\input exos
\fiche{Trigonométrie}

%----------------------------------------------------------------------
\ex Équations trigonométriques

Résoudre les équations suivantes :
\def\qq{\global\advance\qqnum1\hbox to7mm{\hfil\bf\the\qqnum) }\ignorespaces}
\halign{&\hskip\leftskip\qq#\hfil\cr
% ---------------------------------------------- (√6+√2)cosθ + (√6-√2)sinθ = 2
$(√6 + √2)\cosθ + (√6 - √2)\sinθ = 2$
\reponse{$4\cos(θ-\fracπ{12}) = 2 ⇔ θ ≡ ±\fracπ3 + \fracπ{12} \pmod {2π}$.}
% ------------------------------------------- sinθ + sin2θ + sin3θ + sin4θ = 0
&$\sinθ + \sin2θ + \sin3θ + \sin4θ = 0$
\reponse{$\sinθ + … + \sin4θ
              = 2\sinθ\cosθ(4\cos²θ + 2\cosθ - 1)
              = 4\sin(\frac52θ)\cosθ\cos(½θ)$

           $4\cos²θ + 2\cosθ - 1 = 0 ⇔
              \cosθ = \frac {√5 -1}4 = \cos(\frac{2π}5)$
              ou
             $\cosθ = -\frac {√5 +1}4 = \sin(\frac{2π}5)$

           $⇒$ modulo $2π$,
           $θ ∈ ⦃0, π, \fracπ2, \frac{3π}2, \frac{2π}5,
                 \frac{4π}5, \frac{6π}5, \frac{8π}5⦄$. }
\cr
% ---------------------------------------------------- cosθ + cos2θ + cos3θ = 0
$\cosθ + \cos2θ + \cos3θ = 0$
\reponse{$\cosθ ∈ ⦃-½, ±\frac 1{√2}⦄ ⇔ θ ≡ ±\frac{2π}3 \pmod {2π}$
         ou $θ ≡ \fracπ4 \pmod {\fracπ2}$. }
% -------------------------------------------------------- cosθ - cos2θ = sin3θ
&$\cosθ - \cos2θ = \sin3θ$
\reponse{$2\sin(\frac32θ)\sin(½θ) = 2\sin(\frac32θ)\cos(\frac32θ)
         ⇔ θ ≡ 0 \pmod {\frac{2π}3}$ ou
          $θ ≡ \fracπ4 \pmod π$ ou
          $θ ≡ \frac{3π}2 \pmod {2π}$. }
\cr
% -------------------------------------------------------- cosθ + cos7θ = cos4θ
$\cosθ + \cos7θ = \cos4θ$
\reponse{$2\cos4θ\cos3θ = \cos4θ ⇔
         θ ≡ \fracπ8 \pmod {\fracπ4}$ ou $θ ≡ ±\fracπ9 \pmod {\frac{2π}3}$. }

% ----------------------------------------------------- cos2θ + cos12θ = √3cos5θ
&$\cos2θ + \cos12θ = √3\cos5θ$
\reponse{$2\cos7θ\cos5θ = √3\cos5θ ⇔
          θ ≡ \fracπ{10} \pmod {\fracπ5}$ ou
         $θ ≡ ±\fracπ{42} \pmod {\frac{2π}7}$. }
\cr
% -------------------------------------------------------- sin7θ - sinθ = sin3θ
$\sin7θ - \sinθ = \sin3θ$
\reponse{$θ≡ 0\pmod {\fracπ3}$ ou $θ≡±\fracπ{12} \pmod {\fracπ2}$.}
%                                                     3                 3
% ----------------------------------------------- cosθ sin3θ + cos3θsinθ  = 3/4
&$\cos³θ\sin3θ + \cos3θ\sin³θ = ¾$
\reponse{$\cos³θ\sin3θ + \cos3θ\sin³θ = ¾\sin4θ ⇒ θ ≡ \fracπ8 \pmod {\fracπ2}$. }
\cr
% ------------------------------------------------------ sinθsin3θ = sin5θsin7θ
$\sinθ\sin3θ = \sin5θ\sin7θ$
\reponse{$θ ≡ 0 \pmod {\fracπ8}$.}
% --------------------------------------------------------------- 3tanθ = 2cosθ
&$3\tanθ = 2\cosθ$
\reponse{$\sinθ = ½ ⇔ θ ≡ \fracπ6 \pmod {2π}$ ou $θ ≡ \frac{5π}6 \pmod {2π}$.}
\cr
% --------------------------------------------------------------- tan4θ = 4tanθ
$\tan4θ = 4\tanθ$
\reponse{$θ ≡ 0,±\arctan√5 \pmod π$. }
% ------------------------------------------------- cotanθ - tanθ = cosθ + sinθ
&$\cotanθ - \tanθ = \cosθ + \sinθ$
\reponse{$\cos²θ - \sin²θ = \cosθ\sinθ(\cosθ+\sinθ)$.

