Maths · Evaluation of determinants
=\left|\begin{array}{ccc}\cos \frac{\theta}{2} & 1 & 1 \\ 1 & \cos \frac{\theta}
\( \operatorname{tet} f(\theta)=\left|\begin{array}{ccc}\cos \frac{\theta}{2} & 1 & 1 \\ 1 & \cos \frac{\theta}{2} & -\cos \frac{\theta}{2} \\ -\cos \frac{\theta}{2} & 1 & -1\end{array}\right| \) \( f(\pi)+f(-\pi) \) is equal to
- A. The maximum value of \( f(\theta) \)
- B. The minimum value of \( f(\theta) \)
- C. Average value of the range of \( f(\theta) \)
- D. None of these
Step-by-step solution
We computed the determinant as f(θ) = 2(1+cos²(θ/2)). Then f(π) = f(-π) = 2, so f(π)+f(-π)=4. The maximum value of f(θ) occurs when cos²=1, giving 4. Hence, the sum equals the maximum value.
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