The Sum-to-Product and Product-to-Sum Identities

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created September 18, 2025

The Sum-to-Product and Product-to-Sum identities can also be derived without using the Pythagorean Theorem and the Pythagorean Identity. Coupled with the angle sum and angle difference identities, we have another proof of the Pythagorean Identity.

The Sum-to-Product and Product-to-Sum identities use the angle difference and agle-sum identities. From the angle sum identities for the sine function, we have $$ \begin{eqnarray*} \sin(\alpha+\beta) &=& \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \\ \sin(\alpha-\beta) &=& \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \end{eqnarray*} $$ Adding the above together yields \[ \sin(\alpha)\cos(\beta) = \frac{1}{2}\left(\sin(\alpha+\beta)+\sin(\alpha-\beta) \right) \] Let \( p=\alpha+\beta \) and \(q = \alpha - \beta \). We have \( \alpha = \frac{p+q}{2} \) and \( \beta=\frac{p-q}{2} \). Plugging them back into the above gives us \[ \sin(p) + \sin(q) = 2\sin\left(\frac{p+q}{2}\right)\cos\left(\frac{p-q}{2}\right) \]

If we use the angle sum and angle difference identities for the cosine function, we have $$ \begin{eqnarray*} \cos(\alpha+\beta) &=& \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \\ \cos(\alpha-\beta) &=& \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \end{eqnarray*} $$ Adding these two together yields \[ \cos(\alpha)\cos(\beta) = \frac{1}{2}\left(\cos(\alpha+\beta)+\cos(\alpha-\beta) \right) \] We also have \[ \sin(\alpha)\sin(\beta) = \frac{1}{2}\left(\cos(\alpha-\beta)-\cos(\alpha+\beta) \right) \] In summary, we are able to derive all the Sum-to-Product identities $$ \begin{eqnarray*} \sin(u)+\sin(v) &=& 2\sin\left( \frac{u+v}{2}\right) \cos\left( \frac{u-v}{2}\right) \\ \sin(u)-\sin(v) &=& 2\sin\left( \frac{u-v}{2}\right) \cos\left( \frac{u+v}{2}\right) \\ \cos(u)+\cos(v) &=& 2\sin\left( \frac{u+v}{2}\right) \sin\left( \frac{u-v}{2}\right) \\ \cos(u)-\cos(v) &=& -2\sin\left( \frac{u+v}{2}\right) \sin\left( \frac{u-v}{2}\right) \\ \end{eqnarray*} $$ The corresponding Product-to-Sum identities are $$ \begin{eqnarray*} \cos(u)\cos(v) &=& \frac{1}{2}\left( \cos(u-v) + \cos(u+v) \right) \\ \sin(u)\cos(v) &=& \frac{1}{2}\left( \sin(u+v) + \sin(u-v) \right) \\ \sin(u)\sin(v) &=& \frac{1}{2}\left( \cos(u-v) - \cos(u+v) \right) \\ \cos(u)\sin(v) &=& \frac{1}{2}\left( \sin(u+v) - \sin(u-v) \right) \end{eqnarray*} $$ Note that the derivation of the Sum-to-Product and Product-to-Sum identities did not use the Pythagorean Theorem and the Pythagorean Identiy. We only used the angle sum and angle difference identities for the sine and cosine functions, and the derivation of these identities are known to be independent of the Pythagorean Theorem and the Pythagorean Identity. Please see this page for the details.

Yet Another Simple Proof of the Pythagorean Identity

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 \( \sin^2(u) + \cos^2(u) = 1 \).

I believe that I knew this result when I was in high school many decades ago. Thus, proving the Pythagorean Identity using trigonometry is really not something new.

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