Question

\(96\cos\frac{π}{33}\cos \frac{2π}{33}\cos\frac{4π}{33}\cos\frac{8π}{33}\cos\frac{16π}{33}\) is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is C

We are tasked with simplifying the expression using the given formula:

\[ \cos A \cos 2A \cos 2^2 A \dots \cos 2^{n-1} A = \frac{\sin(2^n A)}{2^n \sin A} \]

Step 1: Identify the Parameters

• \( A = \frac{\pi}{33} \)
• \( n = 5 \)

The expression becomes:

\[ 96 \cos \frac{\pi}{33} \cos \frac{2\pi}{33} \cos \frac{4\pi}{33} \cos \frac{8\pi}{33} \cos \frac{16\pi}{33} \]

Using the formula:

\[ 96 \cdot \frac{\sin(2^5 \cdot \frac{\pi}{33})}{2^5 \sin \frac{\pi}{33}} \]

Step 2: Simplify Using the Formula

Substitute \( 2^5 = 32 \):

\[ = 96 \cdot \frac{\sin \frac{32\pi}{33}}{32 \sin \frac{\pi}{33}} \]

Step 3: Evaluate the Terms

Using the trigonometric identity \( \sin(\pi - x) = \sin x \), we know:

\[ \sin \frac{32\pi}{33} = \sin \frac{\pi}{33} \]

Substitute this back into the equation:

\[ = 96 \cdot \frac{\sin \frac{\pi}{33}}{32 \sin \frac{\pi}{33}} \]

The \( \sin \frac{\pi}{33} \) terms cancel out, leaving:

\[ = \frac{96}{32} = 3 \]

Final Answer:

The simplified result is:

\[ \boxed{3} \]