Show that for any positive integers $i$ and $j$ with $ i > j$, we have $T_i (x)T_j(x)= \frac{1}{2}[T_{i+j}(x)T_{i-j}(x)]$

Question

Guys can you explain this demo to me step by step, I don't understand it at all.

Show that for any positive integers $i$ and $j$ with $ i > j$, we have

$$T_i (x)T_j(x)= \frac{1}{2}[T_{i+j}(x)T_{i-j}(x)]$$

I found the solution in a book, but I can't understand it. It appears like this:

If $i >j $, then

$$ \displaystyle \frac{1}{2}(T_{i+j}(x)+ T_{i-j}(x))= \frac{1}{2}\left(\cos\big((i+j)\theta\big) + \cos\big((i-j)\theta\big)\right) = \cos (i \theta) \cos (j \theta) = T_i(x)T_j(x) $$

Answer 1

One way of defining the Chebyshev Polynomials lets $\theta = \arccos x$, and then $T_n = \cos n \theta$. This derivation writes the definition of $T_n$ in terms of this auxiliary variable (which is one-to-one and onto on $x$ in $[-1, 1]$). It then uses a standard trigonometric identity to convert it from a sum to a product. Each multiplier is itself of the same form as the definition, so can immediately be converted back.