1°  ${\displaystyle \sin 3x\cos 5x-\cos 3x\sin 5x=\sin(3x-5x)=-\sin 2x}$.

2°  ${\displaystyle \cos 7x\cos 4x+\sin 7x\sin 4x=\cos(7x-4x)=\cos 3x}$.

${\displaystyle {\frac {\sin b}{\sin a}}-{\frac {\cos b}{\cos a}}={\frac {\sin b\cos a-\cos b\sin a}{\sin a\cos a}}={\frac {\sin(b-a)}{\sin a\cos a}}}$ donc :

1°  ${\displaystyle {\frac {\sin 2a}{\sin a}}-{\frac {\cos 2a}{\cos a}}={\frac {1}{\cos a}}}$ ;

2°  ${\displaystyle {\frac {\sin 3a}{\sin a}}-{\frac {\cos 3a}{\cos a}}=2}$.

${\displaystyle A={\frac {1+\cos 2x}{2}}-\sin 2x-5{\frac {1-\cos 2x}{2}}=-2+3\cos 2x-\sin 2x}$.

${\displaystyle 1=(\sin ^{2}x+\cos ^{2}x)^{3}=\sin ^{6}x+\cos ^{6}x+3\sin ^{2}x\cos ^{2}x(\cos ^{2}x+\sin ^{2}x)=\sin ^{6}x+\cos ^{6}x+3\sin ^{2}x\cos ^{2}x}$

et

${\displaystyle 1=(\sin ^{2}x+\cos ^{2}x)^{2}=\sin ^{4}x+\cos ^{4}x+2\sin ^{2}x\cos ^{2}x}$

donc

${\displaystyle 2(\sin ^{6}x+\cos ^{6}x)-3(\sin ^{4}x+\cos ^{4}x)=2(1-3\sin ^{2}x\cos ^{2}x)-3(1-2\sin ^{2}x\cos ^{2}x)=-1}$.

1°  ${\displaystyle {\frac {\sin x+\cos x}{3\sin x-2\cos x}}={\frac {\tan x+1}{3\tan x-2}}}$.

2°  ${\displaystyle {\frac {\sin x-\cos x}{\sin ^{3}x+2\cos ^{3}x}}=(1+\tan ^{2}x){\frac {\tan x-1}{\tan ^{3}x+2}}}$.

3°  ${\displaystyle {\frac {\sin ^{3}x+\cos x}{\sin x-\cos x}}={\frac {{\frac {\tan ^{3}x}{1+\tan ^{2}x}}+1}{\tan x-1}}}$.

1°  ${\displaystyle {\frac {\sin ^{4}x+\cos ^{4}x}{\sin ^{4}x-\cos ^{4}x}}={\frac {\tan ^{4}x+1}{\tan ^{4}x-1}}}$.

2°  ${\displaystyle {\frac {\sin ^{3}x-\cos ^{3}x}{\sin x-\cos x}}={\frac {\tan ^{3}x-1}{(1+\tan ^{2}x)(\tan x-1)}}}$.

3°  ${\displaystyle {\frac {\sin ^{2}x+\sin x\cos x}{\sin ^{2}x-\cos ^{2}x}}={\frac {\tan ^{2}x+\tan x}{\tan ^{2}x-1}}}$.

4°  ${\displaystyle \cos ^{2}x-\sin x\cos x={\frac {1-\tan x}{1+\tan ^{2}x}}}$.

1°  ${\displaystyle {\frac {\tan a+\tan b}{\cot a+\cot b}}=\tan a\tan b}$.

2°  ${\displaystyle (\sin a+\sin b)^{2}+(\cos a+\cos b)^{2}=2+2\cos(a-b)}$.

3°  ${\displaystyle {\frac {2\sin a-\sin 2a}{2\sin a+\sin 2a}}={\frac {1-\cos a}{1+\cos a}}=\tan ^{2}(a/2)}$.