          $\cosθ + \sinθ = 0 ⇔ θ ≡ -\fracπ4 \pmod π$.

          $\cosθ-\sinθ = \cosθ\sinθ ⇒
           (\cosθ\sinθ)² + 2\cosθ\sinθ = 1 ⇒
           \cases{\cosθ = \frac {√{2√2-1} + √2 - 1}2 \cr
                  \noalign{\vskip 1mm}
                  \sinθ = \frac {√{2√2-1} - √2 + 1}2.}$

          Les valeurs trouvées conviennent.}
\cr
% ----------------------------------------- tan(x) + tan(y) = 1, tan(x+y) = 4/3
$\cases{\tan x + \tan y = 1\cr \tan(x+y) = \frac43}$
\reponse{$\tan x = \tan y = ½$.}
\cr}
%----------------------------------------------------------------------
\ex Inéquations

Résoudre les inéquations suivantes :

% ------------------------------------------------------- cosθ + cos(θ+π/3) > 0
{\bf1)} $\cosθ + \cos(θ + \fracπ3) > 0$.
\reponse{\qq $-\frac{2π}3 < θ\bmod{2π} < \fracπ3$. }
% ------------------------------------------------------------ 2cosθ + sinθ < 2
{\bf2)} $2\cosθ + \sinθ < 2$.
\reponse{\qq $2α < θ \bmod 2π < 2π$ avec
             $\cosα = \frac2{√5}$, $\sinα = \frac1{√5}$.}

%----------------------------------------------------------------------
\ex Linéarisation

Linéariser~:
\def\qq{\global\advance\qqnum1\hbox to7mm{\hfil\bf\the\qqnum) }\ignorespaces}
\halign{&\hskip\leftskip\qq#\hfil\cr
%
 $2\cos²θ$ \reponse{$1 + \cos2θ$}
&$2\sin²θ$ \reponse{$1 - \cos2θ$}
&$4\cos³θ$ \reponse{$3\cosθ + \cos3θ$}
&$4\sin³θ$ \reponse{$3\sinθ - \sin3θ$}
\cr
 $8\cos⁴θ$  \reponse{$3 + 4\cos2θ + \cos4θ$}
&$8\sin⁴θ$  \reponse{$3 - 4\cos2θ + \cos4θ$}
&$32\cos⁶θ$ \reponse{$10 + 15\cos2θ + 6\cos4θ + \cos6θ$}
&$32\sin⁶θ$ \reponse{$10 - 15\cos2θ + 6\cos4θ - \cos6θ$}
\cr
 $32\cos⁴θ\sin²θ$ \reponse{$2 + \cos2θ - 2\cos4θ - \cos6θ$}
&$32\sin⁴θ\cos²θ$ \reponse{$2 - \cos2θ - 2\cos4θ + \cos6θ$}
&$16\cosθ\sin⁴θ$  \reponse{$\cos5θ - 3\cos3θ + 2\cosθ$}
&$16\sinθ\cos⁴θ$  \reponse{$\sin5θ + 3\sin3θ + 2\sinθ$}
\cr
 $4\sinθ\sin(\fracπ3-θ)\sin(\fracπ3+θ)$ \reponse{$\sin3θ$}\hidewidth
\cr}

%----------------------------------------------------------------------
\ex $α+β+γ = π$

Soient $α,β,γ ∈ ℝ$ tels que $α+β+γ = π$.

