Question:
The value of $\cos \frac2π15 · \cos \frac4\pi15 · \cos \frac8\pi15 · \cos \frac16\pi15$ is equal to
The value of $\cos \frac2π15 · \cos \frac4\pi15 · \cos \frac8\pi15 · \cos \frac16\pi15$ is equal to
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The value of $\cos 2\pi15 · \cos \frac4π/15 · \cos 8π/15 · \cos 16π/15$ is equal to
Updated On: Apr 15, 2026
- $1/16$
- $1/32$
- $1/64$
- $1/8$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Use the identity $\prod_{r=0}^{n-1} \cos(2^r \theta) = \frac{\sin(2^n \theta)}{2^n \sin \theta}$.
Step 2: Analysis
Here, $\theta = 2\pi/15$ and $n = 4$.
Step 3: Evaluation
The value is $\frac{\sin(16 \cdot \frac{2\pi}{15})}{2^4 \sin(\frac{2\pi}{15})} = \frac{\sin(32\pi/15)}{16 \sin(2\pi/15)}$.
Step 4: Conclusion
Since $\sin(32\pi/15) = \sin(2\pi + 2\pi/15) = \sin(2\pi/15)$, the terms cancel out, leaving $1/16$.
Final Answer: (a)
Use the identity $\prod_{r=0}^{n-1} \cos(2^r \theta) = \frac{\sin(2^n \theta)}{2^n \sin \theta}$.
Step 2: Analysis
Here, $\theta = 2\pi/15$ and $n = 4$.
Step 3: Evaluation
The value is $\frac{\sin(16 \cdot \frac{2\pi}{15})}{2^4 \sin(\frac{2\pi}{15})} = \frac{\sin(32\pi/15)}{16 \sin(2\pi/15)}$.
Step 4: Conclusion
Since $\sin(32\pi/15) = \sin(2\pi + 2\pi/15) = \sin(2\pi/15)$, the terms cancel out, leaving $1/16$.
Final Answer: (a)
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