1•determiner: tan (x/2) en fonction de tan x 2•on donne tan x=(1/8) tan (x/2)=? 3• la valeur proche de x?


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Question Number 192924 by a.lgnaoui last updated on 31/May/23

$1 ∙ determiner : \tan \frac{x}{2} en fonction de \tan x$ $2 ∙ on donne \tan x = \frac{1}{8} \tan \frac{x}{2} = ?$ $3 ∙ la valeur proche de x ?$
	Answered by a.lgnaoui last updated on 31/May/23
	$1•tan x=((2t)/(1−t^2 )) t=tan( (x/2)) y=tan x ⇒ y(1−t^2 )=2t y+yt^2 =2t t^2 −((2t)/y)+1=0⇒ (t−(1/y))^2 +1−(1/y^2 )=0 (ty−1)^2 =1−y^2 t=(1/y)−((√(1−y^2 ))/y) { ((y≠0 x≠(π/4)+kπ)),((y ≤1)) :} ⇒tan (x/2)=(1/(tan x))−((√(1−tan^2 x))/(tan x)) 2•tan x=(1/8) tan (x/2) = 8 −(√(63)) 3• tan (x/2)=0,062746 ⇒(x/2)=3,6° x=7,2^°$
	$1 ∙ \tan x = \frac{2 t}{1 - t^{2}} t = \tan (\frac{x}{2})$ $y = \tan x \Rightarrow y (1 - t^{2}) = 2 t$ $y + {yt}^{2} = 2 t$ $t^{2} - \frac{2 t}{y} + 1 = 0 \Rightarrow {(t - \frac{1}{y})}^{2} + 1 - \frac{1}{y^{2}} = 0$ ${(ty - 1)}^{2} = 1 - y^{2}$ $t = \frac{1}{y} - \frac{\sqrt{1 - y^{2}}}{y} {\begin{cases} y \neq 0 x \neq \frac{π}{4} + k π \\ y ⩽ 1 \end{cases}$ $\Rightarrow \tan \frac{x}{2} = \frac{1}{\tan x} - \frac{\sqrt{1 - \tan^{2} x}}{\tan x}$ $2 ∙ \tan x = \frac{1}{8}$ $\tan \frac{x}{2} = 8 - \sqrt{63}$ $3 ∙ \tan \frac{x}{2} = 0, 062746 \Rightarrow \frac{x}{2} = 3, 6 °$ $x = 7, 2^{°}$
	Answered by MM42 last updated on 31/May/23

	$t a n x = \frac{2 t a n x}{1 - {t a n}^{2} \frac{x}{2}}$ $\frac{1}{8} = \frac{2 t a n \frac{x}{2}}{1 - {t a n}^{2} \frac{x}{2}} \Rightarrow {t a n}^{2} \frac{x}{2} + 16 t a n \frac{x}{2} - 1 = 0$ $t a n \frac{x}{2} = - 8 \pm \sqrt{65}$
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