\qq Démontrer que : $1 - \cosα + \cosβ + \cosγ = 4\sin(α/2)\cos(β/2)\cos(γ/2)$.
\reponse{$1 - \cosα = 2\sin(α/2) \cos((β+γ)/2)$ et
         $\cosβ + \cosγ = 2\sin(α/2) \cos((β-γ)/2)$.}

\qq Simplifier $\tan(α/2)\tan(β/2)  + \tan(β/2)\tan(γ/2) + \tan(γ/2)\tan(α/2)$.
    \reponse{$=1$.}

%----------------------------------------------------------------------
\ex $\sin²(θ-α), \sin²θ, \sin²(θ+α)$ en progression arithmétique

Montrer qu'il existe $θ ∈ {]0, \fracπ2 [}$ tel que pour tout
$α ∈ ℝ$, les nombres $\sin²(θ-α), \sin²θ, \sin²(θ+α)$
soient en progression arithmétique.
\reponse{$⇔ \cos2θ = 0$.}

%----------------------------------------------------------------------
\ex Calcul de somme

Calculer $\tan p - \tan q$. En déduire la valeur de
$S_n = ⅀_{k=1}ⁿ \dfrac 1{\cos(kθ)\cos((k+1)θ)}$, $θ ∈ ℝ$.
\reponse{$\tan p - \tan q = \dfrac {\sin(p-q)}{\cos p\cos q}$.

         $S_n = \dfrac {\tan((n+1)θ) - \tanθ}{\sinθ}$ si $\sinθ ≠ 0$,
         $S_n = n$ si $θ ≡ 0 \pmod {2π}$,
         $S_n = -n$ si $θ ≡ π \pmod {2π}$.}

%----------------------------------------------------------------------
\ex Calcul de somme

Simplifier $⅀_{k=0}^{n-1} 3^k\sin³\left(\dfracα{3^{k+1}}\right)$.
\reponse{linéariser :
         $Σ = ¼\left(3ⁿ\sin\dfracα{3ⁿ} - \sinα\right)$.}

%----------------------------------------------------------------------
\ex Calcul de somme

Calculer $\cotan x - 2\cotan2x$.
Simplifier $⅀_{k=0}ⁿ \dfrac 1{2^k}\tan\dfracα{2^k}$.
\reponse{$\cotan x - 2\cotan2x = \tan x$,
         $Σ = \dfrac 1{2ⁿ}\cotan\dfracα{2ⁿ} - 2\cotan2α$.}

%----------------------------------------------------------------------
\ex Heptagone régulier

Soit $ABCDEFG$ un heptagone (7 côtés) plan régulier. On pose
$α = AB$, $β = AC$, $γ = AD$ (distances).

Montrer que $\dfrac 1α = \dfrac 1β + \dfrac 1γ$.
\reponse{$θ = \dfrac π7$ :
         $\dfrac 1{\sinθ} - \dfrac 1{\sin3θ} = \dfrac {2\cos2θ}{\sin3θ}
         = \dfrac 1{\sin2θ}$.}


%----------------------------------------------------------------------
\ex lignes trigonométriques algébriques

                               %+-------+
                               %|  π/8  |
                               %+-------+
\qq calculer $\cos\fracπ8$, $\sin\fracπ8$ et $\tan\fracπ8$.
\reponse{%
$\fracπ8 = \fracπ{2×4}  ⇒
 \cos\fracπ8 = \frac{√{2+√2}}2$,
$\sin\fracπ8 = \frac{√{2-√2}}2$,
$\tan\fracπ8 = √2 -1$.}


                               %+--------+
                               %|  π/12  |
                               %+--------+
\qq calculer $\cos\fracπ{12}$, $\sin\fracπ{12}$ et $\tan\fracπ{12}$.
\reponse{%
$\fracπ{12} = \fracπ3 - \fracπ4 ⇒
 \cos\fracπ{12} = \frac{√6+√2}4$,
$\sin\fracπ{12} = \frac{√6-√2}4$,
$\tan\fracπ{12} = 2-√3$.}


                               %+-------+
                               %|  π/5  |
                               %+-------+
\qq calculer $\cos\fracπ5$, $\sin\fracπ5$, $\cos\frac{2π}5$ et
    $\sin\frac{2π}5$.
\reponse{%
$\cos\frac{2π}5 + \cos\frac{3π}5 = 0 ⇒
\left⦃\matrix{
 \cos\fracπ5 = \frac{√5+1}4,
&\sin\fracπ5 = \frac{√{10-2√5}}4.     \cr\noalign{\vskip 2mm}
 \cos\frac{2π}5 = \frac{√5-1}4,
&\sin\frac{2π}5 = \frac{√{10+2√5}}4.  \cr}\right.$}


                               %+--------+
                               %|  π/14  |
                               %+--------+

\qq Montrer que
$\sin\fracπ{14} \sin\frac{3π}{14} \sin\frac {5π}{14} = \frac 18$
(on pourra multiplier les deux membres par $\cos\fracπ{14}$).

\bye
