Question

The value of $\cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right)\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right)$ .

Answer 1

Hint: For these kinds of questions, we need to make use of trigonometric formulae. Specifically, we need to make use of the transformation formulae. These are used to transform addition of two trigonometric functions to product or product of trigonometric function to the sum of those functions. Here , we have to use the transformation formula of $\cos C+\cos D$. We cannot use anything else, since none of the other formulae have two cosines in the product. We might have to apply this twice or maybe make use of another formulae.

Complete step-by-step answer:
The formula of $\cos C+\cos D$ is as follows :
\[\Rightarrow \cos C+\cos D=2\cos \dfrac{C+D}{2}+\cos \dfrac{C-D}{2}\] .
Let us apply this formula for the first two terms and the last two terms.
Let us first multiply and divide it by $2$ since there is a $2$ in the formula and then apply it for the first two terms.
$\Rightarrow \dfrac{1}{2}\times 2\cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right)$
We have $\dfrac{C+D}{2}=\dfrac{\pi }{5},\dfrac{C-D}{2}=\dfrac{2\pi }{5}$ in the question.
$\begin{align}
  & \Rightarrow \dfrac{C+D}{2}=\dfrac{\pi }{5} \\
 & \Rightarrow C+D=\dfrac{2\pi }{5} \\
 & \Rightarrow \dfrac{C-D}{2}=\dfrac{2\pi }{5} \\
 & \Rightarrow C-D=\dfrac{4\pi }{5} \\
\end{align}$
Let us solve $C+D=\dfrac{2\pi }{5},C-D=\dfrac{4\pi }{5}$ and get the value of $C,D$.
Upon doing so, the value we are :
$\Rightarrow C=\dfrac{3\pi }{5},D=\dfrac{\pi }{5}$
So we can $\dfrac{1}{2}\times 2\cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right)$ in the following way :
\[\Rightarrow \dfrac{1}{2}\left( 2\cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right) \right)=\dfrac{1}{2}\left( \cos \left( \dfrac{3\pi }{5} \right)+\cos \left( \dfrac{\pi }{5} \right) \right)\] .
Now, let us apply this formula for the last two terms
Let us first multiply and divide it by $2$ since there is a $2$ in the formula and then apply it for the last two terms.
$\Rightarrow \dfrac{1}{2}\times 2\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right)$
We have $\dfrac{C+D}{2}=\dfrac{4\pi }{5},\dfrac{C-D}{2}=\dfrac{8\pi }{5}$ in the question.
$\begin{align}
  & \Rightarrow \dfrac{C+D}{2}=\dfrac{8\pi }{5} \\
 & \Rightarrow C+D=\dfrac{16\pi }{5} \\
 & \Rightarrow \dfrac{C-D}{2}=\dfrac{4\pi }{5} \\
 & \Rightarrow C-D=\dfrac{8\pi }{5} \\
\end{align}$
Let us solve $C+D=\dfrac{16\pi }{5},C-D=\dfrac{8\pi }{5}$ and get the value of $C,D$.
Upon doing so, the value we are :
$\Rightarrow C=\dfrac{12\pi }{5},D=\dfrac{4\pi }{5}$
So we can $\dfrac{1}{2}\times 2\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right)$ in the following way :
\[\Rightarrow \dfrac{1}{2}\left( 2\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right) \right)=\dfrac{1}{2}\left( \cos \left( \dfrac{12\pi }{5} \right)+\cos \left( \dfrac{4\pi }{5} \right) \right)\]
We all know that $\cos \left( 2\pi +\theta \right)=\cos \left( \theta \right)$ . Let us apply this for $\cos \left( \dfrac{12\pi }{5} \right)$ .
$\Rightarrow \cos \left( \dfrac{12\pi }{5} \right)=\cos \left( 2\pi +\dfrac{2\pi }{5} \right)=\cos \left( \dfrac{2\pi }{5} \right)$ .
Now let us substitute everything we got back into the original equation.
\[\begin{align}
  & = \cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right)\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right) \\
 & = \dfrac{1}{4}\left( \cos \left( \dfrac{\pi }{5} \right)+\cos \left( \dfrac{2\pi }{5} \right)+\cos \left( \dfrac{3\pi }{5} \right)+\cos \left( \dfrac{4\pi }{5} \right) \right) \\
\end{align}\]
Now let us apply the formula of $\cos C+\cos D$. But before applying here, let us group the terms.
Upon grouping terms, we get the following :
\[\begin{align}
  & = \cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right)\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right) \\
 & = \dfrac{1}{4}\left( \cos \left( \dfrac{\pi }{5} \right)+\cos \left( \dfrac{2\pi }{5} \right)+\cos \left( \dfrac{3\pi }{5} \right)+\cos \left( \dfrac{4\pi }{5} \right) \right) \\
 & = \dfrac{1}{4}\left( \cos \left( \dfrac{\pi }{5} \right)+\cos \left( \dfrac{4\pi }{5} \right)+\cos \left( \dfrac{3\pi }{5} \right)+\cos \left( \dfrac{2\pi }{5} \right) \right) \\
 & = \dfrac{1}{4}\left[ \cos \left( \dfrac{\pi +4\pi }{5\times 2} \right)\cos \left( \dfrac{3\pi +2\pi }{5\times 2} \right) \right] \\
 & = \dfrac{1}{4}\left[ \cos \left( \dfrac{5\pi }{5\times 2} \right)\cos \left( \dfrac{5\pi }{5\times 2} \right) \right] \\
 & = \dfrac{1}{4}\left[ \cos \left( \dfrac{\pi }{2} \right)\cos \left( \dfrac{\pi }{2} \right) \right] \\
\end{align}\]
We know that $\cos \dfrac{\pi }{2}=0$.
\[\begin{align}
  & = \cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right)\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right) \\
 & = \dfrac{1}{4}\left( \cos \left( \dfrac{\pi }{5} \right)+\cos \left( \dfrac{2\pi }{5} \right)+\cos \left( \dfrac{3\pi }{5} \right)+\cos \left( \dfrac{4\pi }{5} \right) \right) \\
 & = \dfrac{1}{4}\left( \cos \left( \dfrac{\pi }{5} \right)+\cos \left( \dfrac{4\pi }{5} \right)+\cos \left( \dfrac{3\pi }{5} \right)+\cos \left( \dfrac{2\pi }{5} \right) \right) \\
 & = \dfrac{1}{4}\left[ \cos \left( \dfrac{\pi +4\pi }{5\times 2} \right)\cos \left( \dfrac{3\pi +2\pi }{5\times 2} \right) \right] \\
 & = \dfrac{1}{4}\left[ \cos \left( \dfrac{5\pi }{5\times 2} \right)\cos \left( \dfrac{5\pi }{5\times 2} \right) \right] \\
 & = \dfrac{1}{4}\left[ \cos \left( \dfrac{\pi }{2} \right)\cos \left( \dfrac{\pi }{2} \right) \right] \\
 & = 0 \\
\end{align}\]
$\therefore $ The value of $\cos \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{2\pi }{5} \right)\cos \left( \dfrac{4\pi }{5} \right)\cos \left( \dfrac{8\pi }{5} \right)$ is $0$.

Note: It is very important to know the formulae of trigonometry. It is a very elaborate chapter with lots of formulae. We should be thorough from the basics. There are some manipulations which are to be done such as which terms to group. This can only be built by practice. A lot of practice is required so as to be able to do these quickly in the exam. Also, we should be very careful while solving as there is a huge scope for calculation mistakes. We should also know the value of each trigonometric function.