# Product of Chebyshev polynomials of the second kind?

So Wikipedia has this formula for a product of two Chebyshev polynomials of the second kind evaluated at a fixed $x$ with different indices: $$U_n(x)U_m(x)=\sum_{k=o}^{n}U_{m-n+2k}(x)$$ Which would be really useful for me. The only thing is, I have been looking through the literature for this formula and don't see it anywhere.

It would be nice if someone could shed some light on where I can look for this, why it is true, or why it isn't true.

Thanks!

The sequence $(U_n)$ may be defined by the identity, valid for every $t$ such that $\sin t\ne0$, $$U_n(\cos t)=\frac{\sin((n+1)t)}{\sin t}.$$ Thus, $$\sin t\cdot\sum_{k=0}^nU_{m-n+2k}(\cos t)=\Im\left(\sum_{k=0}^n\mathrm e^{\mathrm i(m-n+2k+1)t}\right)=\Im(z),$$ where $$z=\mathrm e^{\mathrm i(m-n+1)t}\sum_{k=0}^n\mathrm e^{2k\mathrm it}=\mathrm e^{\mathrm i(m-n+1)t}\frac{\mathrm e^{2(n+1)\mathrm it}-1}{\mathrm e^{2\mathrm it}-1}=\mathrm e^{\mathrm i(m+1)t}\frac{\sin((n+1)t)}{\sin t}.$$ Thus, $$\sum_{k=0}^nU_{m-n+2k}(\cos t)=\frac{\sin((m+1)t)}{\sin t}\cdot\frac{\sin((n+1)t)}{\sin t}=U_m(\cos t)\cdot U_n(\cos t).$$