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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

The value of (2 π)/(15) (4 π)/(15) (8 π)/(15) (14 π)/(15) is :

The value of cos⁡2π15cos⁡4π15cos⁡8π15cos⁡14π15\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15} is :

Options

  1. A116\frac{1}{16}
  2. B18\frac{1}{8}
  3. C112\frac{1}{12}
  4. D14\frac{1}{4}

Correct answer

A. 116\frac{1}{16}

Step-by-step solution

cos⁡2π15cos⁡4π15cos⁡8π15cos⁡14π15=cos⁡24∘cos⁡48∘cos⁡96∘cos⁡168∘=−cos⁡12∘cos⁡24∘cos⁡48∘cos⁡96∘=−16sin⁡12∘16sin⁡12∘cos⁡12∘cos⁡24∘cos⁡48∘cos⁡96∘=−8sin⁡24∘cos⁡24∘cos⁡48∘cos⁡96∘16sin⁡12∘=−4sin⁡48∘cos⁡48∘cos⁡96∘16sin⁡12∘=−2sin⁡96∘cos⁡96∘16sin⁡12∘=−sin⁡192∘16sin⁡12∘=sin⁡12∘16sin⁡12∘=116\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}=\cos 24^{{}^{{}^{\circ}}} \cos 48^{{}^{{}^{\circ}}} \cos 96^{{}^{{}^{\circ}}} \cos 168^{{}^{{}^{\circ}}}=-\cos 12^{{}^{{}^{\circ}}} \cos 24^{{}^{{}^{\circ}}} \cos 48^{{}^{{}^{\circ}}} \cos 96^{{}^{{}^{\circ}}}=-\frac{16 \sin 12^{{}^{{}^{\circ}}}}{16 \sin 12^{{}^{{}^{\circ}}}} \cos 12^{{}^{{}^{\circ}}} \cos 24^{{}^{{}^{\circ}}} \cos 48^{{}^{{}^{\circ}}} \cos 96^{{}^{{}^{\circ}}}=\frac{-8 \sin 24^{{}^{{}^{\circ}}} \cos 24^{{}^{{}^{\circ}}} \cos 48^{{}^{{}^{\circ}}} \cos 96^{{}^{{}^{\circ}}}}{16 \sin 12^{{}^{{}^{\circ}}}}=\frac{-4 \sin 48^{{}^{{}^{\circ}}} \cos 48^{{}^{{}^{\circ}}} \cos 96^{{}^{{}^{\circ}}}}{16 \sin 12^{{}^{{}^{\circ}}}}=-\frac{2 \sin 96^{{}^{{}^{\circ}}} \cos 96^{{}^{{}^{\circ}}}}{16 \sin 12^{{}^{{}^{\circ}}}}=-\frac{\sin 192^{{}^{{}^{\circ}}}}{16 \sin 12^{{}^{{}^{\circ}}}}=\frac{\sin 12^{{}^{{}^{\circ}}}}{16 \sin 12^{{}^{{}^{\circ}}}}=\frac{1}{16}
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