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T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x), polynomial, cosn\alpha=T_n(x)

Source: Vietnamese MO (VMO) 1972

August 24, 2018

algebrapolynomialtrigonometryfunctional equation

Problem Statement

Let

\alpha

be an arbitrary angle and let

x = cos\alpha, y = cosn\alpha

(

n \in Z

). i) Prove that to each value

x \in [-1, 1]

corresponds one and only one value of

y

. Thus we can write

y

as a function of

x, y = T_n(x)

. Compute

T_1(x), T_2(x)

and prove that

T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)

. From this it follows that

T_n(x)

is a polynomial of degree

n

. ii) Prove that the polynomial

T_n(x

) has

n

distinct roots in

[-1, 1]

.

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