Multiple choice

If $\displaystyle \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are roots of the equation $8x^3+4x^2-4x-1=0$. The value of $\displaystyle \cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos\frac {3\pi}{7} $ is equal to

  1. $\displaystyle \frac{1}{64}$
  2. $\displaystyle \frac{\sqrt{7}}{8}$
  3. $\displaystyle \frac{7}{64}$
  4. $\displaystyle \frac{1}{8}$
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D Correct answer
Explanation

The product of the cosines is given by the identity cos(pi/7)cos(2pi/7)cos(3pi/7) = 1/8. This is a standard result derived from the product of roots of the associated polynomial.

AI explanation

Using Vieta's formulas for the cubic equation 8x cubed plus 4x squared minus 4x minus 1 equals 0, the sum of the roots is negative 1 half and the product of the roots is 1 eighth. The given roots are cosine of pi over 7, cosine of 3 pi over 7, and cosine of 5 pi over 7, which are numerically equal to cosine of pi over 7, negative cosine of 2 pi over 7, and negative cosine of 3 pi over 7. Equating the product of the roots gives (cosine of pi over 7)(cosine of 2 pi over 7)(cosine of 3 pi over 7) equals 1 eighth.

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