Question Number 152010 by RB95 last updated on 25/Aug/21
Commented by RB95 last updated on 25/Aug/21
$$ \\ $$$${Slt} \\ $$$${Pouviez}\:{vous}\:{m}'{aider}? \\ $$
Answered by puissant last updated on 25/Aug/21
![Exercice 1: 1) cos^4 x=(((e^(ix) +e^(−ix) )/2))^4 =(1/(16))(e^(i4x) +4e^(i3x) e^(−ix) +6e^(i2x) e^(−i2x) +4e^(ix) e^(−i3x) +e^(−i4x) ) =(1/(16))((e^(i4x) +e^(−i4x) )+4(e^(i2x) +e^(−i2x) )+6) ⇒ cos^4 x=(1/8)cos4x+(1/2)cos2x+(3/8) ⇒ cos4x=8(cos^4 x−(1/2)cos2x−(3/8)) ⇒ cos4x=8cos^4 x−4(2cos^2 x−1)−3 ⇒ cos 4x=8cos^4 x−8cos^2 x+1.. 2) linearisons sin^5 x sin^5 x=(((e^(ix) −e^(−ix) )/(2i)))^5 =(1/(32i))(e^(i5x) −5e^(i4x) e^(−ix) +10e^(i3x) e^(−i2x) −10e^(i2x) e^(−i3x) +5e^(ix) e^(−i4x) −e^(−i5x) ) =(1/(32i))((e^(i5x) −e^(i5x) )−5(e^(i3x) −e^(−i3x) )+10(e^(ix) −e^(−ix) )) ⇒ sin^5 x=(1/(16))sin5x−(5/(16))sin3x+(5/8)sinx 3) → sin(a+b)=sinacosb+cosasinb → cos(a+b)=cosacosb−sinasinb →tan(a+b)=((sin(a+b))/(cos(a+b)))=((sinacosb+cosacosb)/(cosacosb−sinasinb)) en divisant par cosacosb, on tan(a+b)=((tana+tanb)/(1−tanatanb)).. →sin(a−b)=sinacosb−cosacosb → de facon analogue, on trouve que tan(a−b)=((tana−tanb)/(1+tanatanb)).. Exercice 2: selon MOIVRE , on sait que (e^(iθ) )^n =e^(inθ) ⇒ (cosθ+isinθ)^n =(cosnθ+isinnθ) sert toi de ca pour repondre aux questions de l′exercie.. Exercice 3: 1) je ne vois pas. mais on sait que arg(z^n )≡narg(z)[2π] 2) En utilisant la factorisation des angles moities, on trouve: z=[1+e^(i((π/2)+α)) ]^n = 2^n cos^n ((π/4)+(α/2))e^(in((π/4)+(α/2))) et donc Re(z)=2^n cos^n ((π/4)+(α/2))cos(((nπ)/4)+((nα)/2)) Re(z)=0 ⇔ α=(π/2)+2kπ ou α=(((2k−1)π)/n)−(π/2).. Exercice 4: 1) z=((√2)/2)+i((√2)/2) nommons w et −w les racines carre^ es de z. w=(√(((√2)+1)/(2(√2))))+i(√(((√2)−1)/(2(√2)))) ⇒ w=(1/2)(√(2+(√2)))+i(1/2)(√(2−(√2))) d′abord z=e^(i(π/4)) on remarque que: (e^(i(π/8)) )^2 =(e^(i((2π)/8)) )=e^(i(π/4)) donc cos((π/8))=(1/2)(√(2+(√2))) et sin((π/8))=(1/2)(√(2−(√2))) (par identification) car e^(i(π/8)) =cos((π/8))+isin((π/8)).. 2) →z=(((1+i(√3))/(1−i)))^n on montre trivialement que 1+i(√3)=2e^(i(π/3)) et 1−i=(√2)e^(−i(π/4)) ⇒ z=((2/( (√2)))e^(i((7π)/(12))) )^n = ((√2))^n e^(i((7nπ)/(12))) le module est ((√2))^n et l′argument est ((7nπ)/(12)) ; θ∈]−π;π[. → z=(1+cosθ+isinθ)^n =(1+e^(iθ) )^n =(e^(i(θ/2)) (e^(i(θ/2)) +e^(−i(θ/2)) ))^n ≪ d′apres la factorisation des angles moitie^ s≫. ⇒ z=(2cos((θ/2))e^(i(θ/2)) )^n = 2^n cos^n ((θ/2))e^(i((nθ)/2)) alors le module de z est 2^n cos^n ((θ/2)) et l′argument est arg(z)=((nθ)/2) , θ∈]−π;π[](https://www.tinkutara.com/question/Q152013.png)
$${Exercice}\:\mathrm{1}: \\ $$$$\left.\mathrm{1}\right) \\ $$$${cos}^{\mathrm{4}} {x}=\left(\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left({e}^{{i}\mathrm{4}{x}} +\mathrm{4}{e}^{{i}\mathrm{3}{x}} {e}^{−{ix}} +\mathrm{6}{e}^{{i}\mathrm{2}{x}} {e}^{−{i}\mathrm{2}{x}} +\mathrm{4}{e}^{{ix}} {e}^{−{i}\mathrm{3}{x}} +{e}^{−{i}\mathrm{4}{x}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left(\left({e}^{{i}\mathrm{4}{x}} +{e}^{−{i}\mathrm{4}{x}} \right)+\mathrm{4}\left({e}^{{i}\mathrm{2}{x}} +{e}^{−{i}\mathrm{2}{x}} \right)+\mathrm{6}\right) \\ $$$$\Rightarrow\:{cos}^{\mathrm{4}} {x}=\frac{\mathrm{1}}{\mathrm{8}}{cos}\mathrm{4}{x}+\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}{x}+\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\Rightarrow\:{cos}\mathrm{4}{x}=\mathrm{8}\left({cos}^{\mathrm{4}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{8}}\right) \\ $$$$\Rightarrow\:{cos}\mathrm{4}{x}=\mathrm{8}{cos}^{\mathrm{4}} {x}−\mathrm{4}\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)−\mathrm{3} \\ $$$$\Rightarrow\:{cos}\:\mathrm{4}{x}=\mathrm{8}{cos}^{\mathrm{4}} {x}−\mathrm{8}{cos}^{\mathrm{2}} {x}+\mathrm{1}.. \\ $$$$\left.\mathrm{2}\right) \\ $$$${linearisons}\:{sin}^{\mathrm{5}} {x} \\ $$$${sin}^{\mathrm{5}} {x}=\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}{i}}\left({e}^{{i}\mathrm{5}{x}} −\mathrm{5}{e}^{{i}\mathrm{4}{x}} {e}^{−{ix}} +\mathrm{10}{e}^{{i}\mathrm{3}{x}} {e}^{−{i}\mathrm{2}{x}} −\mathrm{10}{e}^{{i}\mathrm{2}{x}} {e}^{−{i}\mathrm{3}{x}} +\mathrm{5}{e}^{{ix}} {e}^{−{i}\mathrm{4}{x}} −{e}^{−{i}\mathrm{5}{x}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}{i}}\left(\left({e}^{{i}\mathrm{5}{x}} −{e}^{{i}\mathrm{5}{x}} \right)−\mathrm{5}\left({e}^{{i}\mathrm{3}{x}} −{e}^{−{i}\mathrm{3}{x}} \right)+\mathrm{10}\left({e}^{{ix}} −{e}^{−{ix}} \right)\right) \\ $$$$\Rightarrow\:{sin}^{\mathrm{5}} {x}=\frac{\mathrm{1}}{\mathrm{16}}{sin}\mathrm{5}{x}−\frac{\mathrm{5}}{\mathrm{16}}{sin}\mathrm{3}{x}+\frac{\mathrm{5}}{\mathrm{8}}{sinx} \\ $$$$\left.\mathrm{3}\right) \\ $$$$\rightarrow\:{sin}\left({a}+{b}\right)={sinacosb}+{cosasinb} \\ $$$$\rightarrow\:{cos}\left({a}+{b}\right)={cosacosb}−{sinasinb} \\ $$$$\rightarrow{tan}\left({a}+{b}\right)=\frac{{sin}\left({a}+{b}\right)}{{cos}\left({a}+{b}\right)}=\frac{{sinacosb}+{cosacosb}}{{cosacosb}−{sinasinb}} \\ $$$${en}\:{divisant}\:{par}\:{cosacosb},\:{on} \\ $$$${tan}\left({a}+{b}\right)=\frac{{tana}+{tanb}}{\mathrm{1}−{tanatanb}}.. \\ $$$$\rightarrow{sin}\left({a}−{b}\right)={sinacosb}−{cosacosb} \\ $$$$\rightarrow\:{de}\:{facon}\:{analogue},\:{on}\:{trouve}\:{que} \\ $$$${tan}\left({a}−{b}\right)=\frac{{tana}−{tanb}}{\mathrm{1}+{tanatanb}}.. \\ $$$$ \\ $$$${Exercice}\:\mathrm{2}: \\ $$$$ \\ $$$${selon}\:{MOIVRE}\:,\:{on}\:{sait}\:{que}\:\left({e}^{{i}\theta} \right)^{{n}} ={e}^{{in}\theta} \\ $$$$\Rightarrow\:\left({cos}\theta+{isin}\theta\right)^{{n}} =\left({cosn}\theta+{isinn}\theta\right) \\ $$$${sert}\:{toi}\:{de}\:{ca}\:{pour}\:{repondre}\:{aux}\:{questions}\:{de}\:{l}'{exercie}.. \\ $$$$ \\ $$$${Exercice}\:\mathrm{3}: \\ $$$$\left.\mathrm{1}\right) \\ $$$${je}\:{ne}\:{vois}\:{pas}. \\ $$$${mais}\:{on}\:{sait}\:{que}\:{arg}\left({z}^{{n}} \right)\equiv{narg}\left({z}\right)\left[\mathrm{2}\pi\right] \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$${En}\:{utilisant}\:{la}\:{factorisation}\:{des}\:{angles}\: \\ $$$${moities},\:{on}\:{trouve}: \\ $$$${z}=\left[\mathrm{1}+{e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)} \right]^{{n}} =\:\mathrm{2}^{{n}} {cos}^{{n}} \left(\frac{\pi}{\mathrm{4}}+\frac{\alpha}{\mathrm{2}}\right){e}^{{in}\left(\frac{\pi}{\mathrm{4}}+\frac{\alpha}{\mathrm{2}}\right)} \\ $$$${et}\:{donc}\:{Re}\left({z}\right)=\mathrm{2}^{{n}} {cos}^{{n}} \left(\frac{\pi}{\mathrm{4}}+\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{{n}\pi}{\mathrm{4}}+\frac{{n}\alpha}{\mathrm{2}}\right) \\ $$$${Re}\left({z}\right)=\mathrm{0}\:\Leftrightarrow\:\alpha=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\:{ou}\:\alpha=\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\pi}{{n}}−\frac{\pi}{\mathrm{2}}.. \\ $$$$ \\ $$$${Exercice}\:\mathrm{4}: \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:\:{z}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${nommons}\:{w}\:{et}\:−{w}\:{les}\:{racines}\:{carr}\acute {{e}es}\:{de}\:{z}. \\ $$$${w}=\sqrt{\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}+{i}\sqrt{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\:{w}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}} \\ $$$${d}'{abord}\:{z}={e}^{{i}\frac{\pi}{\mathrm{4}}} \:{on}\:{remarque}\:{que}:\:\left({e}^{{i}\frac{\pi}{\mathrm{8}}} \right)^{\mathrm{2}} =\left({e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{8}}} \right)={e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$${donc}\:{cos}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\:\:{et}\:{sin}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\:\left({par}\:{identification}\right) \\ $$$${car}\:{e}^{{i}\frac{\pi}{\mathrm{8}}} ={cos}\left(\frac{\pi}{\mathrm{8}}\right)+{isin}\left(\frac{\pi}{\mathrm{8}}\right).. \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$$\rightarrow{z}=\left(\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{1}−{i}}\right)^{{n}} \\ $$$${on}\:{montre}\:{trivialement}\:{que}\: \\ $$$$\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{3}}} \:{et}\:\mathrm{1}−{i}=\sqrt{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\Rightarrow\:{z}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{12}}} \right)^{{n}} =\:\left(\sqrt{\mathrm{2}}\right)^{{n}} {e}^{{i}\frac{\mathrm{7}{n}\pi}{\mathrm{12}}} \\ $$$$\left.{le}\:{module}\:{est}\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{et}\:{l}'{argument}\:{est}\:\frac{\mathrm{7}{n}\pi}{\mathrm{12}}\:;\:\theta\in\right]−\pi;\pi\left[.\right. \\ $$$$\rightarrow\:{z}=\left(\mathrm{1}+{cos}\theta+{isin}\theta\right)^{{n}} =\left(\mathrm{1}+{e}^{{i}\theta} \right)^{{n}} \\ $$$$=\left({e}^{{i}\frac{\theta}{\mathrm{2}}} \left({e}^{{i}\frac{\theta}{\mathrm{2}}} +{e}^{−{i}\frac{\theta}{\mathrm{2}}} \right)\right)^{{n}} \ll\:{d}'{apres}\:{la}\:{factorisation}\: \\ $$$${des}\:{angles}\:{moiti}\acute {{e}s}\gg. \\ $$$$\Rightarrow\:{z}=\left(\mathrm{2}{cos}\left(\frac{\theta}{\mathrm{2}}\right){e}^{{i}\frac{\theta}{\mathrm{2}}} \right)^{{n}} =\:\mathrm{2}^{{n}} {cos}^{{n}} \left(\frac{\theta}{\mathrm{2}}\right){e}^{{i}\frac{{n}\theta}{\mathrm{2}}} \\ $$$${alors}\:{le}\:{module}\:{de}\:{z}\:{est}\:\mathrm{2}^{{n}} {cos}^{{n}} \left(\frac{\theta}{\mathrm{2}}\right)\:{et} \\ $$$$\left.{l}'{argument}\:{est}\:{arg}\left({z}\right)=\frac{{n}\theta}{\mathrm{2}}\:,\:\theta\in\right]−\pi;\pi\left[\right. \\ $$