Proof. Let \(x = \cos \theta \). Then \begin {align*} T_n(x) = \cos \left (n\, \cos ^{-1}x\right ) & = \cos n\theta \\ & = \frac {1}{2}\left [e^{in\theta } + e^{-in\theta }\right ]\\ & = \frac {1}{2}\left [\left (e^{i\theta }\right )^n + \left (e^{-i\theta }\right )^n\right ]\\ & = \frac {1}{2}\left [\left \{\cos \theta + i\sin \theta \right \}^n + \left \{\cos \theta - i\sin \theta \right \}^n\right ]\\ & = \frac {1}{2}\left [\left \{x + i \sqrt {1 - x^2}\right \}^n + \left \{x - i\sqrt {1 - x^2}\right \}^n \right ]. \end {align*} □

Proof. From 4.4.2 \begin {align*} T_n(x) & = \frac {1}{2}\left [\left \{x + i\sqrt {1 - x^2}\right \}^n + \left \{x - i\sqrt {1 - x^2}\right \}^n\right ]\\ & = \frac {1}{2}\left [\sum ^{n}_{k = 0} \binom {n}{k}\, x^{n-k}\, \left \{i\sqrt {1 - x^2}\right \}^k \, + \, \sum _{k=0}^n\binom {n}{k}\, x^{n-k}\, \left \{-i\sqrt {1 - x^2}\right \}^k\right ]\\ & = \frac {1}{2}\sum ^n_{k=0}\binom {n}{k}\, x^{n-k}\, (1 - x^2)^{k/2}\, \left [1 + (-1)^k\right ] \, i^k\\ & = \frac {1}{2}\sum ^n_{k=0}\binom {n}{k}x^{n-k}(1-x^2)^{k/2}\left [1 + (-1)^k\right ]\, i^k. \end {align*}

Now when \(k = \) odd then \(1 + (-1)^k = 0\), and when \(k = \) even then \(1 + (-1)^k = 2\). Hence, \[T_n(x) = \frac {1}{2}\sum _{k = \, \text {even}\, \leq n} \binom {n}{k} x^{n-k}(1 - x^2)^{k/2}\, i^k \cdot 2\] But when \(k = \) even, we have \(k = 2s\) for some integers \(s\). \(k \leq n\) means that \(s \leq \frac {n}{2}\). Since \(s\) is an integer, \(s \leq \frac {n}{2}\) means \(s \leq \left [\frac {n}{2}\right ]\). Thus \begin {align*} T_n(x) & = \sum ^{\left [\frac {n}{2}\right ]}_{s = 0} \binom {n}{2s} x^{n-2s} (1 - x^2)^{s}\, i^{2s}\\ & = \sum ^{\left [\frac {n}{2}\right ]}_{s =0} \frac {n!}{(2s)!\, (n-2s)!}x^{n-2s}(1 - x^2)^s(-1)^s. \end {align*} □

Proof. We use the definitions of \(T_n(x)\)

(a).

\(T_n(1) = \cos \left (n\, \cos ^{-1} 1\right ) = \cos (n\, (0)) = 1.\)

(b).

\(T_n(-1) = \cos \left (n\, \cos ^{-1}\, (-1)\right ) = \cos (n\pi ) = (-1)^n.\)

(c).

\(T_n(-x)\) \begin {align*} T_n(-x) & = \frac {1}{2}\left [\left \{-x + i\sqrt {1 - x^2}\right \}^n + \left \{-x - i\sqrt {1 - x^2}\right \}^n\right ]\\ & = \frac {1}{2}\left [(-1)^n\left \{x - i\sqrt {1 - x^2}\right \}^n + (-1)^n\left \{x + i\sqrt {1 - x^2}\right \}^n\right ]\\ & = (-1)^n\, T_n(x). \end {align*}

(d).

\(T_{2n}(0) = \frac {1}{2}\left [(i)^{2n} + (-i)^{2n}\right ] = \frac {1}{2}\left [(-1)^n + (-1)^n\right ] = (-1)^n\,\) by using 4.4.2

(e).

\(T_{n+1}(0) = \frac {1}{2}\left [(i)^{2n+1} + (-i)^{2n+1}\right ] = \frac {1}{2}\left [(-1)^n\, i - (-1)^n\, i\right ] = 0.\)

Proof. Let \(\, x = \cos \theta , \hspace {0.5cm} dx = -\sin \theta \, d\theta \,\,\) and \(\sqrt {1 - x^2} = \sin \theta \), so \begin {align*} \int _{-1}^1\frac {T_m(x)\, T_n(x)}{\sqrt {1 - x^2}}\, dx & = \int _{\pi }^0\frac {T_m(\cos \theta )\, T_n(\cos \theta )\, (-\sin \theta \, d\theta )}{\sin \theta }\\ & = \int ^{\pi }_0\cos n\theta \, \cos m\theta \, d\theta , \end {align*}

\begin {align*} T_n(\cos \theta ) = \cos \left (n\, \cos ^{-1}(\cos \theta )\right ) & = \cos n\theta \\ & = \frac {1}{2}\int _0^{\pi }\left [\cos (m+n)\theta + \cos (n-m)\theta \right ]\, d\theta \\ & = \frac {1}{2}\left [\frac {\sin (m+n)\theta }{m+n} + \frac {\sin (n-m)\theta }{n-m}\right ]^{\pi }_0\\ & = 0, \hspace {0.6cm} m\neq n. \end {align*}

If \(m = n \neq 0\), we have \begin {align*} \int _{-1}^1T_n(x)\, T_n(x)\, dx & = \int _0^{\pi }\cos ^2 n\theta \, d\theta \\ & = \int _0^{\pi }\frac {1}{2}(1 + \cos 2n\theta )\, d\theta \\ & = \frac {\pi }{2}. \end {align*}

If \(m = n = 0,\) \[\int ^1_{-1}\frac {T_0(x)\, T_0(x)}{\sqrt {1 - x^2}}\, dx = \int ^{\pi }_0 d\theta = \pi .\] □