- #1
anemone
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Simplify $\tan x\left(1-\sec \dfrac{x}{2} \right) (1-\sec x)(1-\sec 2x)\cdots(1-\sec 2^{n-1} x)$ at $n=8$.
Simplify a trigonometric expression
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In summary, the Simplify function can simplify an equation containing a secant term at any number of points, but it becomes more difficult as the number of points increases.
- #2
DreamWeaver
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anemone said:
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)
- #3
anemone
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DreamWeaver said:
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
- #4
Saitama
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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)
$$\tan x\left(1-\sec \frac{x}{2}\right)\prod_{r=1}^{8} \left(1-\sec(2^{r-1}x)\right)$$
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)- #5
anemone
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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)
@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)
- #6
Saitama
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anemone said:
No need to apologise, I have got nothing to do these days, working on the problem wasn't a waste of time. :p
- #7
anemone
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Pranav said:
Thank you...that is so kind of you...
Pranav said:
"Nag nag"! If you're pretty free these days, please post some fun problems for our folks to have fun with, hehehe...:p
- #8
Saitama
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anemone said:
Sure but I don't think the community would find them challenging. Anyways, I will post some of them. :)
Related to Simplify a trigonometric expression
What is a trigonometric expression?
A trigonometric expression is an expression that involves one or more trigonometric functions, such as sine, cosine, tangent, etc. These functions are used to represent relationships between the sides and angles of a triangle.
Why is it important to simplify a trigonometric expression?
Simplifying a trigonometric expression can make it easier to understand and manipulate. It can also help to identify patterns and relationships between different trigonometric functions.
How do you simplify a trigonometric expression?
To simplify a trigonometric expression, you can use trigonometric identities, such as the Pythagorean identity, double angle formulas, or sum and difference formulas. You can also use algebraic techniques, such as factoring and combining like terms.
What are some common mistakes when simplifying a trigonometric expression?
Some common mistakes include forgetting to apply the correct trigonometric identity, making errors in algebraic manipulation, and not simplifying fully. It is important to double-check your work and use a calculator to verify your answer.
Are there any tips or tricks for simplifying trigonometric expressions?
One tip is to look for common factors or terms that can be factored out. Another is to use the unit circle to simplify trigonometric functions involving special angles. It can also be helpful to practice and familiarize yourself with common trigonometric identities and their applications.
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