MHB Evaluate Trigonometric Expression.

[anemone]

Evaluate $$\cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$.

 

[chisigma]

Evaluate $$\cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$.

It exists the general formula...

$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k+1}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)

... and setting in (1) $\displaystyle k=0,\ n=5,\ x=\frac{\pi}{65}$ Your product becomes...

$\displaystyle P = \frac{1}{64}\ \frac{\sin \frac{64\ \pi}{65}}{\sin \frac{\pi}{65}} = \frac{1}{64}$ (2)

Probably a more simple way to arrive to the result exists... how to demonstrate (1) is an extra problem... Kind regards $\chi$ $\sigma$

 

[anemone]

It exists the general formula...

$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)

... and setting in (1) $\displaystyle k=0,\ n=5,\ x=\frac{\pi}{65}$ Your product becomes...

$\displaystyle P = \frac{1}{32}\ \frac{\sin \frac{64\ \pi}{65}}{\sin \frac{\pi}{65}} = \frac{1}{32}$ (2)

Probably a more simple way to arrive to the result exists... how to demonstrate (1) is an extra problem... Kind regards $\chi$ $\sigma$

Hi chisigma, thanks for participating in this problem and your answer is of course correct and on the level, I didn't realize there was such a formula exists and that we could just apply it to this particular problem and get its answer so easily...

 

[chisigma]

It exists the general formula...

$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k+1}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)

... and setting in (1) $\displaystyle k=0,\ n=5,\ x=\frac{\pi}{65}$ Your product becomes...

$\displaystyle P = \frac{1}{64}\ \frac{\sin \frac{64\ \pi}{65}}{\sin \frac{\pi}{65}} = \frac{1}{64}$ (2)

Probably a more simple way to arrive to the result exists... how to demonstrate (1) is an extra problem...

The demonstration of the formula...

$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k+1}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)

... is 'easy' and, if I remember correctly, this result was known in the Middle Age...

Let's start from the well known formula...

$\displaystyle \sin 2x = 2\ \sin x \cos x$ (2)

Setting in (2) 4x instead of 2x we obtain...

$\displaystyle \sin 4 x = 2\ \sin 2 x\ \cos 2 x = 4\ \sin x\ \cos x\ \cos 2 x$ (3)

Proceeding in the same way we arrive to...

$\displaystyle \sin (2^{n+1} x) = 2^{n+1}\ \sin x\ \prod_{k=0}^{n} \cos (2^{k}\ x)$ (4)

Kind regards

$\chi$ $\sigma$

 

[Albert1]

Evaluate $$\cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$.

Let:
$$cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=k$$
we have :
$$32\times2\times sin \left(\frac{\pi}{65}\right)\cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right)k$$

$$16\times2\times sin \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$8\times2\times sin \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$4\times2\times sin \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$2\times 2sin \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$2\times sin \left(\frac{32\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$sin \left(\frac{64\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$\therefore k=\dfrac {1}{64}$
Am I wrong ? Why my answer is different from yours ?

 

[Sudharaka]

It exists the general formula...

$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)

Hi chisigma, :)

I think this should be,

\[\prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n\color{red}{+1}-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}\]

 

[anemone]

Let:
$$cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=k$$

$\therefore k=\dfrac {1}{64}$
Am I wrong ? Why my answer is different from yours ?

I am terribly sorry for misleading the readers who have read this thread for saying the answer $$cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=\frac{1}{32}$$ is correct...

I didn't check my answer but I should be able to tell right away (from my approach) why this wasn't correct because the answer depends wholly on the number of terms that the cosine terms exist.

The correct answer for this trigonometric expression is $$\frac{1}{64}$$, as stated by Albert.

My solution:

Let $$P=cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$

and$$2^6Q=2^6(\sin \left(\frac{\pi}{65}\right)\cdot\sin \left(\frac{2\pi}{65}\right)\cdot\sin \left(\frac{4\pi}{65}\right)\cdot\sin \left(\frac{8\pi}{65}\right)\cdot\sin \left(\frac{16\pi}{65}\right)\cdot\sin \left(\frac{32\pi}{65}\right))$$

Multiplying P and $$2^6Q$$ together we obtain

$$2^6PQ=Q$$

$$P=\frac{1}{2^6}=\frac{1}{64}$$

and this approach is essentially the same as what Albert did in his solution...

I am truly sorry for saying the answer was correct without checking it and I promise I won't make this kind of mistake again and will practice the right forum manners on this site in the future.

Sorry...

 

[chisigma]

Hi chisigma, :)

I think this should be,

\[\prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n\color{red}{+1}-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}\]

All right Sudharaka!... I have corrected my past post!... thank You very much!...

Kind regards

$\chi$ $\sigma$