Question

How to prove that \[\cos\left( \dfrac{2\pi}{7} \right) + cos\left( \dfrac{4\pi}{7} \right) + cos\left( \dfrac{6\pi}{7} \right) = - \dfrac{1}{2}\] ?

Answer 1

Hint: In this question, we have to prove that \[\cos\left( \dfrac{2\pi}{7} \right) + cos\left( \dfrac{4\pi}{7} \right) + cos\left( \dfrac{6\pi}{7} \right)\] is equal to \(- \dfrac{1}{2}\) . First we can consider the left hand side of the given expression. In order to prove the given expression ,we need to use the concepts of trigonometric identities. The basic trigonometric functions are sine , cosine and tangent. Cosine function is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle. The value of \[sin(\pi)\] is used to prove the given expression. Mathematically, Pi (\[\pi\]) is Greek letter and is a mathematical constant . In trigonometry, the value of is \[180\] . By using trigonometric identities and functions, we can prove this expression easily.First we can consider the left part of the given expression then we need to multiplying and dividing the given expression by \[2\ \sin\left( \dfrac{\pi}{7} \right)\ \]. Then by using the trigonometric formula, we can find the right part of the given expression.

Complete step by step answer:
Let us consider the left part of the given expression .
\[\Rightarrow \ cos\left( \dfrac{2\pi}{7} \right) + cos\left( \dfrac{4\pi}{7} \right) + cos\left( \dfrac{6\pi}{7} \right)\]
On multiplying and dividing the expression by \[2\ \sin\left( \dfrac{\pi}{7} \right)\ \] ,
We get,
\[\Rightarrow \ \dfrac{2\ \sin\left( \dfrac{\pi}{7} \right)}{2\ sin\left( \dfrac{\pi}{7} \right)}\left( \cos\left( \dfrac{2\pi}{7} \right) + cos\left( \dfrac{4\pi}{7} \right) + cos\left( \dfrac{6\pi}{7} \right) \right)\]
On multiplying the numerator term with the terms inside the parentheses ,
We get,
\[\Rightarrow \ \dfrac{1}{2\ sin\left( \dfrac{\pi}{7} \right)}\left( 2{\ sin}\left( \dfrac{\pi}{7} \right)\cos\left( \dfrac{2\pi}{7} \right) + 2{\ sin}\left( \dfrac{\pi}{7} \right)\cos\left( \dfrac{4\pi}{7} \right) + 2{\ sin}\left( \dfrac{\pi}{7} \right)\cos\left( \dfrac{6\pi}{7} \right) \right)\]
By using the formula
\[2\ sin\ x\ cos\ y\ = \ sin(x\ + \ y)\ + \ sin(x\ -\ y)\] ,
\(\Rightarrow \ \dfrac{1}{2\ sin\left( \dfrac{\pi}{7} \right)}\left( \sin\left( \dfrac{\pi}{7} + \dfrac{2\pi}{7} \right) + sin\left( \dfrac{\pi}{7} - \dfrac{2\pi}{7} \right) + \ sin\left( \dfrac{\pi}{7} + \dfrac{4\pi}{7} \right) + sin\left( \dfrac{\pi}{7} - \dfrac{4\pi}{7} \right) + \text{ sin}\left( \dfrac{\pi}{7} + \dfrac{6\pi}{7} \right) + sin\left( \dfrac{\pi}{7} - \dfrac{6\pi}{7} \right) \right)\)
On simplifying,
We get
\[\Rightarrow \ \dfrac{1}{2\ sin\left( \dfrac{\pi}{7} \right)}\left( \sin\left( \dfrac{3\pi}{7} \right) + sin\left( \dfrac{- \pi}{7} \right) + \ sin\left( \dfrac{5\pi}{7} \right) + sin\left( \dfrac{- 3\pi}{7} \right) + \ sin\left( \dfrac{7\pi}{7} \right) + sin\left( \dfrac{- 5\pi}{7} \right) \right)\]
By rewriting the terms,
We get,
\[\Rightarrow \ \dfrac{1}{2\ sin\left( \dfrac{\pi}{7} \right)}\left( \sin\left( \dfrac{3\pi}{7} \right) – sin\left( \dfrac{\pi}{7} \right) + \ sin\left( \dfrac{5\pi}{7} \right) – sin\left( \dfrac{3\pi}{7} \right) + \ sin\left( \pi \right) – sin\left( \dfrac{5\pi}{7} \right) \right)\]
On further simplifying,
We get,
\[\Rightarrow \dfrac{- sin\left( \dfrac{\pi}{7} \right) + sin\left( \pi \right)}{2\ sin\left( \dfrac{\pi}{7} \right)}\]
We know that the value of \[\pi\] \[sin(\pi)\] is \[0\]
Thus we get,
\[\Rightarrow \dfrac{- sin\left( \dfrac{\pi}{7} \right) + 0}{2\ sin\left( \dfrac{\pi}{7} \right)}\]
On simplifying,
We get,
\[\Rightarrow \dfrac{- 1}{2\ }\]
Hence proved.
Thus we have proved that \[\cos\left( \dfrac{2\pi}{7} \right) + cos\left( \dfrac{4\pi}{7} \right) + cos\left( \dfrac{6\pi}{7} \right) = - \dfrac{1}{2}\]
We have proved that \[\cos\left( \dfrac{2\pi}{7} \right) + cos\left( \dfrac{4\pi}{7} \right) + cos\left( \dfrac{6\pi}{7} \right) = - \dfrac{1}{2}\]

Note: The concept used to prove the given expression is trigonometric identities. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. Sine is nothing but a ratio of the opposite side of a right angle to the hypotenuse of the right angle. The common technique used in this problem is the substitution rule with the use of trigonometric functions. We need to know that \[sin(\pi)\] lies in the second quadrant. Hence the value of \[sin(\pi)\] is non-negative.