Two Very Simple Proofs of the Pythagorean Identity

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created October 6, 2025

A Proof Uses the Half-Angle Identities

We have shown on the Half-Angle page that the half-angle identities for $ \sin() $ and $ \cos() $ are independent of the Pythagorean Theorem and the Pythagorean Identity: $$ \begin{eqnarray*} \sin^2\left( \frac{x}{2} \right) &=& \frac{1-\cos(x)}{2} \mbox{\ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ } \sin\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1-\cos(x)}{2}} \\ \cos^2\left( \frac{x}{2} \right) &=& \frac{1+\cos(x)}{2} \mbox{\ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ } \cos\left( \frac{x}{2} \right) = \pm\sqrt{\frac{1+\cos(x)}{2}} \end{eqnarray*} $$ We also note that the Pythagorean Identity can be obtained very easily. We just repeat the same here: \[ \sin^2\left( \frac{x}{2} \right) + \cos^2\left( \frac{x}{2} \right) = \frac{1-\cos(x)}{2} + \frac{1+\cos(x)}{2} = 1 \]
Another Proof Uses the Product-to-Sum Identities
From the Product-to-Sum identities, we have the following $$ \begin{eqnarray*} \cos(u)\cos(v) &=& \frac{1}{2}\left( \cos(u-v) + \cos(u+v) \right) \\ \sin(u)\sin(v) &=& \frac{1}{2}\left( \cos(u-v) - \cos(u+v) \right) \end{eqnarray*} $$ Setting $ u = v $ we have $$ \begin{eqnarray*} \cos^2(u) &=& \frac{1}{2}\left( \cos(0) + \cos(2u) \right) = \frac{1}{2}(1+\cos(2u)) \\ \sin^2(u) &=& \frac{1}{2}\left( \cos(0) - \cos(2u) \right) = \frac{1}{2}(1-\cos(2u)) \end{eqnarray*} $$ Adding the above together we have $ \sin^2(u) + \cos^2(u) = 1 $.
This is actually another proof of the Half-Angle identites.
I believe that I knew both results when I was in high school many decades ago. However, at that time there was no discussions of whether the Pythagorean Theorem or the Pythagorean Identity can be (or cannot be) proved using trigonometry. Thus, proving the Pythagorean Identity using trigonometry is really not something new.
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