How to determine Tchebysheff polynomial general expression

[EmilyRuck]

Hello!
Tchebysheff polynomials are often defined with trigonometric functions:

T_m (x) = <br /> \begin{cases} \cos(m \arccos (x)) &amp; -1 \le x \le 1\\<br /> \mathrm{cosh} (m \mathrm{arccosh} (x)) &amp; x &gt; 1\\(-1)^m \mathrm{cosh} (m \mathrm{arccosh} |x|) &amp; x &lt; 1<br /> \end{cases}<br />

But they are also polynomials, and for m even their definition could be

<br /> T_m (x) = \sum_{n = 0}^{m/2} (-1)^{m/2 - n} \frac{m/2}{m/2 + n} \binom{m/2 + n}{2n}(2x)^{2n}<br />

How could one derive the first expression from the latter? That is, how could we pass from a polynomial with powers of x to a \cos (m \arccos (x)) function?
I searched several times in the web for this demonstration, but I never found it. If you can suggest a link or a book instead of the demonstration itself I thank you so much anyway!
Bye :)

Emily

 

[Office_Shredder]

cos(my) can be written as a polynomial in cos(y) of degree m. If y = arccos(x), then cos(my) is a polynomial in cos(arccos(x)) = x.

As a simple example the double angle formula, cos(2y) = cos²(y) - sin²(y) = 2cos²(y) - 1.

Triple angle, quartic angle etc. formulas exist as well. I think the easiest way is to use Euler's formula:

cos(my) + i sin(my) = e^{imy} = \left(e^{iy}\right)^n = \left( cos(y) + i sin(y) \right)^{m}
Expanding the right hand side gives cos(my) and sin(my) as polynomials in cos(y) and sin(y) (from which you can write it in terms of only cosine or sine)