Simplify a trigonometric expression

[anemone]

Simplify $\tan x\left(1-\sec \dfrac{x}{2} \right) (1-\sec x)(1-\sec 2x)\cdots(1-\sec 2^{n-1} x)$ at $n=8$.

 

[DreamWeaver]

Simplify $\tan x\left(1-\sec \dfrac{x}{2} \right) (1-\sec x)(1-\sec 2x)\cdots(1-\sec 2^{n-1} x)$ at $n=8$.

Sorry, everyone. I know this isn't a chat board, but I must just rudely interrupt anyway; this is a superb thread, Anemone! (Yes) I can't wait to see the replies... (Dance)(Dance)(Dance)

 

[anemone]

Sorry, everyone. I know this isn't a chat board, but I must just rudely interrupt anyway; this is a superb thread, Anemone! (Yes)

Hi DreamWeaver, :)

I think it would be interesting to know how would you rate this problem, on the two scales between $n!$ and $\left(\dfrac{n}{e} \right)^n$ for all natural number $n$? :p

 

[Saitama]

Spoiler

We have to find:
$$\tan x\left(1-\sec \frac{x}{2}\right)\prod_{r=1}^{8} \left(1-\sec(2^{r-1}x)\right)$$
Write:
$$1-\sec(2^{r-1}x)=1-\frac{2}{z^{2^{r-1}}+z^{-2^{r-1}}}=1-\frac{2z^{2^{r-1}}}{1+z^{2^r}}=\frac{\left(1-z^{2^{r-1}}\right)^2}{1+z^{2^r}}$$
where $z=e^{ix}$.
$$\Rightarrow \tan x\left(1-\sec \frac{x}{2}\right)\prod_{r=1}^{8} \left(1-\sec(2^{r-1}x)\right)=\tan x\left(1-\sec \frac{x}{2}\right)\dfrac{\displaystyle \left(\prod_{r=1}^{8} \left(1-z^{2^{r-1}}\right)\right)^2}{\displaystyle \prod_{r=1}^{8} \left(1+z^{2^r}\right)}$$
$$\large \prod_{r=1}^{8} \left(1-z^{2^{r-1}}\right)=1-z+z^2-z^3+\cdots -z^{2^8-1}=\frac{1-z^{2^8}}{1+z}$$
$$\large \prod_{r=1}^{8} \left(1+z^{2^r}\right)=1+z^2+z^4+\cdots+z^{2(2^8-1)}=\frac{\left(1-z^{2^9}\right)}{1-z^2}=\frac{\left(1-z^{2^8}\right)\left(1+z^{2^8}\right)}{(1-z)(1+z)}$$
Hence,
$$\tan x\left(1-\sec \frac{x}{2}\right)\dfrac{\displaystyle \left(\prod_{r=1}^{8} \left(1-z^{2^{r-1}}\right)\right)^2}{\displaystyle \prod_{r=1}^{8} \left(1+z^{2^r}\right)}=\tan x\left(1-\sec \frac{x}{2}\right)\frac{1-z}{1+z}\frac{1-z^{2^8}}{1+z^{2^8}}$$
Substitute back $z=e^{ix}$ and simplify in the following way:
$$\frac{1-z}{1+z}=\frac{1-e^{ix}}{1+e^{ix}}=\frac{e^{-ix/2}-e^{ix/2}}{e^{-ix/2}+e^{ix/2}}=-i\tan\frac{x}{2}$$
Similarly,
$$\frac{1-z^{2^8}}{1+z^{2^8}}=-i\tan(2^7x)$$
$$\Rightarrow \tan x\left(1-\sec \frac{x}{2}\right)\prod_{r=1}^{8} \left(1-\sec(2^{r-1}x)\right)=-\tan x\left(1-\sec \frac{x}{2}\right)\tan\frac{x}{2}\tan(2^7x)$$
Use:
$$1-\sec\frac{x}{2}=-\frac{2\sin^2\frac{x}{4}}{\cos\frac{x}{2}}$$
to obtain:
$$2\frac{\sin^2\frac{x}{4}}{\cos\frac{x}{2}}\tan\frac{x}{2}\tan x\tan(2^7x)$$
Do I have to simplify further? (Doh)

 

[anemone]

It now dawned on me that I had made a terrible mistake(I don't think I want to talk about it)...and this problem isn't a good problem too, as it has no good answer...

@Pranav, please accept my sincere apology for the inexcusable oversight...:(. Sorry too that you've wasted your valuable time on this problem...

@DreamWeaver, sorry that I've posted a weak problem here that made you thought that this is a superb thread, when it is not...(Worried)

 

[Saitama]

@Pranav, please accept my sincere apology for the inexcusable oversight...:(. Sorry too that you've wasted your valuable time on this problem...

No need to apologise, I have got nothing to do these days, working on the problem wasn't a waste of time. :p

 

[anemone]

No need to apologise,

Thank you...that is so kind of you...

I have got nothing to do these days, working on the problem wasn't a waste of time. :p

"Nag nag"! If you're pretty free these days, please post some fun problems for our folks to have fun with, hehehe...:p

 

[Saitama]

"Nag nag"! If you're pretty free these days, please post some fun problems for our folks to have fun with, hehehe...:p

Sure but I don't think the community would find them challenging. Anyways, I will post some of them. :)