How Do You Solve This Trigonometry Challenge Involving Cosine Powers?

[anemone]

Evaluate $2\cos^3 \dfrac{\pi}{7}-\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}$.

 

[kaliprasad]

Evaluate $2\cos^3 \dfrac{\pi}{7}-\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}$.

Spoiler

$2\cos^3 \dfrac{\pi}{7}-\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}$

= $\cos \dfrac{\pi}{7}(2\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}-1)$

= - $\cos \dfrac{\pi}{7}(\cos \dfrac{\pi}{7} - (2\cos^2 \dfrac{\pi}{7}-1))$

= - $\cos \dfrac{\pi}{7}(\cos \dfrac{\pi}{7} - \cos \dfrac{2\pi}{7})$

= -$ 2 \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14} \sin\dfrac{\pi}{14}$

= -$\dfrac{ 2 \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14} \sin\dfrac{\pi}{14} \cos \dfrac{\pi}{14}}{ \cos \dfrac{\pi}{14}}$

= -$\dfrac{ \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14} \sin\dfrac{\pi}{7}}{ \cos \dfrac{\pi}{14}}$

= -$\dfrac{ 2\cos \dfrac{\pi}{7}\sin \dfrac{\pi}{7} \sin\dfrac{3\pi}{14}}{2 \cos \dfrac{\pi}{14}}$

= -$\dfrac{ \sin \dfrac{2\pi}{7} \sin\dfrac{3\pi}{14}}{2 \cos \dfrac{\pi}{14}}$

=- $\dfrac{ \sin \dfrac{2\pi}{7} \cos (\dfrac{\pi}{2} - \dfrac{3\pi}{14})}{2 \cos \dfrac{\pi}{14}}$

= - $\dfrac{ \sin \dfrac{2\pi}{7} \cos \dfrac{2\pi}{7}}{2 \cos \dfrac{\pi}{14}}$

= - $\dfrac{ 2 \sin \dfrac{2\pi}{7} \cos \dfrac{2\pi}{7}}{4 \cos \dfrac{\pi}{14}}$

= -$\dfrac{ \sin \dfrac{4\pi}{7}}{4 \cos \dfrac{\pi}{14}}$

= - $\dfrac{ \cos( \dfrac{\pi}{2}- \dfrac{4\pi}{7})}{4 \cos \dfrac{\pi}{14}}$

= -$\dfrac{ \cos \dfrac{-\pi}{14}}{4 \cos \dfrac{\pi}{14}}$

= - $\dfrac{ \cos \dfrac{\pi}{14}}{4 \cos \dfrac{\pi}{14}}$

= - $\dfrac{1}{4}$

 

[anemone]

Well done, kaliprasad,my friend!(Yes) And thanks for participating!

My solution:

Spoiler

If we let $\cos \dfrac{\pi}{7}=x$, then we're actually asked to evaluate the expression $2x^3-x^2-x$.

From the well-known identity $\cos \dfrac{2\pi}{7}+\cos \dfrac{4\pi}{7}+\cos \dfrac{6\pi}{7}=-\dfrac{1}{2}$, we can rewrite it as $\cos \dfrac{2\pi}{7}+\cos \left(\pi-\dfrac{3\pi}{7}\right)+\cos \left(\pi-\dfrac{\pi}{7}\right)=-\dfrac{1}{2}$ $\cos \dfrac{2\pi}{7}-\cos \dfrac{3\pi}{7}-\cos \dfrac{\pi}{7}=-\dfrac{1}{2}$

Note that $\begin{align*}\cos \dfrac{2\pi}{7}&=2\cos^2 \dfrac{\pi}{7}-1\\&=2x^2-1 \end{align*}$ and $\begin{align*}\cos \dfrac{3\pi}{7}&=4\cos^3 \dfrac{\pi}{7}-3\cos \dfrac{\pi}{7}\\&=\cos \dfrac{\pi}{7}\left(4\cos^2 \dfrac{\pi}{7}-3\right)\\&=x(4x^2-3)\end{align*}$

$\therefore \cos \dfrac{2\pi}{7}-\cos \dfrac{3\pi}{7}-\cos \dfrac{\pi}{7}=-\dfrac{1}{2}$ becomes $2x^2-1-x(4x^2-3)-x=-\dfrac{1}{2}$, and it's then purely algebraic work to show that $2x^3-x^2-x=-\dfrac{1}{4}$.