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Mathematics Question on Trigonometric Functions

The value of cos⁡π3.cos⁡π7.cos⁡2π7cos⁡4π7:\cos \frac{\pi}{3}. \cos \frac{\pi}{7}. \cos \frac{2\pi}{7}\cos \frac{4\pi}{7} :

A

16

B

−116\frac{-1}{16}

C

1116\frac{11}{16}

D

1611\frac{16}{11}

Answer

−116\frac{-1}{16}

Explanation

Solution

Consider cos⁡π3.cos⁡π7.cos⁡2π7cos⁡4π7\cos \frac{\pi}{3}. \cos \frac{\pi}{7}. \cos \frac{2\pi}{7}\cos \frac{4\pi}{7} Let A=π7A = \frac{\pi}{7} then cos⁡π3.cos⁡π7.cos⁡2π7cos⁡4π7\cos \frac{\pi}{3}. \cos \frac{\pi}{7}. \cos \frac{2\pi}{7}\cos \frac{4\pi}{7} =12cos⁡Acos⁡2Acos⁡4A = \frac{1}{2} \cos A \cos2A \cos4A =12cos⁡Acos⁡2Acos⁡22A = \frac{1}{2} \cos A \cos 2 A \cos2^{2} A =12.sin⁡23A23sin⁡A=12sin⁡8A8sin⁡A = \frac{1}{2} . \frac{\sin2^{3} A}{2^{3} \sin A} = \frac{1}{2} \frac{\sin8A}{8 \sin A} (usingcos⁡Acos⁡2Acos⁡22A.......cos⁡2n−1A=sin⁡2nA2nsin⁡A) \left(\text{using} \cos A \cos2A \cos2^{2} A ....... \cos 2^{n-1} A = \frac{ \sin2^{n }A}{2^{n} \sin A}\right) =116sin⁡(7A+A)sin⁡A=116sin⁡(π+A)sin⁡A =\frac{1}{16} \frac{\sin\left(7A +A\right)}{\sin A} = \frac{1}{16} \frac{\sin\left(\pi+A\right)}{\sin A} =−sin⁡A16sin⁡A=−116.= - \frac{\sin A}{16 \sin A} = - \frac{1}{16} .