Some 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
Updated November 10, 2025

This page discusses several proofs of the Pythagorean Identity using basic trigonometric identities. Some of them (e.g., the first three) were known to me when I was in high school, while others seem to be known for sometime and can be found in books and on the web. Of course, when I was in high school, I did not know Loomis' claim and proving the Pythagorean Identity with trigonometry was not an issue.

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.

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: \[ \sin^2\left( \frac{x}{2} \right) + \cos^2\left( \frac{x}{2} \right) = \frac{1-\cos(x)}{2} + \frac{1+\cos(x)}{2} = 1 \]

A 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.

A Proof Uses the Sum-to-Product Identities

Here is a proof using 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 functions. Because \( \sin(0) = 0\) and \( \cos(0)=1\), we have \( \sin^2(x)+\cos^2(x) =1\).

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