Consider the following statements in respect of the determinant \(\left| {\begin{array}{} {{{\cos }^2}\frac{\alpha }{2}}&{{{\sin }^2}\frac{\alpha }{2}}\\ {{{\sin }^2}\frac{\beta }{2}}&{{{\cos }^2}\frac{\beta }{2}} \end{array}} \right|\) Where α, β are complementary angles 1. The value of the determinant is \(\frac{1}{{√ 2 }}\cos \left( {\frac{{\alpha - \beta }}{2}} \right)\;\) 2. The maximum value of the determinant is \(\frac{1}{\sqrt2}\) Which of the above statements is/are correct?
Both 1 and 2
Understanding the Problem: Determinant with Complementary Angles
The question asks us to consider a determinant whose elements involve trigonometric functions of half angles, specifically \(\cos^2\frac{\alpha}{2}\), \(\sin^2\frac{\alpha}{2}\), \(\sin^2\frac{\beta}{2}\), and \(\cos^2\frac{\beta}{2}\). We are given that \(\alpha\) and \(\beta\) are complementary angles, which means their sum is \(\frac{\pi}{2}\) radians, or 90 degrees (\(\alpha + \beta = \frac{\pi}{2}\)). We need to evaluate two statements about the value and maximum value of this determinant.
The given determinant is:
$$ D = \left| {\begin{array}{} {{{\cos }^2}\frac{\alpha }{2}}&{{{\sin }^2}\frac{\alpha }{2}}\\ {{{\sin }^2}\frac{\beta }{2}}&{{{\cos }^2}\frac{\beta }{2}} \end{array}} \right| $$
Calculating the Determinant Value
For a \(2 \times 2\) determinant \(\left| {\begin{array}{} a&b\\ c&d \end{array}} \right|\), the value is calculated as \(ad - bc\).
Applying this to our determinant:
$$ D = \left({\cos^2\frac{\alpha}{2}}\right)\left({\cos^2\frac{\beta}{2}}\right) - \left({\sin^2\frac{\alpha}{2}}\right)\left({\sin^2\frac{\beta}{2}}\right) $$
We can use the half-angle identities: \(\cos^2 x = \frac{1 + \cos(2x)}{2}\) and \(\sin^2 x = \frac{1 - \cos(2x)}{2}\).
Let's substitute these identities into the determinant expression:
$$ D = \left(\frac{1 + \cos \alpha}{2}\right)\left(\frac{1 + \cos \beta}{2}\right) - \left(\frac{1 - \cos \alpha}{2}\right)\left(\frac{1 - \cos \beta}{2}\right) $$
$$ D = \frac{1}{4} \left[ (1 + \cos \alpha)(1 + \cos \beta) - (1 - \cos \alpha)(1 - \cos \beta) \right] $$
Expand the terms inside the square brackets:
$$ (1 + \cos \alpha)(1 + \cos \beta) = 1 + \cos \beta + \cos \alpha + \cos \alpha \cos \beta $$
$$ (1 - \cos \alpha)(1 - \cos \beta) = 1 - \cos \beta - \cos \alpha + \cos \alpha \cos \beta $$
Now, subtract the second expanded expression from the first:
$$ (1 + \cos \alpha + \cos \beta + \cos \alpha \cos \beta) - (1 - \cos \alpha - \cos \beta + \cos \alpha \cos \beta) $$
$$ = 1 + \cos \alpha + \cos \beta + \cos \alpha \cos \beta - 1 + \cos \alpha + \cos \beta - \cos \alpha \cos \beta $$
$$ = (1 - 1) + (\cos \alpha + \cos \alpha) + (\cos \beta + \cos \beta) + (\cos \alpha \cos \beta - \cos \alpha \cos \beta) $$
$$ = 0 + 2 \cos \alpha + 2 \cos \beta + 0 = 2 \cos \alpha + 2 \cos \beta $$
Substitute this back into the determinant expression:
$$ D = \frac{1}{4} (2 \cos \alpha + 2 \cos \beta) = \frac{2(\cos \alpha + \cos \beta)}{4} = \frac{1}{2}(\cos \alpha + \cos \beta) $$
Using Complementary Angles Property
We are given that \(\alpha\) and \(\beta\) are complementary angles, so \(\alpha + \beta = \frac{\pi}{2}\). This implies \(\beta = \frac{\pi}{2} - \alpha\).
