Question

Evaluate the value of following:
$\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}\cos \dfrac{{4\pi }}{7}$
${\text{A}}{\text{.}}$ 0
${\text{B}}{\text{.}}$ $\dfrac{1}{2}$
${\text{C}}{\text{.}}$ $\dfrac{1}{4}$
${\text{D}}{\text{.}}$ $ - \dfrac{1}{8}$

Answer 1

Hint: Assume $\dfrac{\pi }{7} = A$, then the given equation will become as, $\cos A\cos 2A\cos 4A$. Then multiply the equation by $2\sin A$ in numerator and denominator and then solve to obtain the answer.

Complete step-by-step answer:
We have been given the equation $\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}\cos \dfrac{{4\pi }}{7} - (1)$.
Let $\dfrac{\pi }{7} = A$. Put in equation (1), the equation transforms to-
$\Rightarrow \cos A\cos 2A\cos 4A - (2)$
Now, multiply by $2\sin A$ in the numerator and denominator of equation (2), we get-
$\Rightarrow \dfrac{1}{{2\sin A}}.[2\sin A.\cos A.\cos 2A.\cos 4A] - (3)$
Now we know that, $2\sin A\cos A = \sin 2A$, put it in equation (3), it will become-
$\Rightarrow \dfrac{1}{{2\sin A}}.[\sin 2A.\cos 2A.\cos 4A]$
Multiplying by 2 in numerator and denominator, we get-
$\Rightarrow \dfrac{1}{{2\sin A}}.\dfrac{1}{2}[2\sin 2A.\cos 2A.\cos 4A] - (4)$
Again, using the result, $2\sin 2A\cos 2A = \sin 4A$, now the equation (4) will transform to-
$\Rightarrow \dfrac{1}{{2\sin A}}.\dfrac{1}{2}[\sin 4A.\cos 4A] - (5)$
Again, multiplying by 2 in numerator and denominator in equation (5),
$\Rightarrow \dfrac{1}{{2\sin A}}.\dfrac{1}{2}.\dfrac{1}{2}[2\sin 4A.\cos 4A] - (6)$
Again, using the result, $2\sin 4A\cos 4A = \sin 8A$ in equation (6), we get-
$
  \Rightarrow \dfrac{1}{{2\sin A}}.\dfrac{1}{2}.\dfrac{1}{2}[\sin 8A] \\
   \Rightarrow \dfrac{1}{{8\sin A}}.\sin 8A \\
 $
Solving further put $\dfrac{\pi }{7} = A$, we get
$
  \dfrac{1}{{8\sin \dfrac{\pi }{7}}}.\sin \dfrac{{8\pi }}{7} \\
   = \dfrac{{\sin (\pi + \dfrac{\pi }{7})}}{{8\sin \dfrac{\pi }{7}}} [\text{as we know} sin(\pi + x) = - sinx ] \\
 = - \dfrac{{\sin \dfrac{\pi }{7}}}{{8\sin \dfrac{\pi }{7}}} \\
   = - \dfrac{1}{8} \\
$
Therefore, the value of $\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}\cos \dfrac{{4\pi }}{7} = - \dfrac{1}{8}$.
Hence, the correct option is D.

Note: When solving such types of questions, always see whether some standard result can be obtained by multiplying with some value or not. As mentioned in the solution by multiplying in numerator and denominator by $2\sin A$, the equation becomes easier to solve and by using the result, $2\sin A\cos B = \sin 2A$, the whole equation reduced in a simpler form and easier to find the answer.