\(96\cos\frac{π}{33}\cos \frac{2π}{33}\cos\frac{4π}{33}\cos\frac{8π}{33}\cos\frac{16π}{33}\) is equal to
- 1
- 2
- 3
- 4
The Correct Option is C
Solution and Explanation
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:
\[ \boxed{3} \]
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