Substitute \(\beta\) in terms of \(\alpha\) into the determinant value:
$$ D = \frac{1}{2}(\cos \alpha + \cos (\frac{\pi}{2} - \alpha)) $$
Using the trigonometric identity \(\cos(\frac{\pi}{2} - \theta) = \sin \theta\), we have \(\cos (\frac{\pi}{2} - \alpha) = \sin \alpha\).
So, the value of the determinant is:
$$ D = \frac{1}{2}(\cos \alpha + \sin \alpha) $$
Evaluating Statement 1: Value of the Determinant
Statement 1 says the value of the determinant is \(\frac{1}{\sqrt{2}}\cos \left( {\frac{{\alpha - \beta }}{2}} \right)\). Let's try to express our result \(D = \frac{1}{2}(\cos \alpha + \sin \alpha)\) in this form.
We know \(\alpha + \beta = \frac{\pi}{2}\). We can write \(\alpha = \frac{\alpha + \beta}{2} + \frac{\alpha - \beta}{2} = \frac{\pi}{4} + \frac{\alpha - \beta}{2}\) and \(\beta = \frac{\alpha + \beta}{2} - \frac{\alpha - \beta}{2} = \frac{\pi}{4} - \frac{\alpha - \beta}{2}\).
Let \(x = \frac{\alpha - \beta}{2}\). Then \(\alpha = \frac{\pi}{4} + x\) and \(\beta = \frac{\pi}{4} - x\).
Our determinant value is \(D = \frac{1}{2}(\cos \alpha + \cos \beta)\).
Substitute the new expressions for \(\alpha\) and \(\beta\):
$$ D = \frac{1}{2}\left(\cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right)\right) $$
Using the sum-to-product identity: \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\).
Here, \(A = \frac{\pi}{4} + x\) and \(B = \frac{\pi}{4} - x\).
$$ \frac{A+B}{2} = \frac{(\frac{\pi}{4} + x) + (\frac{\pi}{4} - x)}{2} = \frac{\frac{\pi}{2}}{2} = \frac{\pi}{4} $$
$$ \frac{A-B}{2} = \frac{(\frac{\pi}{4} + x) - (\frac{\pi}{4} - x)}{2} = \frac{2x}{2} = x $$
So, \(\cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) = 2 \cos\left(\frac{\pi}{4}\right)\cos(x)\).
Substitute this back into the determinant expression:
$$ D = \frac{1}{2} \left[ 2 \cos\left(\frac{\pi}{4}\right)\cos(x) \right] = \cos\left(\frac{\pi}{4}\right)\cos(x) $$
We know \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\) and \(x = \frac{\alpha - \beta}{2}\).
$$ D = \frac{1}{\sqrt{2}} \cos\left(\frac{\alpha - \beta}{2}\right) $$
This matches the expression given in statement 1. Thus, statement 1 is correct.
Evaluating Statement 2: Maximum Value of the Determinant
The value of the determinant is \(D = \frac{1}{\sqrt{2}} \cos\left(\frac{\alpha - \beta}{2}\right)\).
To find the maximum value of \(D\), we need to consider the range of the cosine function. The maximum value of \(\cos \theta\) for any real angle \(\theta\) is 1.
Therefore, the maximum value of \(\cos\left(\frac{\alpha - \beta}{2}\right)\) is 1.
The maximum value of the determinant \(D\) is then:
$$ D_{max} = \frac{1}{\sqrt{2}} \times 1 = \frac{1}{\sqrt{2}} $$
Statement 2 says the maximum value of the determinant is \(\frac{1}{\sqrt{2}}\). This matches our calculation. Thus, statement 2 is correct.
