En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Simplification d'expressionsTrigonométrie/Exercices/Simplification d'expressions », n'a pu être restituée correctement ci-dessus.
Simplifiez :
1°
sin
3
x
cos
5
x
−
cos
3
x
sin
5
x
{\displaystyle \sin 3x\cos 5x-\cos 3x\sin 5x}
2°
cos
7
x
cos
4
x
+
sin
7
x
sin
4
x
{\displaystyle \cos 7x\cos 4x+\sin 7x\sin 4x}
Solution
1°
sin
3
x
cos
5
x
−
cos
3
x
sin
5
x
=
sin
(
3
x
−
5
x
)
=
−
sin
2
x
{\displaystyle \sin 3x\cos 5x-\cos 3x\sin 5x=\sin(3x-5x)=-\sin 2x}
2°
cos
7
x
cos
4
x
+
sin
7
x
sin
4
x
=
cos
(
7
x
−
4
x
)
=
cos
3
x
{\displaystyle \cos 7x\cos 4x+\sin 7x\sin 4x=\cos(7x-4x)=\cos 3x}
Simplifiez :
1°
sin
2
a
sin
a
−
cos
2
a
cos
a
{\displaystyle {\frac {\sin 2a}{\sin a}}-{\frac {\cos 2a}{\cos a}}}
2°
sin
3
a
sin
a
−
cos
3
a
cos
a
{\displaystyle {\frac {\sin 3a}{\sin a}}-{\frac {\cos 3a}{\cos a}}}
Solution
sin
b
sin
a
−
cos
b
cos
a
=
sin
b
cos
a
−
cos
b
sin
a
sin
a
cos
a
=
sin
(
b
−
a
)
sin
a
cos
a
{\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}}}
1°
sin
2
a
sin
a
−
cos
2
a
cos
a
=
1
cos
a
{\displaystyle {\frac {\sin 2a}{\sin a}}-{\frac {\cos 2a}{\cos a}}={\frac {1}{\cos a}}}
2°
sin
3
a
sin
a
−
cos
3
a
cos
a
=
2
{\displaystyle {\frac {\sin 3a}{\sin a}}-{\frac {\cos 3a}{\cos a}}=2}
Exprimer l'expression suivante en fonction de
cos
2
x
{\displaystyle \cos 2x}
sin
2
x
{\displaystyle \sin 2x}
A
=
cos
2
x
−
2
sin
x
cos
x
−
5
sin
2
x
{\displaystyle A=\cos ^{2}x-2\sin x\cos x-5\sin ^{2}x}
Solution
A
=
1
+
cos
2
x
2
−
sin
2
x
−
5
1
−
cos
2
x
2
=
−
2
+
3
cos
2
x
−
sin
2
x
{\displaystyle A={\frac {1+\cos 2x}{2}}-\sin 2x-5{\frac {1-\cos 2x}{2}}=-2+3\cos 2x-\sin 2x}
Simplifier l'expression :
2
(
sin
6
x
+
cos
6
x
)
−
3
(
sin
4
x
+
cos
4
x
)
{\displaystyle 2(\sin ^{6}x+\cos ^{6}x)-3(\sin ^{4}x+\cos ^{4}x)}
Solution
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
{\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
1
=
(
sin
2
x
+
cos
2
x
)
2
=
sin
4
x
+
cos
4
x
+
2
sin
2
x
cos
2
x
{\displaystyle 1=(\sin ^{2}x+\cos ^{2}x)^{2}=\sin ^{4}x+\cos ^{4}x+2\sin ^{2}x\cos ^{2}x}
donc
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
{\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}
Montrer que les expressions :
1°
sin
x
+
cos
x
3
sin
x
−
2
cos
x
{\displaystyle {\frac {\sin x+\cos x}{3\sin x-2\cos x}}}
2°
sin
x
−
cos
x
sin
3
x
+
2
cos
3
x
{\displaystyle {\frac {\sin x-\cos x}{\sin ^{3}x+2\cos ^{3}x}}}
3°
sin
3
x
+
cos
x
sin
x
−
cos
x
{\displaystyle {\frac {\sin ^{3}x+\cos x}{\sin x-\cos x}}}
