Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created October 6, 2025
Updated November 10, 2025
Modified April 5, 2026
This page discusses another very simple proof of the Pythagorean Identity using basic trigonometric identities. There are some rather complex and sophisticated proofs using various trigonometric identities. I personally did not like these not so elegant proofs and won't include them on this page. On the other hand, I will update this page when new, short and elegant proofs will be found in the future. We shall use the Sum-to-Product identities.
Recall the following Sum-to-Product idwentities: $$ \begin{eqnarray*} \cos(x) + \cos(y) &=& 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \cos(x) - \cos(y) &=& -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) \\ \sin(x) + \sin(y) &=& 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \sin(x) - \sin(y) &=& 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) \end{eqnarray*} $$ Multiplying the first two gives \[ \cos^2(x) - \cos^2(y) = -4\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x-y}{2}\right) \] Multiplying the last two gives \[ \sin^2(x) - \sin^2(y) = 4\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x-y}{2}\right) \] Hence, we have \[ \cos^2(x)-\cos^2(y) = -(\sin^2(x) - \sin^2(y) ) \mbox{\ \ \ \ \ for every \(x\) and \(y\)} \] Rearranging yields \[ \sin^2(x)+\cos^2(x) = \sin^2(y) + \cos^2(y) \mbox{\ \ \ \ \ for every \(x\) and \(y\)} \] This means that \( \sin^2(x)+\cos^2(x) \) is a constant function. Because \( \sin(0) = 0\) and \( \cos(0)=1\), we have \( \sin^2(x)+\cos^2(x) =1\).
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