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

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,
$\begin{aligned} & \Rightarrow \cos \left( \dfrac{\pi }{7} \right)\cdot \cos \left( \dfrac{4\cdot \pi }{7} \right)\cdot \cos \left( \dfrac{5\cdot \pi }{7} \right) \\\ & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{4\pi }{7}\cdot \cos \dfrac{5\pi }{7}......\left( 2 \right) \\\ \end{aligned}$
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.
$\begin{aligned} & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{4\pi }{7}\cdot \cos \dfrac{5\pi }{7} \\\ & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \left( \pi -\dfrac{3\pi }{7} \right)\cdot \cos \left( \pi -\dfrac{2\pi }{7} \right).....\left( 3 \right) \\\ \end{aligned}$
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.
$\begin{aligned} & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \left( \pi -\dfrac{3\pi }{7} \right)\cdot \cos \left( \pi -\dfrac{2\pi }{7} \right) \\\ & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{3\pi }{7}\cdot \cos \dfrac{2\pi }{7}.....\left( 4 \right) \\\ \end{aligned}$
We should rearrange the cosines in the increasing order of their angle in identity (4).
$\begin{aligned} & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{3\pi }{7}\cdot \cos \dfrac{2\pi }{7} \\\ & \Rightarrow \cos \dfrac{\pi }{7}\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7}.....\left( 5 \right) \\\ \end{aligned}$
Multiply $2\sin \dfrac{\pi }{7}$ in the numerator and the denominator of the identity (5).
$\begin{aligned} & \Rightarrow \dfrac{2\sin \dfrac{\pi }{7}}{2\sin \dfrac{\pi }{7}}\left( \cos \dfrac{\pi }{7}\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \left( 2\sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7} \right)\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right).....\left( 6 \right) \\\ \end{aligned}$
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:
$\begin{aligned} & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \left( \sin \dfrac{2\pi }{7} \right)\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( 2\sin \dfrac{2\pi }{7}\cdot \cos \dfrac{2\pi }{7} \right)\cdot \cos \dfrac{3\pi }{7} \right).....\left( 7 \right) \\\ \end{aligned}$
Now, applying the same formula in the identity (7) as explained in the above steps, we get:
$\begin{aligned} & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( 2\sin \dfrac{2\pi }{7}\cdot \cos \dfrac{2\pi }{7} \right)\cdot \cos \dfrac{3\pi }{7} \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( \sin \dfrac{2\left( 2\pi \right)}{7} \right)\cdot \cos \dfrac{3\pi }{7} \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\left( \sin \dfrac{4\pi }{7} \right)\cdot \cos \dfrac{3\pi }{7} \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\cdot \dfrac{1}{2}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \right)......\left( 8 \right) \\\ \end{aligned}$

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:
$\begin{aligned} & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\cdot \dfrac{1}{2}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \dfrac{3\pi }{7} \right) \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{2}\cdot \dfrac{1}{2}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \left( \pi -\dfrac{4\pi }{7} \right) \right) \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{4}\left( 2\sin \dfrac{4\pi }{7}\cdot \cos \dfrac{4\pi }{7} \right) \right) \\\ & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{4}\left( \sin \dfrac{8\pi }{7} \right) \right).....\left( 9 \right) \\\ \end{aligned}$
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:
$\begin{aligned} & \Rightarrow \dfrac{1}{2\sin \dfrac{\pi }{7}}\left( \dfrac{1}{4}\left( \sin \left( \pi +\dfrac{\pi }{7} \right) \right) \right) \\\ & \Rightarrow \dfrac{1}{8\sin \dfrac{\pi }{7}}\left( \sin \dfrac{\pi }{7} \right) \\\ & \Rightarrow \dfrac{1}{8} \\\ \end{aligned}$
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.