Question

Prove that $2\cos \dfrac{\pi }{13}\cos \dfrac{9\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13}=0$.

Answer 1

Hint: Change $2\cos \dfrac{\pi }{13}\cos \dfrac{9\pi }{13}$ into the sum of cosines of required angles by using the product to sum formula of cosine given by: $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$. Now, use the supplementary angle rule given by: $\cos \theta =-\cos \left( \pi -\theta \right)$ to change the angle of cosine obtained by product to sum rule. Cancel all the terms to get the answer 0.

Complete step-by-step answer:
We have been given,
$L.H.S=2\cos \dfrac{\pi }{13}\cos \dfrac{9\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13}$
Using the product to sum formula of cosine: $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$, we get,
$\begin{align}
  & L.H.S=\cos \left( \dfrac{\pi }{13}+\dfrac{9\pi }{13} \right)+\cos \left( \dfrac{\pi }{13}-\dfrac{9\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} \\
 & =\cos \left( \dfrac{10\pi }{13} \right)+\cos \left( \dfrac{-8\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} \\
\end{align}$
We know that, $\cos \left( -\theta \right)=\cos \theta $,
$L.H.S=\cos \left( \dfrac{10\pi }{13} \right)+\cos \left( \dfrac{8\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13}$
Now, using supplementary angle change formula, given by: $\cos \theta =-\cos \left( \pi -\theta \right)$, for $\cos \left( \dfrac{10\pi }{13} \right)$ and $\cos \left( \dfrac{8\pi }{13} \right)$, we get,
$\begin{align}
  & L.H.S=-\cos \left( \pi -\dfrac{10\pi }{13} \right)+\left( -\cos \left( \pi -\dfrac{8\pi }{13} \right) \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} \\
 & =-\cos \left( \dfrac{13\pi -10\pi }{13} \right)-\cos \left( \dfrac{13\pi -8\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} \\
 & =-\cos \dfrac{3\pi }{13}-\cos \dfrac{5\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} \\
\end{align}$
Now, cancelling the common terms, we get,
$\begin{align}
  & L.H.S=\cos \dfrac{3\pi }{13}-\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13}-\cos \dfrac{5\pi }{13} \\
 & =0 \\
 & =R.H.S \\
\end{align}$

Note: One may note that there are certain other approaches to solve the above question. Here, we have changed the product of cosines into the sum of cosines leaving the two last terms as it is. One may also apply the reverse process. Take the sum of $\cos \dfrac{3\pi }{13}$ and $\cos \dfrac{5\pi }{13}$ by using the formula: $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$, while leaving the product of cosines given in the first term as it is. Now, again apply the supplementary angle rule given by: $\cos \theta =-\cos \left( \pi -\theta \right)$ to change the angle of cosine obtained by the sum to the product rule. Cancel all the terms and get the value of the expression equal to 0.