Writing 1+1= 2 in a complicated way

Question

I am learning Unit Circle at the moment and I am using this source as an education tool Trigonometry: Unit Circle (Starts at 20:00).

The author solves these simple equations like below:

1 + 1 = 2
1 + 3 = 4
3 + 1 = 4

After he introduces this formula:

(√x)² = x

The first equation is then reintroduced to:

 (√1)² +  (√1)² =  (√2)²
 (√1)² +  (√3)² =  (√4)²
 (√3)² +  (√1)² =  (√4)²

I understand this, since the formula above says the square root of any number squared will equal the number.

The author then divides the first two numbers x and y by the answer and he is given the answer 1.

During some more "Clean up" as the author describes, this is the final out come;

I understand steps one and two, but when it comes to the third step, where he "cleans up" again, I get lost in thought. I would Google this, but I do not know the technical term, hence the title saying "Writing 1+1= 2 in a complicated way".

Can someone please give me the technical term, for this and simplify the learning curve. Or do I not need this understanding, in order to move on to the next stage where Pythagoras theorem is used to find generated points in a circle and so on.

Answer 1

Given the frames you've posted, it appears the video starts with an equation $$ a + b = c $$ involving positive numbers, and after some algebraic manipulation arrives at a point on the "unit circle": $$ \left(\frac{\sqrt{a}}{\sqrt{c}}\right)^{2} + \left(\frac{\sqrt{b}}{\sqrt{c}}\right)^{2} = 1. $$ I don't know of a specific name for this, but in case it helps, here are the steps: \begin{align*} a + b &= c && \\ (\sqrt{a})^{2} + (\sqrt{b})^{2} &= (\sqrt{c})^{2} && \text{$x = (\sqrt{x})^{2}$ if $x \geq 0$;} \\ \frac{(\sqrt{a})^{2}}{(\sqrt{c})^{2}} + \frac{(\sqrt{a})^{2}}{(\sqrt{c})^{2}} &= 1 && \text{divide through by $(\sqrt{c})^{2}$;} \\ \left(\frac{\sqrt{a}}{\sqrt{c}}\right)^{2} + \left(\frac{\sqrt{b}}{\sqrt{c}}\right)^{2} &= 1 && \frac{(\sqrt{a})^{2}}{(\sqrt{c})^{2}} = \left(\frac{\sqrt{a}}{\sqrt{c}}\right)^{2}. \end{align*} Introducing $$ x = \frac{\sqrt{a}}{\sqrt{c}},\quad y = \frac{\sqrt{b}}{\sqrt{c}}, $$ you have $x^{2} + y^{2} = 1$.