Relationship between Chebyshev Polynomials of first and second types

Question

How do I prove this relation between the first and second kinds of Chebyshev Polynomials:

Given $U_n(x) = \frac{\sin ((n+1) arccos(x))}{\sin (arccos(x))} $

Show that $T_{n+1}(x) = (n+1)U_n(x)$

Answer 1

$T_2(x) = 2x^2-1 \neq 4x = 2 \cdot U_1(x)$

But you can show that $\frac{d}{dx}cos((n + 1) \cdot arccos(x)) = \frac{(n + 1)sin((n + 1) \arccos(x))}{\sqrt{1 - x^2}} = (n + 1) \cdot U_{n+1}(x)$