MHB Finding this recurrence relation for stuck-together right-angle triangles

[nacho-man]

Given the image:
http://i.stack.imgur.com/EJ3ax.jpgand that $x_0 = 1, y_0=0$ and $\text{angles} \space θ_i
, i = 1, 2, 3, · · ·$ can be arbitrarily picked.

How can I derive a recurrence relationship for $x_{n+1}$ and $x_n$?

I actually know what the relationship is, however, don't know how to derive it.

My first attempt was to try and see some geometrical pattern, so I did the following:

http://i.stack.imgur.com/4BNyq.jpg

But still couldn't find a derivation.

Although it is obvious to see that $x_0 = x_1$

so for $n = 0$

We have $x_{1} = x_{0}+0 $

any tips, hints or help very much appreciated.

 

[chisigma]

Given the image:
http://i.stack.imgur.com/EJ3ax.jpgand that $x_0 = 1, y_0=0$ and $\text{angles} \space θ_i
, i = 1, 2, 3, · · ·$ can be arbitrarily picked.

How can I derive a recurrence relationship for $x_{n+1}$ and $x_n$?

I actually know what the relationship is, however, don't know how to derive it.

My first attempt was to try and see some geometrical pattern, so I did the following:

http://i.stack.imgur.com/4BNyq.jpg

But still couldn't find a derivation.

Although it is obvious to see that $x_0 = x_1$

so for $n = 0$

We have $x_{1} = x_{0}+0 $

any tips, hints or help very much appreciated.

Setting $\displaystyle \alpha_{n}= \theta_{0} + \sum_{i =1}^{n} \theta_{n}$ is $x_{n} = \cos \alpha_{n}$ and $y_{n} = \sin \alpha_{n}$, so that, applying the 'angle sum identities', the requested recursive relation are...

$\displaystyle x_{n+1} = x_{n}\ \cos \theta_{n} - y_{n}\ \sin \theta_{n}$

$\displaystyle y_{n+1} = x_{n}\ \sin \theta_{n} + y_{n}\ \cos \theta_{n}\ (1)$

Kind regards

$\chi$ $\sigma$

 

[nacho-man]

Setting $\displaystyle \alpha_{n}= \theta_{0} + \sum_{i =1}^{n} \theta_{n}$ is $x_{n} = \cos \alpha_{n}$ and $y_{n} = \sin \alpha_{n}$, so that, applying the 'angle sum identities', the requested recursive relation are...

$\displaystyle x_{n+1} = x_{n}\ \cos \theta_{n} - y_{n}\ \sin \theta_{n}$

$\displaystyle y_{n+1} = x_{n}\ \sin \theta_{n} + y_{n}\ \cos \theta_{n}\ (1)$

Kind regards

$\chi$ $\sigma$

Still helpful as ever chi sigma!

Thank you my friend.

Could I just clarify which formula you used to get the recursive relationship?
can it be found on this page anywhere?:
Trigonometric Identities

I am thinking it is this one:

$\cos(α + β) = \cos(α)\cos(β) – \sin(α)\sin(β) $Although, the recurrence relationship I am given for this is that

$x_{n+1} = x_n - y_n \tan(\theta_{n+1})$

and

$y_{n+1} = y_n + x_n \tan(\theta_{n+1})$

 

[chisigma]

Still helpful as ever chi sigma!

Thank you my friend.

Could I just clarify which formula you used to get the recursive relationship?
can it be found on this page anywhere?:
Trigonometric Identities

I am thinking it is this one:

$\cos(α + β) = \cos(α)\cos(β) – \sin(α)\sin(β) $Although, the recurrence relationship I am given for this is that

$x_{n+1} = x_n - y_n \tan(\theta_{n+1})$

and

$y_{n+1} = y_n + x_n \tan(\theta_{n+1})$

If the $x_{n}$ are 'cosines' and the $y_{n}$ are 'sines' the trigonometric identities to be used are...

$\displaystyle \cos (\alpha + \beta) = \cos \alpha\ \cos \beta - \sin \alpha\ \sin \beta\ (1)$

$\displaystyle \sin (\alpha + \beta) = \cos \alpha\ \sin \beta + \sin \alpha\ \cos \beta\ (2)$

Regarding the relation You have found, I think it is in some way critical: what does it happen if fon one n is $\theta_{n} = \frac{\pi}{2}$?(Speechless)...

Kind regards

$\chi$ $\sigma$

 

[nacho-man]

If the $x_{n}$ are 'cosines' and the $y_{n}$ are 'sines' the trigonometric identities to be used are...

$\displaystyle \cos (\alpha + \beta) = \cos \alpha\ \cos \beta - \sin \alpha\ \sin \beta\ (1)$

$\displaystyle \sin (\alpha + \beta) = \cos \alpha\ \sin \beta + \sin \alpha\ \cos \beta\ (2)$

Regarding the relation You have found, I think it is in some way critical: what does it happen if fon one n is $\theta_{n} = \frac{\pi}{2}$?(Speechless)...

Kind regards

$\chi$ $\sigma$

$\chi$ the relation that I put was the one we are given to derive. So I guess the assumption is $\theta < \frac{\pi}{2}$, which is also consistent with the diagram given.

I.e We are told the recurrence relation is :$x_{n+1} = x_n - y_n \tan(\theta_{n+1})$

and

$y_{n+1} = y_n + x_n \tan(\theta_{n+1})$

and I need to derive it.

It seems almost similar to what you got, so I wonder where the difference comes about?