4°  ${\displaystyle 1+\cos x\pm \sin x=2\cos {\frac {x}{2}}\left(\cos {\frac {x}{2}}\pm \sin {\frac {x}{2}}\right)}$ donc (cf. exercice 4-4) ${\displaystyle {\frac {1-\sin x+\cos x}{1+\sin x+\cos x}}=\tan \left({\frac {\pi }{4}}-{\frac {x}{2}}\right)}$.

1°  ${\displaystyle {\frac {\cos ^{2}a-\cos ^{2}b}{\sin(a+b)}}={\frac {\left(2\cos {\frac {a+b}{2}}\cos {\frac {a-b}{2}}\right)\left(-2\sin {\frac {a+b}{2}}\sin {\frac {a-b}{2}}\right)}{2\sin {\frac {a+b}{2}}\cos {\frac {a+b}{2}}}}=\sin(b-a)}$.

2°  ${\displaystyle {\frac {\sin(a-b)}{\sin a+\sin b}}={\frac {2\sin {\frac {a-b}{2}}\cos {\frac {a-b}{2}}}{2\sin {\frac {a+b}{2}}\cos {\frac {a-b}{2}}}}={\frac {\sin {\frac {a-b}{2}}}{\sin {\frac {a+b}{2}}}}}$ et ${\displaystyle {\frac {\sin(a-b)}{\sin a-\sin b}}={\frac {2\sin {\frac {a-b}{2}}\cos {\frac {a-b}{2}}}{2\sin {\frac {a-b}{2}}\cos {\frac {a+b}{2}}}}={\frac {\cos {\frac {a-b}{2}}}{\cos {\frac {a+b}{2}}}}}$.

3°  ${\displaystyle {\frac {\sin ^{2}a-\sin ^{2}b}{(\cos a+\cos b)^{2}}}=\tan {\frac {a+b}{2}}\tan {\frac {a-b}{2}}}$.

4°  ${\displaystyle {\frac {\sin a-\sin b}{\tan a-\tan b}}={\frac {2\sin {\frac {a-b}{2}}\cos {\frac {a+b}{2}}}{2\sin {\frac {a-b}{2}}\cos {\frac {a-b}{2}}}}\cos a\cos b={\frac {\cos {\frac {a+b}{2}}}{\cos {\frac {a-b}{2}}}}\cos a\cos b}$.

1°  ${\displaystyle {\frac {\sin a+\sin 4a+\sin 7a}{\cos a+\cos 4a+\cos 7a}}={\frac {\sin 4a+2\sin 4a\cos 3a}{\cos 4a+2\cos 4a\cos 3a}}=\tan 4a}$ (voir aussi l'exercice suivant).

2°  ${\displaystyle {\frac {\sin 2a+2\sin 3a+\sin 4a}{\sin 3a+2\sin 4a+\sin 5a}}={\frac {2\sin 3a+2\sin 3a\cos a}{2\sin 4a+2\sin 4a\cos a}}={\frac {\sin 3a}{\sin 4a}}}$.

1°  a)

• La formule ${\displaystyle \sin a\sin b={\frac {\cos(a-b)-\cos(a+b)}{2}}}$ donne

  ${\displaystyle \sin(x+kr)\sin {\frac {r}{2}}={\frac {\cos(x+(k-{\frac {1}{2}})r)-\cos(x+(k+{\frac {1}{2}})r)}{2}}}$

  donc (en sommant de ${\displaystyle k=0}$ à ${\displaystyle k=n-1}$ et en remarquant que les termes intermédiaires s'éliminent 2 par 2) :

  ${\displaystyle S\sin {\frac {r}{2}}={\frac {\cos(x-{\frac {r}{2}})-\cos(x+(n-{\frac {1}{2}})r)}{2}}}$ ;

  puis, la formule ${\displaystyle \cos a-\cos b=2\sin {\frac {a+b}{2}}\sin {\frac {b-a}{2}}}$ donne

  ${\displaystyle S\sin {\frac {r}{2}}=\sin {\frac {2x+(n-1)r}{2}}\sin {\frac {nr}{2}}}$.

