The Chebyshev polynomials of first kind are defined by the simple formula $$T_n(x)=\cos(nt),$$ where $x=\cos t$ .

It is an example of a trigonometric polynomial.

This can be seen to be a polynomial by expressing $\cos(kt)$ as a polynomial of $\cos(t)$ , by using the formula for cosine of angle-sum:

\begin{eqnarray*} \cos(1t)&=&\cos(t)\\ \cos(2t)&=&\cos(t)\cos(t) - \sin(t)\sin(t) = 2(\cos(t))^2-1\\ \cos(3t)&=&4(\cos(t))^3-3\cos(t)\\ &\vdots& \end{eqnarray*} So we have \begin{eqnarray*} T_0(x)&=&1\\ T_1(x)&=&x\\ T_2(x)&=&2x^2-1\\ T_3(x)&=&4x^3-3x\\ &\vdots& \end{eqnarray*} These polynomials obey the recurrence relation: $$T_{n+1}(x) \;=\; 2xT_n(x)-T_{n-1}(x)$$ for $n = 1,\,2,\,\ldots$

Related are the Chebyshev polynomials of the second kind that are defined as $$U_{n-1}(\cos t) = \frac{\sin(n t)}{\sin (t)},$$ which can similarly be seen to be polynomials through either a similar process as the above or by the relation $U_{n-1}(t) = n T_n'(t)$ .

The first few are: \begin{eqnarray*} U_0(x)&=&1\\ U_1(x)&=&2x\\ U_2(x)&=&4x^2-1\\ U_3(x)&=&8x^3-4x\\ &\vdots& \end{eqnarray*} The same recurrence relation also holds for $U$ : $$U_{n+1}(x) \;=\; 2xU_n(x)-U_{n-1}(x)$$ for $n = 1,\,2,\,\ldots$ .