Linéarisation de cosnθ, sinnθ
\(1^{\textrm{er}} \textrm{ cas}\) : \(n\) est pair : \(n = 2p\) , \(p \in \mathbb{N}^{\ast}\)
\(\varepsilon = + 1 :\)
\(2^{2p} \cos^{2p} \theta = \left(\underline{z} + \frac{1}{\underline{z}}\right)^{2p} = 2 \cos 2 p \theta + 2 C_{2p}^{1} \cos2(p-1) \theta + ... + 2 C_{2p}^{p-1} \cos 2 \theta + C_{2p}^{p}\)
\(\Rightarrow \cos^{2p} \theta = \frac{1}{2^{2p-1}} \left(\cos 2p \theta + C_{2p}^{1} \cos 2(p-1) \theta + ... +C_{2p}^{p-1} \cos 2 \theta \right) + \frac{1}{2^{2p}} C^{p}_{2p}\qquad \mathbf{(1)}\)
\(\varepsilon = - 1 :\)
\(2^{2p} j^{2p} \sin^{2p} \theta = \left(\underline{z} - \frac{1}{\underline{z}}\right)^{2p} = 2 \cos 2 p \theta - 2 C_{2p}^{1} \cos2(p-1) \theta + ... + 2(-1)^{p-1} C_{2p}^{p-1} \cos 2 \theta + (-1)^{p}C_{2p}^{p}\)
\(\Rightarrow \sin^{2p} \theta = \frac{(-1)^{p}}{2^{2p-1}} \left(\cos 2p \theta - C_{2p}^{1} \cos 2(p-1) \theta + ... +(-1)^{p-1}2C_{2p}^{p-1} \cos 2 \theta \right) + \frac{1}{2^{2p}} C^{p}_{2p}\qquad \mathbf{(2)}\)
\(2^{\textrm{\`eme}} \textrm{ cas}\) : \(n\) est impair : \(n = 2p + 1\) , \(p \in \mathbb{N}\)
\(\varepsilon = + 1 :\)
\(2^{2p+1} \cos^{2p+1} \theta = \left(\underline{z} + \frac{1}{\underline{z}}\right)^{2p+1} = 2 \cos (2 p+1) \theta + 2 C_{2p+1}^{1} \cos(2p-1) \theta + ... + 2 C_{2p+1}^{p} \cos \theta\)
\(\Rightarrow \cos^{2p+1} \theta = \frac{1}{2^{2p}} \left(\cos (2p+1) \theta + C_{2p+1}^{1} \cos (2p-1) \theta + ... +2C_{2p+1}^{p} \cos \theta \right) \qquad \mathbf{(3)} \)
\(\varepsilon = - 1 :\)
\(2^{2p+1} j^{2p+1} \sin^{2p+1} \theta = \left(\underline{z} - \frac{1}{\underline{z}}\right)^{2p+1} = 2 j\sin (2 p+1) \theta - 2j C_{2p+1}^{1} \sin(2p-1) \theta + ... +(-1)^{p} 2j C_{2p+1}^{p} \sin \theta\)
\(\Rightarrow \sin^{2p+1} \theta = \frac{(-1)^{p}}{2^{2p}} \left(\sin(2p+1) \theta - C_{2p+1}^{1} \sin (2p-1) \theta + ... +(-1)^{p}2C_{2p+1}^{p} \sin \theta \right) \qquad \mathbf{(4)}\)
Exemple :
Linéarisation de \(\mathbf{\cos^{3}\theta, \sin^{3}\theta, \cos^{4}\theta, \sin^{4}\theta}\)
par application des formules \(\mathbf{(1)}\) à \(\mathbf{(4)}\) nous obtenons :
\(\cos^{3}\theta = \frac{1}{2^{2}} \left(\cos 3 \theta + C ^{1}_{3} \cos \theta \right) = \frac{1}{4} \left(\cos 3 \theta + 3 \cos \theta\right)\)
\(\sin^{3} \theta = \frac{(-1)}{2^{2}} \left(\sin 3 \theta - C^{1}_{3} \sin \theta \right)= -\frac{1}{4} \left(\sin 3 \theta - 3 \sin \theta\right)\)
\(\cos^{4}\theta = \frac{1}{2^{3}} \left(\cos 4 \theta + C ^{1}_{4} \cos 2\theta \right) + \frac{1}{2^{4}}C_{4}^{2}= \frac{1}{8} \left(\cos 4 \theta + 4 \cos 2\theta + 3\right)\)
\(\sin^{4}\theta = \frac{(-1)^{2}}{2^{3}} \left(\cos 4 \theta - C ^{1}_{4} \cos 2\theta \right) + \frac{1}{2^{4}}C_{4}^{2}= \frac{1}{8} \left(\cos 4 \theta - 4 \cos 2\theta + 3\right)\)