Conclusion
Based on our step-by-step calculation and analysis using the properties of determinants, trigonometric identities, and complementary angles, both statements 1 and 2 are found to be correct.
Statement 1: The value of the determinant is indeed \(\frac{1}{\sqrt{2}}\cos \left( {\frac{{\alpha - \beta }}{2}} \right)\).
Statement 2: The maximum value of the determinant, which occurs when \(\cos\left(\frac{\alpha - \beta}{2}\right) = 1\), is \(\frac{1}{\sqrt{2}}\).
| Statement | Description | Evaluation |
|---|---|---|
| 1 | The value of the determinant is \(\frac{1}{{√ 2 }}\cos \left( {\frac{{\alpha - \beta }}{2}} \right)\). | Correct |
| 2 | The maximum value of the determinant is \(\frac{1}{\sqrt2}\). | Correct |
Revision Table: Key Concepts for Determinants and Trigonometry
| Concept | Formula/Description |
|---|---|
| Determinant of \(2 \times 2\) Matrix | For \(\left| {\begin{array}{} a&b\\ c&d \end{array}} \right|\), value is \(ad - bc\). |
| Half-Angle Identity for \(\cos^2 x\) | \(\cos^2 x = \frac{1 + \cos(2x)}{2}\) |
| Half-Angle Identity for \(\sin^2 x\) | \(\sin^2 x = \frac{1 - \cos(2x)}{2}\) |
| Complementary Angles | Angles \(\alpha, \beta\) are complementary if \(\alpha + \beta = \frac{\pi}{2}\) or 90°. |
| Complementary Angle Identity | \(\cos(\frac{\pi}{2} - \theta) = \sin \theta\) |
| Sum-to-Product Identity | \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\) |
| Maximum Value of \(\cos \theta\) | The maximum value of \(\cos \theta\) is 1. |
Additional Information: Exploring Complementary Angles and Determinants
Complementary Angles: In geometry, complementary angles are two angles that add up to 90 degrees (\(\frac{\pi}{2}\) radians). These angles are often seen together in right triangles. The relationship \(\sin \theta = \cos(90^\circ - \theta)\) is a fundamental property stemming from complementary angles.
Determinants: Determinants are scalar values calculated from the elements of a square matrix. They have various applications in linear algebra, such as determining if a matrix is invertible, finding the area of a triangle (for 2x2 matrices), or the volume of a parallelepiped (for 3x3 matrices). The determinant of a 2x2 matrix \(\left| {\begin{array}{} a&b\\ c&d \end{array}} \right|\) is \(ad-bc\).
In this problem, the determinant value \(D = \frac{1}{\sqrt{2}} \cos\left(\frac{\alpha - \beta}{2}\right)\) shows how the determinant depends on the difference between the complementary angles \(\alpha\) and \(\beta\). While their sum is fixed (\(\frac{\pi}{2}\)), their difference (\(\alpha - \beta\)) can vary. For example, if \(\alpha = \frac{\pi}{2}, \beta = 0\), then \(\alpha + \beta = \frac{\pi}{2}\) and \(\alpha - \beta = \frac{\pi}{2}\). If \(\alpha = \frac{\pi}{4}, \beta = \frac{\pi}{4}\), then \(\alpha + \beta = \frac{\pi}{2}\) and \(\alpha - \beta = 0\). The maximum value occurs when \(\cos\left(\frac{\alpha - \beta}{2}\right) = 1\), which happens when \(\frac{\alpha - \beta}{2}\) is a multiple of \(2\pi\). For angles typically considered in this context (e.g., positive acute angles), this happens when \(\frac{\alpha - \beta}{2} = 0\), i.e., \(\alpha = \beta\). Since \(\alpha + \beta = \frac{\pi}{2}\) and \(\alpha = \beta\), we get \(\alpha = \beta = \frac{\pi}{4}\). In this case, \(\alpha - \beta = 0\), \(\frac{\alpha - \beta}{2} = 0\), \(\cos(0) = 1\), and \(D = \frac{1}{\sqrt{2}}\).
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