L'énoncé
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Question 1
Que vaut $cos(2a)$?
$cos^2(a)+sin^2(a)$
$cos^2(a)-sin^2(a)$
$sin^2(a)-cos^2(a)$
$2cos(a)sin(a)$
$cos(2a)=cos(a+a)$
$cos(2a)=cos(a)cos(a)-sin(a)sin(a)$
$cos(2a)=cos^2(a)-sin^2(a)$
Question 2
Que vaut $sin(2a)$?
$cos^2(a)+sin^2(a)$
$cos^2(a)-sin^2(a)$
$sin^2(a)-cos^2(a)$
$2sin(a)cos(a)$
$sin(2a)=sin(a+a)$
$sin(2a)=sin(a)cos(a)+sin(a)cos(a)$
$sin(2a)=2sin(a)cos(a)$
Question 3
Que vaut $cos(4a)$
$cos^4(a)+4cos^2(a)sin^2(a)+2sin^4(a)$
$2cos^4(a)+6cos^2(a)sin^2(a)+2sin^4(a)$
$cos^4(a)-6cos^2(a)sin^2(a)+sin^4(a)$
$cos^4(a)-6cos^3(a)sin(a)+sin^4(a)$
On a:
$cos(4a)=Re(cos(4a)+i.sin(4a))=Re((cos(a)+i.sin(a))^4)$ D'après la formule de Moivre.
En utilisant les coefficients binomiaux:
$(cos(a)+i.sin(a))^4=cos^4(a)+4.cos^3(a).i.sin(a)+6.cos^2(a).(i.sin(a))^2+4.cos(a).(i.sin(a))^3+(i.sin(a))^4$
$(cos(a)+i.sin(a))^4=cos^4(a)-6cos^2(a)sin^2(a)+sin^4(a)+i(4.cos^3(a)sin(a)-4.cos(a)sin^3(a))$
Ainsi:
$cos(4a)=Re((cos(a)+i.sin(a))^4)=cos^4(a)-6cos^2(a)sin^2(a)+sin^4(a)$
Question 4
Que vaut $sin(4a)$?
$3.cos^3(a)sin(a)-3.cos(a)sin^3(a)$
$4.cos^2(a)sin^2(a)-4.cos(a)sin^3(a)$
$4.cos^4(a)-4.sin^4(a)$
$4.cos^3(a)sin(a)-4.cos(a)sin^3(a)$
Nous avons vu à la question précédente que:
$(cos(a)+i.sin(a))^4=cos^4(a)-6cos^2(a)sin^2(a)+sin^4(a)+i(4.cos^3(a)sin(a)-4.cos(a)sin^3(a))$
Donc $sin(4a)=Im((cos(a)+i.sin(a))^4)=4.cos^3(a)sin(a)-4.cos(a)sin^3(a)$
Question 5
Calculer $(1+i)^4$:
$\sqrt{3}(cos(\frac{3\pi}{2})+i.sin(\frac{3\pi}{2}))$
$\sqrt{2}(cos(\frac{\pi}{3})+i.sin(\frac{\pi}{3}))$
$4(\frac{\sqrt{2}}{3})+i.\frac{\sqrt{3}}{2})$
$-4$
On a:
$(1+i)=\sqrt{2}(\frac{1}{\sqrt{2}}+i.\frac{1}{\sqrt{2}})=\sqrt{2}(\frac{\sqrt{2}}{2}+i.\frac{\sqrt{2}}{2}=\sqrt{2}(cos(\frac{\pi}{4})+i.sin(\frac{\pi}{4})$
Donc $(1+i)^4=(\sqrt{2}(cos(\frac{\pi}{4})+i.sin(\frac{\pi}{4}))^4=\(\sqrt{2})^4(cos(\frac{\pi}{4})+i.sin(\frac{\pi}{4})^4=4.(cos(\frac{4\pi}{4})+i.sin(\frac{4\pi}{4})=-4$