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Question Number 198018 by obia last updated on 07/Oct/23

	Answered by a.lgnaoui last updated on 08/Oct/23
	$Z=(√3) +1+ i((√3) −1) •1): Z^2 =4+2(√3) −(4−2(√3) )+2i((√3) +1)((√3) − 1) =4(√3) +4i =4((√3) +i) 8(((√3)/2)+(1/2)i) =8(cos (𝛑/6)+ isin (𝛑/6)) ⇒Module (Z^2 )=8 Argument=(𝛑/6)[2𝛑] •2): Module de Z ese alors : 2(√2) et argument de Z=(𝛑/(12))+k𝛑 soit: Z=2(√2) (cos (𝛑/(12))+isin (𝛑/(12))) Z^2 =8[(cos^2 (𝛑/(12))−sin^2 (𝛑/(12)))+i(2sin (𝛑/(12))cos (𝛑/(12))) ⇒ { ((8(2cos^2 (𝛑/(12))−1)=((√3)/2))),((16sin (𝛑/(12))cos (π/(12))=(1/2))) :} •3): cos (𝛑/(12)) et sin (𝛑/(12)) sont deduites de: ⇒ { ((cos (π/(12)) = ((√((√6) +16(√2)))/8))),((sin (𝛑/(12))= (1/8)(√(64−((√6) +16(√2))) ))) :} •4):Resolution de l equation: ((√3) +1)cos x+((√3) −1)sin x=(√2) − methode de tangente cos x[((√3) +1)+((√3) −1)tan x=(√2) ((√3) +1)+((√3) −1)tan x=((√2)/(cos x)) (1/(cos^2 x))=1+tan^2 x ⇒ (1/(cos x))=(√(1+tan^2 x^2 )) ⇒ ((√3) +1)+ ((√3) −1) tan x=(√(2(1+tan^2 x))) posins t=tan x ((√3) +1)^2 +((√3) −1)^2 t^2 +4t=2t^2 +2 (2−2(√3) )t^2 +4t +(2+2(√3) ) =0 t^2 −(((2+(√3) )t)/2)−2−(√3) =0 △=((4+3+4(√3))/4)+8+4(√3) =((7+32+20(√3))/4)=((39+4(√3))/4) x=((2+(√3))/4)±((√(39+4(√3)))/4)=((1/2)+((√3)/2))±((√(39+4(√3)))/4) { ((t_1 =((2+(√3))/4) −((√(39+4(√3)))/4))),((t_2 =((2+(√3))/4) +((√(39+4(√3)))/4))) :} Calcul de x tan x_1 =((2+(√3))/4)+((√(39+4(√3)))/4)=2,62727 x_1 ≈69 , t_2 =−0,7612 x_2 =−37°=$
	$Z = \sqrt{3} + 1 + i (\sqrt{3} - 1)$ $∙ 1) : Z^{2} = 4 + 2 \sqrt{3} - (4 - 2 \sqrt{3}) + 2 i (\sqrt{3} + 1) (\sqrt{3} - 1)$ $= 4 \sqrt{3} + 4 i = 4 (\sqrt{3} + i)$ $8 (\frac{\sqrt{3}}{2} + \frac{1}{2} i) = 8 (\cos \frac{π}{6} + i \sin \frac{π}{6})$ $\Rightarrow Module (Z^{2}) = 8 Argument = \frac{π}{6} [2 π]$ $∙ 2) : Module de Z ese alors : 2 \sqrt{2}$ $et argument de Z = \frac{π}{12} + k π$ $soit : Z = 2 \sqrt{2} (\cos \frac{π}{12} + i \sin \frac{π}{12})$ $Z^{2} = 8 [(\cos^{2} \frac{π}{12} - \sin^{2} \frac{π}{12}) + i (2 \sin \frac{π}{12} \cos \frac{π}{12})$ $\Rightarrow {\begin{cases} 8 (2 \cos^{2} \frac{π}{12} - 1) = \frac{\sqrt{3}}{2} \\ 16 \sin \frac{π}{12} \cos \frac{π}{12} = \frac{1}{2} \end{cases}$ $∙ 3) : \cos \frac{π}{12} et \sin \frac{π}{12} sont deduites de :$ $\Rightarrow {\begin{cases} \cos \frac{π}{12} = \frac{\sqrt{\sqrt{6} + 16 \sqrt{2}}}{8} \\ \sin \frac{π}{12} = \frac{1}{8} \sqrt{64 - (\sqrt{6} + 16 \sqrt{2}}) \end{cases}$ $∙ 4) : R e s o l u t i o n d e l e q u a t i o n :$ $(\sqrt{3} + 1) \cos x + (\sqrt{3} - 1) \sin x = \sqrt{2}$ $- m e t h o d e d e t a n g e n t e$ $\cos x [(\sqrt{3} + 1) + (\sqrt{3} - 1) \tan x = \sqrt{2}$ $(\sqrt{3} + 1) + (\sqrt{3} - 1) \tan x = \frac{\sqrt{2}}{\cos x}$ $\frac{1}{\cos^{2} x} = 1 + \tan^{2} x \Rightarrow \frac{1}{\cos x} = \sqrt{1 + \tan^{2} x^{2}}$ $\Rightarrow$ $(\sqrt{3} + 1) + (\sqrt{3} - 1) \tan x = \sqrt{2 (1 + \tan^{2} x)}$ $posins t = \tan x$ ${(\sqrt{3} + 1)}^{2} + {(\sqrt{3} - 1)}^{2} t^{2} + 4 t = 2 t^{2} + 2$ $(2 - 2 \sqrt{3}) t^{2} + 4 t + (2 + 2 \sqrt{3}) = 0$ $t^{2} - \frac{(2 + \sqrt{3}) t}{2} - 2 - \sqrt{3} = 0$ $△ = \frac{4 + 3 + 4 \sqrt{3}}{4} + 8 + 4 \sqrt{3}$ $= \frac{7 + 32 + 20 \sqrt{3}}{4} = \frac{39 + 4 \sqrt{3}}{4}$ $x = \frac{2 + \sqrt{3}}{4} \pm \frac{\sqrt{39 + 4 \sqrt{3}}}{4} = (\frac{1}{2} + \frac{\sqrt{3}}{2}) \pm \frac{\sqrt{39 + 4 \sqrt{3}}}{4}$ ${\begin{cases} t_{1} = \frac{2 + \sqrt{3}}{4} - \frac{\sqrt{39 + 4 \sqrt{3}}}{4} \\ t_{2} = \frac{2 + \sqrt{3}}{4} + \frac{\sqrt{39 + 4 \sqrt{3}}}{4} \end{cases}$ $Calcul de x$ $\tan x_{1} = \frac{2 + \sqrt{3}}{4} + \frac{\sqrt{39 + 4 \sqrt{3}}}{4} = 2, 62727$ $x_{1} \approx 69$ $, t_{2} = - 0, 7612 x_{2} = - 37 ° =$
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