For every integer $n \in \mathbb{N}$, we set $F_n(x) = \cos(n \arccos x)$.
a) Show that the functions $F_n$ are defined on the same domain $D$ which should be specified.
b) Calculate $F_1(x), F_2(x)$ and $F_3(x)$ for all $x \in D$.
c) Calculate $F_n(1), F_n(0)$ and $F_n(-1)$ for all $n \in \mathbb{N}$.
d) Specify the parity properties of $F_n$ as a function of $n$.

For every integer $n \in \mathbb{N}$, we set $F_n(x) = \cos(n \arccos x)$.
Calculate $F_{n+1}(x) + F_{n-1}(x)$ for all $n \in \mathbb{N}^*$ and all $x \in D$.

Let $n \in \mathbb{N}$. Explicitly give a polynomial $P_n \in \mathbb{R}[X]$ such that, for all $\theta \in \mathbb{R}$, $$\sin((2n+1)\theta) = \sin(\theta) P_n\left(\sin^2(\theta)\right).$$ Hint: you may expand $(\cos(\theta) + i\sin(\theta))^{2n+1}$.

Let $n \in \mathbb{N}$ and let $P_n \in \mathbb{R}[X]$ be such that, for all $\theta \in \mathbb{R}$, $\sin((2n+1)\theta) = \sin(\theta) P_n\left(\sin^2(\theta)\right)$, and $$P_n(x) = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{x}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$
Deduce that, for all $x \in \mathbb{R}$, $$\sin(\pi x) = (2n+1)\sin\left(\frac{\pi x}{2n+1}\right) \prod_{k=1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$

Let $n \in \mathbb{N}^*$ and $$T _ { n } ( X ) = \sum _ { p = 0 } ^ { \lfloor n / 2 \rfloor } ( - 1 ) ^ { p } \binom { n } { 2 p } X ^ { n - 2 p } \left( 1 - X ^ { 2 } \right) ^ { p }.$$ Show that $T _ { n }$ is the unique polynomial with real coefficients satisfying the relation $$\forall \theta \in \mathbb { R } , \quad T _ { n } ( \cos ( \theta ) ) = \cos ( n \theta ).$$

We consider two strictly positive real numbers $a$ and $b$, and we set $\rho = \frac{b-a}{b+a}$. We call $\Psi$ the application from $\mathbf{R}$ to $\mathbf{R}$ defined by: $$\forall x \in \mathbf{R}, \Psi(x) = \ln(a^2 \cos^2 x + b^2 \sin^2 x)$$
Deduce that for all $x \in \mathbf{R}$, $$\Psi(x) = 2\ln\left(\frac{a+b}{2}\right) - 2\sum_{k=1}^{+\infty} \frac{\cos(2kx)}{k}\rho^k$$

We fix a pair $( p , q ) \in E _ { 3 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p > q \right\}$, and set $\theta _ { k } := ( 2 k + 1 ) \dfrac { \pi } { p }$.
Show that $$\sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } \cos \left( q \theta _ { k } \right) = \begin{cases} 0 & \text{if } p \text{ is even} \\ \dfrac { ( - 1 ) ^ { q + 1 } } { 2 } & \text{if } p \text{ is odd} \end{cases}$$

We denote by $J_n^{(\mathrm{s})}$ the matrix of $\mathcal{M}_n(\mathbb{R})$ defined by $$\forall (i,j) \in \llbracket 1,n \rrbracket^2, \quad J_n^{(\mathrm{S})}(i,j) = \frac{2}{\sqrt{2n+1}} \sin\left(\frac{2\pi ij}{2n+1}\right).$$
Show that, for all $p \in \mathbb{N}^*$ and $x \in \mathbb{R} \backslash \pi\mathbb{Z}$, $$\sum_{k=1}^{p} \cos(2kx) = \frac{1}{2}\left(\frac{\sin((2p+1)x)}{\sin(x)} - 1\right)$$

The value of $$\sum _ { k = 0 } ^ { 202 } ( - 1 ) ^ { k } \binom { 202 } { k } \cos \left( \frac { k \pi } { 3 } \right)$$ equals
(A) $\quad \sin \left( \frac { 202 } { 3 } \pi \right)$.
(B) $- \sin \left( \frac { 202 } { 3 } \pi \right)$.
(C) $\quad \cos \left( \frac { 202 } { 3 } \pi \right)$.
(D) $\cos ^ { 202 } \left( \frac { \pi } { 3 } \right)$.

For every real number $x$, the number $A$ is defined as $$\sum _ { k = 2 } ^ { 4 } \cos ( 2 k x ) = A$$ Accordingly, $$\sum _ { k = 2 } ^ { 4 } \cos ^ { 2 } ( k x )$$ What is the equivalent of the expression in terms of A?\ A) $A + 2$\ B) $A + 4$\ C) $\frac { \mathrm { A } + 1 } { 2 }$\ D) $\frac { A + 2 } { 2 }$\ E) $\frac { A + 3 } { 2 }$

$$\frac{\cos(2x+y) + \sin(2x-y)}{\cos(2x) + \sin(2x)}$$
Which of the following is the simplified form of this expression?
A) $\cos y - \sin y$ B) $\cos y + \sin y$ C) $\cos x - \sin y$ D) $\sin x - \cos y$ E) $\sin x - \cos x$