Question

If \(m\) and \(n\) respectively are the numbers of positive and negative values of \(\theta\) in the interval \([-\pi, \pi]\) that satisfy the equation \(\cos 2 \theta \cos \frac{\theta}{2}=\cos 3 \theta \cos \frac{99}{2}\), then \(m n\) is equal to__

Answer 1

Correct Answer: 25

The given equation is:

\( \cos 2\theta \cdot \cos \frac{\theta}{2} = \cos 30 \cdot \cos \frac{90}{2}. \)

Simplify using trigonometric identities:

\( 2 \cos 2\theta \cdot \cos \frac{\theta}{2} = \cos \frac{50}{2} + \cos \frac{150}{2}. \)

Solve for \(\theta\), leading to:

\( \theta = \frac{2\pi}{5}, \quad \theta = \frac{k}{5}. \)

There are \(m = 5\) positive and \(n = 5\) negative values:

\( mn = 5 \cdot 5 = 25. \)

Final Answer:

\( \boxed{25}. \)

Answer 2

The correct answer is 25.
$\cos 2\theta \cdot \cos \frac{\theta }{2}=\cos 3\theta \cdot \cos \frac{9\theta }{2}$
$\Rightarrow 2\cos 2\theta \cdot \cos \frac{\theta }{2}=2\cos \frac{9\theta }{2}\cdot \cos 3\theta$
$\Rightarrow \cos \frac{5\theta }{2}+\cos \frac{3\theta }{2}=\cos \frac{15\theta }{2}+\cos \frac{3\theta }{2}$
$\Rightarrow \cos \frac{15\theta }{2}=\cos \frac{5\theta }{2}$
$\Rightarrow \frac{15\theta }{2}=2k\pi \pm \frac{5\theta }{2}$
$5\theta =2k\pi \text{or}10\theta =2k\pi$
$\theta =\frac{2k\pi }{5}\theta =\frac{k\pi }{5}$
$\therefore \theta =\left\{-\pi ,\frac{-4\pi }{5},\frac{-3\pi }{5},\frac{-2\pi }{5},\frac{-\pi }{5},0,\frac{\pi }{5},\frac{2\pi }{5},\frac{3\pi }{5},\frac{4\pi }{5},\pi \right\}$
$m=5,n=5$
$\therefore m\cdot n=25$