Question

How do you evaluate $\cos \left( \dfrac{\pi }{7} \right)\cdot \cos \left( \dfrac{4\cdot \pi }{7} \right)\cdot \cos \left( \dfrac{5\cdot \pi }{7} \right)$?

Answer 1

Hint: If we want to solve the trigonometric identity$\cos \left( \dfrac{\pi }{7} \right)\cdot \cos \left( \dfrac{4\cdot \pi }{7} \right)\cdot \cos \left( \dfrac{5\cdot \pi }{7} \right)$then convert the given identity into simplified form. After we have converted the identity into the simplified form then divide \[2\sin \dfrac{\pi }{7}\]in the numerators and denominators to simplify the identity and reduce it.

Complete step by step solution:
We have our given identity that is $\cos \left( \dfrac{\pi }{7} \right)\cdot \cos \left( \dfrac{4\cdot \pi }{7} \right)\cdot \cos \left( \dfrac{5\cdot \pi }{7} \right)......\left( 1 \right)$.
We have to write the identity into the simplified form such that,
$\begin{align}
  & \Rightarrow \cos \left( \dfrac{\pi }{7} \right)\cdot \cos \left( \dfrac{4\cdot \pi }{7} \right)\cdot \cos \left( \dfrac{5\cdot \pi }{7} \right) \\
 & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{4\pi }{7}\cdot \cos \dfrac{5\pi }{7}......\left( 2 \right) \\
\end{align}$
Now, we have the identity in simplified form in identity (2). The \[\cos \dfrac{4\pi }{7}\]can be written as\[\cos \left( \pi -\dfrac{3\pi }{7} \right)\] and \[\cos \dfrac{4\pi }{7}\] can be written as$\cos \left( \pi -\dfrac{2\pi }{7} \right)$. The identities are converted in this form because it will be easier to solve.
$\begin{align}
  & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{4\pi }{7}\cdot \cos \dfrac{5\pi }{7} \\
 & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \left( \pi -\dfrac{3\pi }{7} \right)\cdot \cos \left( \pi -\dfrac{2\pi }{7} \right).....\left( 3 \right) \\
\end{align}$
Now, we have obtained the identity (3). We know that \[\cos \left( \pi -\theta \right)\]is equal to\[\cos \theta \]. So apply it in the identity.
$\begin{align}
  & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \left( \pi -\dfrac{3\pi }{7} \right)\cdot \cos \left( \pi -\dfrac{2\pi }{7} \right) \\
 & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{3\pi }{7}\cdot \cos \dfrac{2\pi }{7}.....\left( 4 \right) \\
\end{align}$
We should rearrange the cosines in the increasing order of their angle in identity (4).
$\begin{align}
  & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{3\pi }{7}\cdot \cos \dfrac{2\pi }{7} \\
 & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7}.....\left( 5 \right) \\
\end{align}$
Multiply \[2\sin \dfrac{\pi }{7}\] in the numerator and the denominator of the identity (5).
$\begin{align}
  & \Rightarrow \dfrac{2\sin \dfrac{\pi }{7}}{2\sin \dfrac{\pi }{7}}\left( \cos \dfrac{\pi }{7}\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \left( 2\sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7} \right)\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right).....\left( 6 \right) \\
\end{align}$
Now, we know that the formula for\[2\sin x\cos x\]will be\[\sin 2x\]so applying this formula in identity (6), we get:
$\begin{align}
  & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \left( \sin \dfrac{2\pi }{7} \right)\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( 2\sin \dfrac{2\pi }{7}\cdot \cos \dfrac{2\pi }{7} \right)\cdot \cos \dfrac{3\pi }{7} \right).....\left( 7 \right) \\
\end{align}$
Now, applying the same formula in the identity (7) as explained in the above steps, we get:
$\begin{align}
  & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( 2\sin \dfrac{2\pi }{7}\cdot \cos \dfrac{2\pi }{7} \right)\cdot \cos \dfrac{3\pi }{7} \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( \sin \dfrac{2\left( 2\pi \right)}{7} \right)\cdot \cos \dfrac{3\pi }{7} \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( \sin \dfrac{4\pi }{7} \right)\cdot \cos \dfrac{3\pi }{7} \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\cdot \dfrac{1}{2}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \right)......\left( 8 \right) \\
\end{align}$

We know that \[\cos \dfrac{3\pi }{7}\] can also be written as \[\cos \left( \pi -\dfrac{4\pi }{7} \right)\] and we also know that \[\cos \left( \pi -\theta \right)\] is equal to \[\cos \theta \]. So applying this in the identity (8), we get:
$\begin{align}
  & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\cdot \dfrac{1}{2}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\cdot \dfrac{1}{2}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \left( \pi -\dfrac{4\pi }{7} \right) \right) \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{4}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \dfrac{4\pi }{7} \right) \right) \\
 & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{4}\left( \sin \dfrac{8\pi }{7} \right) \right).....\left( 9 \right) \\
\end{align}$
As we know, \[\sin \left( \dfrac{8\pi }{7} \right)\] can also be written as \[\sin \left( \pi +\dfrac{\pi }{7} \right)\] and since \[\sin \left( \pi +x \right)=\sin x\], apply this in identity (9), we get:
 $\begin{align}
  & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{4}\left( \sin \left( \pi +\dfrac{\pi }{7} \right) \right) \right) \\
 & \Rightarrow \dfrac{1}{8\sin \dfrac{\pi }{7}}\left( \sin \dfrac{\pi }{7} \right) \\
 & \Rightarrow \dfrac{1}{8} \\
\end{align}$
Now, we have obtained the solution to the problem i.e. \[\dfrac{1}{8}\].

Note: We should always keep in mind that while solving a trigonometric problem, we should have learned all the formulas before solving the question. To solve trigonometry the main requirement is the formulas.