Using series to prove hypothesis of right triangle's < limits

[BreCheese]

Homework Statement

I need to mathematically prove that the center angle(s) (labeled as "A" in the photo below) approach what I believe to be 60 degrees (but never reach 60 degrees). We are given the values of all longer legs of each right triangle. Furthermore, the value of the length of each longer leg (corresponding to each right triangle) can be found by the formula 2^(n-1).
https://scontent.fsnc1-1.fna.fbcdn.net/hphotos-xlp1/v/t1.0-9/13010688_10209437230275493_1200339292039179261_n.jpg?oh=58d8cc3f924d6bbe7d649d6789a9a05e&oe=5775B755

Homework Equations

Pythagorean Theorem
Finding angles with Trigonometry
General knowledge of sequences and series.

The Attempt at a Solution

step one:
First I noticed that triangle A1P1P2 is a 45-45-90 degree triangle, thus angle A1 is 45° and the length of each leg is 1.
Then I found the hypotenuse of triangle A1P1P2 using the Pythagorean Theorem; the hypotenuse is root 2 (as I depicted in the photo above).
step two:
To find the value of the angle P2A2P3 I used trigonometry:
Tanθ=2/root 2. θ≈54.7356. (as I depicted in the photo as "∠2=54.7356")
step three:
I again used the Pythagorean Theorem to find the hypotenuse of triangle A2P2P3; the hypotenuse is root 6 (as I depicted in the photo).

I repeated the steps above multiple times and found that ∠A sub n approaches 60°. Intuitively, this makes sense, but now I need to prove it using concepts from series and sequences.

Things I know:
⋅The hypotenuse of the first triangle, becomes the shortest leg of the next triangle
What I know is true about 30°-60°-90° triangles:
⋅The shortest leg is always ½ of its hypotenuse; alternatively, the hypotenuse is always 2 times longer than the shortest leg
⋅The longest leg is always equal to the shortest leg multiplied by root 3
Question I have:
I'm not sure how to incorporate the given formula for the lengths of the longer legs, 2^(n-1). I think I need to somehow come up with a series that calculates the ∠An for all values of n.
∈ tanθ=(2^(n-1)) / (shortest leg) Where "θ" is ∠A sub n.

Can anyone help me come up with a formula to replace "(shortest leg)" in the summation written above?

Can anyone shed some insights?

Lastly, my professor provided us with a hint:

∈4^k ; where the lower index is k=1, and the upper index is (n-1).
My professor also noted that the series equals (4-4^n) / (1-4).
Does this mean the series converges to (4-4^n) / (-3)?

 

[mfb]

Ignore the angles for a while. Can you find a sequence for the hypotenuse lengths? What can you say about their ratios?

 

[geoffrey159]

One way to do this is to work in the complex plane.
You can start with the following sequence defined for $n\ge 1$ by :

$P_{n+1} - P_n = r_n e^{-i\frac{\pi}{2}} (A - P_n)$
$P_1$ given

Taking the modulus, you find the following expression for $r_n = \frac{2^{n-1}}{|A-P_n|}$

You can then rewrite things as $P_{n+1} - A = (P_n - A) (1 + i r_n)$, the angle $\theta_n$ for which you want to find the limit being the argument of $(1+ir_n)$, so you have to find an explicit expression for $r_n$

This leads you to solve the sequence $U_{n+1} = | U_n + i 2^{n-1}|$, where $U_n := |P_n - A|$ and $U_1 = 1$, which has solution $U_n = \sqrt{\frac{4^{n-1} + 2}{3}}$, giving you the explicit expression of $r_n$.

You have $\cos(\theta_n) = 1 / |1+ir_n| \to 1/2$ and $\sin(\theta_n) = r_n / |1+ir_n| \to \sqrt{3}/2$. So the limit angle is $\pi/3$ by continuity of cosine and sine functions.

 

[BreCheese]

Thank you both so much for your help! I appreciate it greatly :)
Below is what I came up with. However, the challenge now is finding the limit properly, instead of just observing the expansion of the sequence expression.
https://scontent.fsnc1-1.fna.fbcdn.net/hphotos-xaf1/v/t1.0-9/13076794_10209443351308515_2493838683025935097_n.jpg?oh=ee983b4e0e976cc2eace11109f8d9a31&oe=57A38C68

 

[mfb]

Where does the expression for the angle come from?

No, your approach does not work. The calculator has to round, so apparently for n=100 and n=150 it is rounded to 60 degrees. But you cannot prove a limit by plugging in large numbers!

Getting the explicit expression is most of the work, finding the limit can be done with standard rules for limits.

 

[BreCheese]

Where does the expression for the angle come from?

No, your approach does not work. The calculator has to round, so apparently for n=100 and n=150 it is rounded to 60 degrees. But you cannot prove a limit by plugging in large numbers!

Getting the explicit expression is most of the work, finding the limit can be done with standard rules for limits.

Because the hypotenuse becomes the proceeding triangle's shortest leg, and because we were given the expression of the longest leg. I thought that I'd use trigonometry to define the angle:
tanθ=((2^(n-1)) / (root((4^(n-1)+2) / 3)) ; where the numerator is the expression for the longest legs, and the denominator is the expression for the shortest leg.
Therefore, θ= arctan ((2^(n-1)) / (root((4^(n-1)+2) / 3)).
I agree I need to properly evaluate the limit, instead of just evaluating it at high numbers of n. (that's my next challenge). I also need to explain how I got the expression for the denominator (which is an expression for the hypotenuse, and an expression for the shortest leg). (Full disclosure, geoffrey159's post has the expression for the denominator, and how he arrived at it; it wasn't my work).

 

[BreCheese]

Ignore the angles for a while. Can you find a sequence for the hypotenuse lengths? What can you say about their ratios?

https://scontent.fsnc1-1.fna.fbcdn.net/v/t1.0-9/13062253_10209444697822177_4361847450617446643_n.jpg?oh=8d9fd11e5dbb18b8bf498f0a370dce5e&oe=57BA672C

 

[geoffrey159]

The transformation of point $A$ to point $P_{n+1}$ is for every $n\ge 1$ determined by a similarity of center $P_n$, angle $-\pi / 2$, and ratio of similarity $r_n$, which translates in complex coordinates by the sequence

$P_{n+1} - P_n = r_n e^{-i\frac{\pi}{2}} (A - P_n)$.

After you find $r_n = 2^{n-1} / |P_n - A |$, and rewrite the formula above as $P_{n+1} - A = (P_n - A) (1 + i r_n)$, you can see that point $P_{n+1}$ is deduced from point $P_n$ by the similarity of center $A$, angle $\theta_n := \text{Arg}(1+ir_n)$, and ratio $|1+ir_n|$.

At this point, and as MFB said, you can see that finding the expression of the sequence of hypothenuses $U_n := |P_n - A |$ is the knot of the problem, but once you have it, you quickly reach to the conclusion.

 

[BreCheese]

I can't recall learning much about the complex plane. My math experience is only up through Calc 2 (I'm 3/4 of the way through completion of college Calc 2). What level of mathematics did you learn in depth about the complex plane that you're referring to?

 

[geoffrey159]

You just need to understand the geometric interpretation of complex transformation $z\to az+b$ when $a$ and $b$ are complex numbers. When it is not a translation, it is a similarity (find its center, its angle, its ratio). Then you will understand the geometric interpretation of $(z' - z_0) = a (z - z_0)$.