Question

If $S = \{ \theta \in [-\pi, \pi] : \cos\theta \cos\frac{5\theta}{2} = \cos 7\theta \cos\frac{7\theta}{2} \}$, then $n(S)$ is equal to :}

Answer 1

Correct Answer: 21

Step 1: Understanding the Concept:
We use trigonometric product-to-sum identities to simplify the given equation and then solve for $\theta$ in the specified interval.
: Key Formula or Approach:
$2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.
Step 2: Detailed Explanation:
$\cos \theta \cos \frac{5\theta}{2} = \cos 7\theta \cos \frac{7\theta}{2}$
Multiply by 2:
$\cos( \theta + \frac{5\theta}{2} ) + \cos( \frac{5\theta}{2} - \theta ) = \cos( 7\theta + \frac{7\theta}{2} ) + \cos( 7\theta - \frac{7\theta}{2} )$
$\cos \frac{7\theta}{2} + \cos \frac{3\theta}{2} = \cos \frac{21\theta}{2} + \cos \frac{7\theta}{2}$
$\cos \frac{3\theta}{2} = \cos \frac{21\theta}{2}$
$\frac{21\theta}{2} = 2n\pi \pm \frac{3\theta}{2}$
Case 1: $\frac{21\theta}{2} = 2n\pi + \frac{3\theta}{2} \implies \frac{18\theta}{2} = 2n\pi \implies 9\theta = 2n\pi \implies \theta = \frac{2n\pi}{9}$
Values in $[-\pi, \pi]$: $n = 0, \pm 1, \pm 2, \pm 3, \pm 4$. (9 values).
Case 2: $\frac{21\theta}{2} = 2n\pi - \frac{3\theta}{2} \implies \frac{24\theta}{2} = 2n\pi \implies 12\theta = 2n\pi \implies \theta = \frac{n\pi}{6}$
Values in $[-\pi, \pi]$: $n = 0, \pm 1, \dots, \pm 6$. (13 values).
Total distinct values:
$\theta = \{ 0, \pm \frac{2\pi}{9}, \pm \frac{4\pi}{9}, \pm \frac{6\pi}{9}, \pm \frac{8\pi}{9} \}$ and $\{ 0, \pm \frac{\pi}{6}, \pm \frac{2\pi}{6}, \pm \frac{3\pi}{6}, \pm \frac{4\pi}{6}, \pm \frac{5\pi}{6}, \pm \frac{6\pi}{6} \}$.
Note: $\pm \frac{6\pi}{9} = \pm \frac{2\pi}{3} = \pm \frac{4\pi}{6}$ (Common values).
Counting distinct elements: $1 + 8 + 12 = 21$.
Step 3: Final Answer:
The number of elements in set $S$ is 21.