Trigonometric Equations Question 1
Question 1 - 25 January - Shift 2
If $m$ and $n$ respectively are the numbers of positive and negative value of $\theta$ in the interval $[-\pi, \pi]$ that satisfy the equation $\cos 2 \theta \cos \frac{\theta}{2}=\cos 3 \theta \cos \frac{9 \theta}{2}$, then $m\cdot n$ is equal to
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Answer: 25
Solution:
Formula: Product identities, General solution of Trigonometric Equations
$\cos 2 \theta \cdot \cos \frac{\theta}{2}=\cos 3 \theta \cdot \cos \frac{9 \theta}{2}$
$\Rightarrow 2 \cos 2 \theta \cdot \cos \frac{\theta}{2}=2 \cos \frac{9 \theta}{2} \cdot \cos 3 \theta$
$\Rightarrow \cos \frac{5 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2}+\cos \frac{3 \theta}{2}$
$\Rightarrow \cos \frac{15 \theta}{2}=\cos \frac{5 \theta}{2}$
$\Rightarrow \frac{15 \theta}{2}=2 k \pi \pm \frac{5 \theta}{2}$
$5 \theta=2 k \pi$ or $10 \theta=2 k \pi$
$\theta=\frac{2 k \pi}{5} \quad \theta=\frac{k \pi}{5}$
$\therefore \theta={-\pi, \frac{-4 \pi}{5}, \frac{-3 \pi}{5}, \frac{-2 \pi}{5}, \frac{-\pi}{5}, 0, \frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi}$
$m=5, n=5$
$\therefore m . n=25$