• De même, la formule ${\displaystyle \cos a\sin b={\frac {\sin(a+b)-\sin(a-b)}{2}}}$ donne

  ${\displaystyle \cos(x+kr)\sin {\frac {r}{2}}={\frac {-\sin(x+(k-{\frac {1}{2}})r)+\sin(x+(k+{\frac {1}{2}})r)}{2}}}$

  donc

  ${\displaystyle S'\sin {\frac {r}{2}}={\frac {\sin(x+(n-{\frac {1}{2}})r)-\sin(x-{\frac {r}{2}})}{2}}}$ ;

  puis, la formule ${\displaystyle \sin a-\sin b=2\cos \left({\frac {a+b}{2}}\right)\sin \left({\frac {a-b}{2}}\right)}$ donne

  ${\displaystyle S'\sin {\frac {r}{2}}=\cos {\frac {2x+(n-1)r}{2}}\sin {\frac {nr}{2}}}$.

• Finalement, on a trouvé

  ${\displaystyle S={\frac {\sin {\frac {2x+(n-1)r}{2}}\sin {\frac {nr}{2}}}{\sin {\frac {r}{2}}}}{\text{ et }}S'={\frac {\cos {\frac {2x+(n-1)r}{2}}\sin {\frac {nr}{2}}}{\sin {\frac {r}{2}}}}}$,

  sauf bien sûr si ${\displaystyle \sin {\frac {r}{2}}=0}$, c'est-à-dire ${\displaystyle r}$ multiple de ${\displaystyle 2\pi }$, auquel cas ${\displaystyle S=n\sin x}$ et ${\displaystyle S'=n\cos x}$.

b)  Lorsque ${\displaystyle r={\frac {2\pi }{n}}}$, ${\displaystyle \sin {\frac {nr}{2}}=0}$ donc ${\displaystyle S=S'=0}$.

2°  a)  Dans le cas ${\displaystyle x=r=a}$, ${\displaystyle \sin a+\sin 2a+\cdots +\sin na=S={\frac {\sin {\frac {(n+1)a}{2}}\sin {\frac {na}{2}}}{\sin {\frac {a}{2}}}}}$.

b)  Le cas ${\displaystyle x=0}$ et ${\displaystyle r=a}$ donne (en remplaçant ${\displaystyle n}$ par ${\displaystyle n+1}$ dans ${\displaystyle S'}$) ${\displaystyle 1+\cos a+\cos 2a+\cdots +\cos na={\frac {\cos {\frac {na}{2}}\sin {\frac {(n+1)a}{2}}}{\sin {\frac {a}{2}}}}}$.

c)  ${\displaystyle \sin ^{2}a+\sin ^{2}2a+\cdots +\sin ^{2}na={\frac {1-\cos 2a}{2}}+{\frac {1-\cos 4a}{2}}+\dots +{\frac {1-\cos 2na}{2}}={\frac {n+1}{2}}-{\frac {\cos na\sin(n+1)a}{2\sin a}}}$.

d)  ${\displaystyle 1+\cos ^{2}a+\cos ^{2}2a+\cdots +\cos ^{2}na=1+{\frac {1+\cos 2a}{2}}+{\frac {1+\cos 4a}{2}}+\dots +{\frac {1+\cos 2na}{2}}={\frac {n+1}{2}}+{\frac {\cos na\sin(n+1)a}{2\sin a}}}$ (on peut vérifier que c) + d) = n + 1).

3°  La question 1, appliquée à ${\displaystyle x=a}$ et ${\displaystyle r=2a}$, donne

${\displaystyle {\frac {\sin a+\sin 3a+\sin 5a+\cdots +\sin(2n-1)a}{\cos a+\cos 3a+\cos 5a+\cdots +\cos(2n-1)a}}={\frac {S}{S'}}=\tan(na)}$.