1. Trouver une équation cartésienne de la tangente à la fonction f au point d'abscisse a.

n=0;
Clear[a];
n++;f[x_,n]:=ArcSin[2x+3];a[n]=-3/2; 
Print[n,") f(x) = ",f[x,n],"      et a = ",a[n]];

n++;f[x_,n]:=2 ArcCos[x^2-1/4];a[n]=1/2; 
Print[n,") f(x) = ",f[x,n],"      et a = ",a[n]];

n++;f[x_,n]:=Sqrt[ArcSin[3x]];a[n]=1/6; 
Print[n,") f(x) = ",f[x,n],"      et a = ",a[n]];

n++;f[x_,n]:=ArcTan[x-1];a[n]=2; 
Print[n,") f(x) = ",f[x,n],"      et a = ",a[n]];

n++;f[x_,n]:=ArcSin[(x-1)/(x+1)];a[n]=1/3; 
Print[n,") f(x) = ",f[x,n],"      et a = ",a[n]];

Solutions

Do[Print[i,") "];
reponse=DetailTangente[f[x,i],x,a[i]],{i,n}];

2. Rechercher les éventuelles asymptotes de cette fonction

f[x_]:=x^2 ArcTan[1/(1+x)];
Presentef[f[x],x]

Solution

AV[f[x],x];

AO[f[x],x];

3. Démontrer que:

1)  ∀ x ∈ [-1,1]: Arcsin(x) + Arccos(x) = (π/2)
2)  ∀ x ∈ (SuperPlus[ℝ])\{0}:  Arctg(1/x) = Arccotg x
3) Arcsin (3/5)+Arcsin (4/5) = (π/2)
4) ∀ a ∈ [1,->  :  Arcsin(((a - 1)/(a+1))) = Arccos(((2(Sqrt[a]))/(a+1)))
5) Arctg(1/3)+Arctg (1/2)=(π/4)
6) Arctg (1/2)+Arctg (1/5) +Arctg (1/8)=(π/4)

Solutions

1)  ∀ x ∈ [-1,1] : 
	Arcsin(x) + Arccos(x) = π/2
	⟺
	Arcsin(x)==(π/2)-Arccos(x)
	⟺
	sin(Arcsin(x))==sin((π/2)-Arccos(x))
	⟺
	x = cos(Arccos(x)) = x

2) ∀ x ∈ (SuperPlus[ℝ])\{0}:  Arctg((1/x)) = Arccotg x
	Arctg((1/x))=Arccotg(x) 
	⟺  
	x=cotg(Arctg((1/x)))
	 ⟺	(sachant que  Arctg((1/x)) ∈ ]0,->  )
	 x=(1/(tg(Arctg((1/x))))) =0

3)	Arcsin(3/5) + Arcsin(4/5) = (π/2)
	Arcsin((3/5))==(π/2)-Arcsin((4/5))
	sin(Arcsin((3/5)))==sin((π/2)-Arcsin((4/5)))
	(3/5)==cos(Arcsin((4/5)))
	(3/5)==(Sqrt[1-(sin^2)(Arcsin((4/5)))])
	(3/5)==(Sqrt[1-((((4/5)))^2)])

4)	∀ a ∈ [1,->  :  Arcsin(((a-1)/(a+1))) = Arccos(((2(Sqrt[a]))/(a+1)))

Si a≥1, ((a-1)/(a+1))≥0 et donc 0≤Arcsin(((a-1)/(a+1)))≤(π/2)
Si a≥1,  0<((2(Sqrt[a]))/(a+1))≤1 et donc 0≤Arccos(((2(Sqrt[a]))/(a+1)))<(π/2)

	sin(Arcsin(((a-1)/(a+1))))==sin(Arccos(((2 (Sqrt[a]))/(a+1))))
	
	((a-1)/(a+1))==(Sqrt[1-((4 a)/(((a+1))^2))])
	
	((a-1)/(a+1))==(Sqrt[(((a^2)-2 a+1)/(((a+1))^2))])

5)	Arctg((1/2))+Arctg((1/3))==(π/4)
remarque: Arctg((1/2)) est compris entre 0 et (π/4), de même pour (π/4)-Arctg((1/3))
	tg(Arctg((1/2)))==tg((π/4)-Arctg((1/3)))
	
	on sait que ∀ x ∈ ℝ: tg(Arctg(x)) = x
						( par contre Arctg(tg(((3π)/4))) ≠ ((3π)/4) !)

	(1/2)==((tg((π/4))-(1/3))/((1/3) tg((π/4))+1))
	
	(1/2)==((1-(1/3))/(1+(1/3)))

6)	Arctg((1/2))+Arctg((1/5))+Arctg((1/8))==(π/4)
vérifier les conditions d’équivalence..
	tg(Arctg((1/2)))==tg((π/4)-Arctg((1/8))-Arctg((1/5)))
	
	(1/2)==((tg((π/4))-tg(Arctg((1/8))+Arctg((1/5))))/(tg((π/4)) tg(Arctg((1/8))+Arctg((1/5)))+1))
	
	(1/2)==((1-tg(Arctg((1/8))+Arctg((1/5))))/(tg(Arctg((1/8))+Arctg((1/5)))+1))	(1)
	
	on a :
	
	tg(Arctg((1/8))+Arctg((1/5)))==(((1/5)+(1/8))/(1-(1/40)))
	
	et donc  (1) devient
	(1/2)==((1-(1/3))/(1+(1/3)))
	(1/2)==(1/2)