A Simple Proof with L'Hôpital's Rule

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created September 30, 2025

We proved the following on Other Trigonometric Fundamental Formulae: \[ \sin^2(x) + \cos^2(x) = \left( \sin^2\left(\frac{x}{2^n}\right) + \cos^2\left(\frac{x}{2^n}\right) \right)^{2^n} \mbox{\ \ \ \ \ \ \ \ \ \ \ \ for every \( n > 0 \) } \] We shall prove the following here: \[ \sin^2(x) + \cos^2(x) = \lim_{n\rightarrow\infty} \left( \sin^2\left(\frac{x}{2^n}\right) + \cos^2\left(\frac{x}{2^n}\right) \right)^{2^n} = 1\] Hence, the Pythagorean Identity holds.

For convenience, let \( k = \frac{1}{2^n} \). The above becomes \[ \sin^2(x) + \cos^2(x) = \lim_{k\rightarrow 0} \left( \cos^2(kx) + \sin^2(kx) \right)^{\frac{1}{k}} = 1\] Therefore, we have $$ \begin{eqnarray*} \left( \cos^2(kx) + \sin^2(kx) \right)^{\frac{1}{k}} &=& \exp\left( \ln\left( \left( \sin^2(kx) + \cos^2(kx) \right)^{\frac{1}{k}} \right) \right) \\ &=& \exp\left( \frac{\ln\left( \sin^2(kx) + \cos^2(kx)\right)}{k} \right) \end{eqnarray*} $$

As \( k\rightarrow 0 \), the above approaches \( \exp(0/0) \), and hence L'Hôpital's Rule is used to compute the limit. Differentiate the numerator and denominator yields $$ \begin{eqnarray*} \lim_{k\rightarrow 0} \left( \cos^2(kx) + \sin^2(kx) \right)^{\frac{1}{k}} &=& \lim_{k\rightarrow 0}\exp\left( \ln\left( \left( \sin^2(kx) + \cos^2(kx) \right)^{\frac{1}{k}} \right) \right) \\ &=& \exp\left( \lim_{k\rightarrow 0} \frac{\ln\left( \sin^2(kx) + \cos^2(kx)\right)}{k} \right) \\ &=& \exp\left( \lim_{k\rightarrow 0} \frac{\left( \frac{2\sin(kx)\cos(kx)\cdot k+2\cos(kx)(-\sin(kx))\cdot k}{\sin^2(kx)+\cos^2(kx)}\right) }{1} \right) \\ &=& \exp(0) = 1 \end{eqnarray*} $$ In this way, we proved the Pythagorean Identity using the derivatives of \( \sin() \) and \( \cos(x) \) and L'Hôpital's Rule. Because we have shown that computing the derivatives of the sine and cosines is independent of the Pythagorean Theorem and the Pythagorean Identiy, the above proof has no circular reasoning.

Remarks

I came up with this proof in late 2023 when I started to have some interesting in proving the Pythagorean Identity using trigonometry without circular reasoning.

The next step was trying to remove the use of L'Hôpital's Rule. My first try was using the Bionomial Expansion, but was not successful. On the other hand, I did have a very long and complex proof using the \( \epsilon-\delta \) notation; however, I abandoned it as its ugly. The next attempt was using the concept of a dense set in \( \left[ 0,\frac{1}{2}\pi \right] \), which was successful. Please see The Use of a Dense Set for the details.

Finally, I retunred to the Binomial Theorem idea and finished it. Please refer to A Proof Using Limits for the details.

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