Exercice
On rappelle que, pour tout nombre réel a, on a:
\(\cos^{2}a=\frac{1+cos(2a)}{2}.\)
Question
1. Déterminer l'expression exacte de \(\cos\left(\dfrac{\pi}{12}\right)\).
Solution
\(\cos^{2}\left(\dfrac{\pi}{12}\right)=\frac{1+cos(2 \times \dfrac{\pi}{12})}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{12}\right)=\frac{1+cos(\dfrac{\pi}{6})}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{12}\right)=\frac{1+\sqrt{3}{2}}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{12}\right)=\frac{2+\sqrt{3}}{4}\)
\(\iff \cos\left(\dfrac{\pi}{12}\right)=\sqrt{\frac{2+\sqrt{3}}{4}}\) ou \(\cos\left(\dfrac{\pi}{12}\right)=-\sqrt{\frac{2+\sqrt{3}}{4}}\)
\(\iff \cos\left(\dfrac{\pi}{12}\right)=\frac{\sqrt{2+\sqrt{3}}}{2}\) ou \(\cos\left(\dfrac{\pi}{12}\right)=-\frac{\sqrt{2+\sqrt{3}}}{2}\)
or \(\dfrac{\pi}{12} \in [0 ;\frac{\pi}{2}]\) donc \(cos(\dfrac{\pi}{12})\ge 0\)
finalement \(\cos\left(\dfrac{\pi}{12}\right)=\frac{\sqrt{2+\sqrt{3}}}{2}\)
Question
2. a. Quelle est la valeur de \(\cos\left(\dfrac{\pi}{2}\right)\)
Solution
\(cos\left(\dfrac{\pi}{2}\right)=0\)
Question
b. Sachant que \(\dfrac{1}{1024}\)=\(\dfrac{1}{2^{10}}\), écrire un algorithme donnant une valeur approchée de \(\cos\left(\dfrac{\pi}{1024}\right).\)
Solution
\(\cos^{2}\left(\dfrac{\pi}{4}\right)=\frac{1+cos(2\left(\dfrac{\pi}{4}\right))}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{4}\right)=\frac{1+cos(\left(\dfrac{\pi}{2}\right))}{2}\)
\(\iff \cos\left(\dfrac{\pi}{4}\right)=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
\(\cos^{2}\left(\dfrac{\pi}{8}\right)=\frac{1+cos(2\left(\dfrac{\pi}{8}\right))}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{8}\right)=\frac{1+cos(\left(\dfrac{\pi}{4}\right))}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{8}\right)=\frac{1+\frac{\sqrt{2}}{2}}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{8}\right)=\frac{2+\sqrt{2}}{4}\)
\(\iff \cos\left(\dfrac{\pi}{8}\right)=\sqrt{\frac{2+\sqrt{2}}{4}}\)
\(\iff \cos\left(\dfrac{\pi}{8}\right)=\frac{\sqrt{2+\sqrt{2}}}{2}\)
\(\cos^{2}\left(\dfrac{\pi}{16}\right)=\frac{1+cos(2\left(\dfrac{\pi}{16}\right))}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{16}\right)=\frac{1+cos(\left(\dfrac{\pi}{8}\right))}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{16}\right)=\frac{1+\frac{\sqrt{2+\sqrt{2}}}{2}}{2}\)
\(\iff \cos^{2}\left(\dfrac{\pi}{16}\right)=\frac{2+\sqrt{2+\sqrt{2}}}{4}\)
\(\iff \cos\left(\dfrac{\pi}{16}\right)=\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\)
Ainsi de suite donc :
\(\iff \cos\left(\dfrac{\pi}{10^n}\right)=\frac{\sqrt{2+\sqrt{...+\sqrt{2+\sqrt{2}}}}}{2}\) (n-1 termes 2)
from math import *
U=0
for i in range(2):
U=(1+sqrt(U))/2
print(U)