MHB Evaluate trigonometric expression

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The discussion revolves around evaluating the trigonometric expression $\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7}\cos \dfrac{4\pi}{7}$ without a calculator. Participants share various methods to approach the problem, highlighting the diversity of solutions in trigonometry. One user mentions an additional solution they encountered, suggesting that there are multiple effective strategies for tackling such expressions. The conversation emphasizes the collaborative nature of problem-solving in mathematics. Overall, the thread showcases the richness of mathematical exploration in evaluating trigonometric functions.
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Without the help of calculator, evaluate $\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7}\cos \dfrac{4\pi}{7}$.
 
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My solution:

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

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

$\begin{align*}\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7}\cos \dfrac{4\pi}{7}&=\dfrac{\left(2\sin \dfrac{\pi}{7}\cos \dfrac{\pi}{7} \right) \left(2\sin \dfrac{2\pi}{7}\cos \dfrac{2\pi}{7} \right) \left(2\sin \dfrac{4\pi}{7}\cos \dfrac{4\pi}{7} \right)}{8\sin \dfrac{\pi}{7}\sin \dfrac{2\pi}{7}\sin \dfrac{4\pi}{7}}\\&=\dfrac{\sin \dfrac{2 \pi}{7}\sin \dfrac{4\pi}{7}\sin \dfrac{8\pi}{7}}{8\sin \dfrac{\pi}{7}\sin \dfrac{2\pi}{7}\sin \dfrac{4\pi}{7}}\\&=\dfrac{\sin \left(\pi+\dfrac{\pi}{7} \right)}{8\sin \dfrac{\pi}{7}}\\&=-\dfrac{1}{8}\end{align*}$
 
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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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