Luzia's 2015 Second Proof

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created October 1, 2025

In Luzia's 2015 unpublished article, he offered two proofs, one of which used simple trigonometry (discussed here) and the other used the concept of limit. We shall present the second proof here on this page.

We proved the following on the Luzia's 2015 Proof page: $$ \begin{eqnarray*} \sin(x) &=& 2\sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right) \\ \cos(x) &=& 1-2\sin^2\left( \frac{x}{2} \right) \end{eqnarray*} $$ From these two identities, we have \[ \sin^2(x) + \cos^2(x) = 1 + \left( 2\sin^2\left( \frac{x}{2} \right) \right)^2 \left[ \sin^2\left( \frac{x}{2}\right) + \cos^2\left( \frac{x}{2} \right) -1 \right] \] Replacing \( x \) by \( x/2 \) gives \[ \sin^2\left( \frac{x}{2} \right) + \cos^2\left( \frac{x}{2} \right) = 1 + \left( 2\sin^2\left( \frac{x}{2^2} \right) \right)^2 \left[ \sin^2\left( \frac{x}{2^2}\right) + \cos^2\left( \frac{x}{2^2} \right) -1 \right] \] Plugguing this into the result of \( \sin^2(x) + \cos^2(x) \) yields $$ \begin{eqnarray*} \sin^2(x) + \cos^2(x) &=& 1 + \left( 2\sin^2\left( \frac{x}{2} \right) \right)^2 \left[ \sin^2\left( \frac{x}{2}\right) + \cos^2\left( \frac{x}{2} \right) -1 \right] \\ &=& 1 + \left( 2\sin^2\left( \frac{x}{2} \right) \right)^2 \left[ \left( 1+\left( 2\sin\left( \frac{x}{2^2} \right) \right)^2 \left[ \sin^2\left(\frac{x}{2^2}\right)+\cos^2\left( \frac{x}{2^2} \right)-1\right) \right] -1\right] \\ &=& 1 + \left( 2\sin^2\left( \frac{x}{2} \right) \right)^2 \left( 2\sin^2\left( \frac{x}{2} \right) \right)^2 \left[ \sin^2\left(\frac{x}{2^2} \right) + \cos^2\left( \frac{x}{2^2} \right) - 1\right] \end{eqnarray*} $$ Repeating this process, we have the following: \[ \sin^2(x) + \cos^2(x) = 1 + \prod_{i=1}^n \left( 2\sin\left( \frac{x}{2^i} \right) \right)^2 \left[ \sin^2\left( \frac{x}{2^n} \right) + \cos^2\left( \frac{x}{2^n} \right) - 1 \right] \label{EQN} \tag{1} \] You may proved the above using mathematical induction easily.

As \( n \) approaches infinity, \( \sin\left( \frac{x}{2^n} \right) \) and \( \cos\left( \frac{x}{2^n} \right) \) approach 0 and 1, respectively. Hence, the last term approaches 0. Because \( |\sin\left( \frac{x}{2^i} \right)| < 1 \) and \( \sin^2\left( \frac{x}{2^i} \right) \leq |\sin\left( \frac{x}{2^i} \right)| < 1 \) hold, any value multiplied by \( \sin^2\left( \frac{x}{2^i} \right) \) becomes smallewr. As a result, \( \prod_{i=1}^n \left( 2\sin\left( \frac{x}{2^i} \right) \right)^2 \) approaches 0. This means that Eqn (\ref{EQN}) is equal to 1, which is the Pythagorean Identity.

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