Evaluate trigonometric expression

[anemone]

Without the help of calculator, evaluate $\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7}\cos \dfrac{4\pi}{7}$.

 

[Prove It]

My solution:

Spoiler

Using the fact that \displaystyle \begin{align*} \sin{ (2X)} \equiv 2\sin{(X)}\cos{(X)} \end{align*} we can see that \displaystyle \begin{align*} \cos{(X)} \equiv \frac{\sin{(2X)}}{2\sin{(X)}} \end{align*}. From here

\displaystyle \begin{align*} \cos{(x)} &\equiv \frac{\sin{(2x)}}{2\sin{(x)}} \\ \cos{(2x)} &\equiv \frac{\sin{(4x)}}{2\sin{(2x)}} \\ \cos{(4x)} &\equiv \frac{\sin{(8x)}}{2\sin{(4x)}} \\ \vdots \\ \cos{\left( 2^j x \right) } &\equiv \frac{\sin{ \left( 2^{j + 1}x \right) }}{2\sin{ \left( 2^j x \right) }} \end{align*}

and so

\displaystyle \begin{align*} \cos{(x)} \cdot \cos{(2x)} \cdot \cos{(4x)} \cdot \dots \cdot \cos{ \left( 2^j x \right) } &\equiv \frac{\sin{(2x)}}{2\sin{(x)}} \cdot \frac{\sin{(4x)}}{2\sin{(2x)}} \cdot \frac{\sin{(8x)}}{2\sin{(4x)}} \cdot \dots \cdot \frac{\sin{ \left( 2^{j + 1}x \right) }}{2 \sin{ \left( 2^j x \right) }} \\ &\equiv \frac{ \sin{ \left( 2^{j + 1}x \right) } }{ 2^{j + 1} \sin{(x)} } \end{align*}

thereby giving the identity \displaystyle \begin{align*} \prod_{j = 0}^{k - 1}{\cos{\left( 2^j x \right) }} \equiv \frac{\sin{ \left( 2^k x \right) }}{2^k \sin{(x)}} \end{align*}, and setting \displaystyle \begin{align*} x = \frac{\pi}{7} \end{align*} we have

\displaystyle \begin{align*} \cos{ \left( \frac{\pi}{7} \right) } \cos{ \left( \frac{2\pi}{7} \right) } \cos{ \left( \frac{4\pi}{7} \right) } &= \prod _{j = 0}^2{\cos{ \left( 2^j \cdot \frac{\pi}{7} \right) } } \\ &= \frac{ \sin{ \left( 2^3 \cdot \frac{\pi}{7} \right) }}{ 2^3 \sin{ \left( \frac{\pi}{7} \right) } } \\ &= \frac{\sin{\left( \frac{8\pi}{7} \right) }}{8\sin{ \left( \frac{\pi}{7} \right) }} \\ &= \frac{\sin{ \left( \pi + \frac{\pi}{7} \right) }}{8\sin{ \left( \frac{\pi}{7} \right) }} \\ &= \frac{-\sin{\left( \frac{\pi}{7} \right) }}{8\sin{ \left( \frac{\pi}{7} \right) }} \\ &= -\frac{1}{8} \end{align*}

 

[kaliprasad]

Spoiler

multiply by 8 $\sin(\pi/7) t$o get
$8\sin (\pi/7) cos (\pi/7)\ cos ( 2\pi/7) \cos(4\pi/7)$
=$4\ sin (2\pi/7) \cos ( 2\pi/7) \cos(4\pi/7)$
=$2 \sin (4\pi/7) \cos(4\pi/7)$
=$ \sin (8\pi/7)$
=$ -\sin (\pi/7) $
hence $ \cos (\pi/7) \cos ( 2\pi/7) \cos(4\pi/7)= -1/8$

 

[I like Serena]

Without the help of calculator, evaluate $\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7}\cos \dfrac{4\pi}{7}$.

Spoiler

Using the substitution $\cos x = \frac 1 2 \left(e^{ix} + e^{-ix}\right)$, we find:

$$\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}
=\frac 1 8 \left(e^{i\pi/7} + e^{-i\pi/7}\right) \left(e^{i 2\pi/7} + e^{-i 2\pi/7}\right)
\left(e^{i 4\pi/7} + e^{-i 4\pi/7}\right)$$

Define $z=e^{i\pi/7}$ and this yields:

$$\frac 1 8 (z+z^{-1})(z^2+z^{-2})(z^4+z^{-4})
=\frac 1 8 (z^{-7} + z^{-5} + z^{-3} + z^{-1} + z^{1} + z^{3} + z^{5} + z^{7})$$
Using the formula for a geometric series, this rolls up into:
$$\frac 1 8 z^{-7} \frac{1-(z^2)^8}{1-z^2}$$

From the definition of $z$ it follows that $z^7 = -1$.
Consequently the expression simplifies to just $$-\frac 1 8$$.

In other words:
$$\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} = -\frac 1 8$$

 

[anemone]

Hey Prove It, kaliprasad and I like Serena,

Thank you so much for participating! The solutions provided by the three of you suggest that there are many ways to tackle a math problem, and this is especially true in trigonometric problem!:)

Another solution that I saw somewhere which I think is good to share it here:

Spoiler

$\begin{align*}\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7}\cos \dfrac{4\pi}{7}&=\dfrac{\left(2\sin \dfrac{\pi}{7}\cos \dfrac{\pi}{7} \right) \left(2\sin \dfrac{2\pi}{7}\cos \dfrac{2\pi}{7} \right) \left(2\sin \dfrac{4\pi}{7}\cos \dfrac{4\pi}{7} \right)}{8\sin \dfrac{\pi}{7}\sin \dfrac{2\pi}{7}\sin \dfrac{4\pi}{7}}\\&=\dfrac{\sin \dfrac{2 \pi}{7}\sin \dfrac{4\pi}{7}\sin \dfrac{8\pi}{7}}{8\sin \dfrac{\pi}{7}\sin \dfrac{2\pi}{7}\sin \dfrac{4\pi}{7}}\\&=\dfrac{\sin \left(\pi+\dfrac{\pi}{7} \right)}{8\sin \dfrac{\pi}{7}}\\&=-\dfrac{1}{8}\end{align*}$