Question:
Evaluate
\[
\cos\frac{6\pi}{17}\cos\frac{10\pi}{17}\cos\frac{12\pi}{17}\cos\frac{14\pi}{17}.
\]
Evaluate
\[
\cos\frac{6\pi}{17}\cos\frac{10\pi}{17}\cos\frac{12\pi}{17}\cos\frac{14\pi}{17}.
\]
Show Hint
Use $\cos(\pi-\theta)=-\cos\theta$ first to convert angles greater than $\frac{\pi}{2}$ into acute angles before applying product identities.
Updated On: Jun 3, 2026
- $-\dfrac{1}{16}$
- $\dfrac{1}{16}$
- $-16$
- $\dfrac{1}{4}$
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Use trigonometric identities involving products of cosine terms and symmetry relations.
Step 2: Meaning
Using \[ \cos(\pi-\theta)=-\cos\theta, \] we obtain \[ \cos\frac{10\pi}{17} =-\cos\frac{7\pi}{17}, \] \[ \cos\frac{12\pi}{17} =-\cos\frac{5\pi}{17}, \] \[ \cos\frac{14\pi}{17} =-\cos\frac{3\pi}{17}. \] Hence \[ P=\cos\frac{6\pi}{17} \cos\frac{10\pi}{17} \cos\frac{12\pi}{17} \cos\frac{14\pi}{17} \] becomes \[ P = -\cos\frac{3\pi}{17} \cos\frac{5\pi}{17} \cos\frac{6\pi}{17} \cos\frac{7\pi}{17}. \]
Step 3: Analysis
A standard trigonometric product identity states that \[ \cos\frac{3\pi}{17} \cos\frac{5\pi}{17} \cos\frac{6\pi}{17} \cos\frac{7\pi}{17} = \frac{1}{16}. \] Substituting this value, \[ P=-\frac{1}{16}. \]
Step 4: Conclusion
Therefore, \[ \cos\frac{6\pi}{17} \cos\frac{10\pi}{17} \cos\frac{12\pi}{17} \cos\frac{14\pi}{17} = -\frac{1}{16}. \]
Final Answer: (A)
Use trigonometric identities involving products of cosine terms and symmetry relations.
Step 2: Meaning
Using \[ \cos(\pi-\theta)=-\cos\theta, \] we obtain \[ \cos\frac{10\pi}{17} =-\cos\frac{7\pi}{17}, \] \[ \cos\frac{12\pi}{17} =-\cos\frac{5\pi}{17}, \] \[ \cos\frac{14\pi}{17} =-\cos\frac{3\pi}{17}. \] Hence \[ P=\cos\frac{6\pi}{17} \cos\frac{10\pi}{17} \cos\frac{12\pi}{17} \cos\frac{14\pi}{17} \] becomes \[ P = -\cos\frac{3\pi}{17} \cos\frac{5\pi}{17} \cos\frac{6\pi}{17} \cos\frac{7\pi}{17}. \]
Step 3: Analysis
A standard trigonometric product identity states that \[ \cos\frac{3\pi}{17} \cos\frac{5\pi}{17} \cos\frac{6\pi}{17} \cos\frac{7\pi}{17} = \frac{1}{16}. \] Substituting this value, \[ P=-\frac{1}{16}. \]
Step 4: Conclusion
Therefore, \[ \cos\frac{6\pi}{17} \cos\frac{10\pi}{17} \cos\frac{12\pi}{17} \cos\frac{14\pi}{17} = -\frac{1}{16}. \]
Final Answer: (A)
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