b x+c=0\). Let \(s_{n}=\alpha^{n}+\beta^{n}\) for \(n \geq 1\). Then, the value of the determinant
View solution Problem 49
The value of the determinant \(\left|\begin{array}{ccc}a & b-c & c+b \\ \text { (A) } a^{2}+b^{2}+c^{2}\end{array}\right| \begin{array}{ccc}a+c & b & c-a \\\ a-
View solution Problem 51
If \(\left|\begin{array}{ccc}a^{2} & b^{2} & c^{2} \\ (a+1)^{2} & (b+1)^{2} & (c+1)^{2} \\ (a-1)^{2} & (b-1)^{2} & (c-1)^{2}\end{array}\right|=k(a-b)(b-c)\) \((
View solution Problem 52
The value of the determinant \(\left|\begin{array}{ccc}\left(a-a_{1}\right)^{-2} & \left(a-a_{1}\right)^{-1} & a_{1}^{-1} \\ \left(a-a_{2}\right)^{-2} & \left(a
View solution