peuvent s'exprimer à l'aide de la seule fonction
tan
x
{\displaystyle \tan x}
Solution
1°
sin
x
+
cos
x
3
sin
x
−
2
cos
x
=
tan
x
+
1
3
tan
x
−
2
{\displaystyle {\frac {\sin x+\cos x}{3\sin x-2\cos x}}={\frac {\tan x+1}{3\tan x-2}}}
2°
sin
x
−
cos
x
sin
3
x
+
2
cos
3
x
=
(
1
+
tan
2
x
)
tan
x
−
1
tan
3
x
+
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°
sin
3
x
+
cos
x
sin
x
−
cos
x
=
tan
3
x
1
+
tan
2
x
+
1
tan
x
−
1
{\displaystyle {\frac {\sin ^{3}x+\cos x}{\sin x-\cos x}}={\frac {{\frac {\tan ^{3}x}{1+\tan ^{2}x}}+1}{\tan x-1}}}
Montrer que les expressions :
1°
sin
4
x
+
cos
4
x
sin
4
x
−
cos
4
x
{\displaystyle {\frac {\sin ^{4}x+\cos ^{4}x}{\sin ^{4}x-\cos ^{4}x}}}
2°
sin
3
x
−
cos
3
x
sin
x
−
cos
x
{\displaystyle {\frac {\sin ^{3}x-\cos ^{3}x}{\sin x-\cos x}}}
3°
sin
2
x
+
sin
x
cos
x
sin
2
x
−
cos
2
x
{\displaystyle {\frac {\sin ^{2}x+\sin x\cos x}{\sin ^{2}x-\cos ^{2}x}}}
4°
cos
2
x
−
sin
x
cos
x
{\displaystyle \cos ^{2}x-\sin x\cos x}
peuvent s'exprimer à l'aide de la seule fonction
tan
x
{\displaystyle \tan x}
Solution
1°
sin
4
x
+
cos
4
x
sin
4
x
−
cos
4
x
=
tan
4
x
+
1
tan
4
x
−
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°
sin
3
x
−
cos
3
x
sin
x
−
cos
x
=
tan
3
x
−
1
(
1
+
tan
2
x
)
(
tan
x
−
1
)
{\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°
sin
2
x
+
sin
x
cos
x
sin
2
x
−
cos
2
x
=
tan
2
x
+
tan
x
tan
2
x
−
1
{\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°
cos
2
x
−
sin
x
cos
x
=
1
−
tan
x
1
+
tan
2
x
{\displaystyle \cos ^{2}x-\sin x\cos x={\frac {1-\tan x}{1+\tan ^{2}x}}}
Simplifier les expressions :
1°
tan
a
+
tan
b
cos
a
+
cos
b
{\displaystyle {\frac {\tan a+\tan b}{\cos a+\cos b}}}
2°
(
sin
a
+
sin
b
)
2
+
(
cos
a
+
cos
b
)
2
{\displaystyle (\sin a+\sin b)^{2}+(\cos a+\cos b)^{2}}
3°
2
sin
a
−
sin
2
a
2
sin
a
+
sin
2
a
{\displaystyle {\frac {2\sin a-\sin 2a}{2\sin a+\sin 2a}}}
4°
1
−
sin
x
+
cos
x
1
+
sin
x
+
cos
x
{\displaystyle {\frac {1-\sin x+\cos x}{1+\sin x+\cos x}}}
Solution
1°
tan
a
+
tan
b
cos
a
+
cos
b
=
tan
a
tan
b
{\displaystyle {\frac {\tan a+\tan b}{\cos a+\cos b}}=\tan a\tan b}
2°
(
sin
a
+
sin
b
)
2
+
(
cos
a
+
cos
b
)
2
=
2
+
2
cos
(
a
−
b
)
{\displaystyle (\sin a+\sin b)^{2}+(\cos a+\cos b)^{2}=2+2\cos(a-b)}
3°
2
sin
a
−
sin
2
a
2
sin
a
+
sin
2
a
=
1
−
cos
a
1
+
cos
a
=
tan
2
(
a
/
2
)
{\displaystyle {\frac {2\sin a-\sin 2a}{2\sin a+\sin 2a}}={\frac {1-\cos a}{1+\cos a}}=\tan ^{2}(a/2)}
4°
1
+
cos
x
±
sin
x
=
2
cos
x
2
(
cos
x
2
±
sin
x
2
)
{\displaystyle 1+\cos x\pm \sin x=2\cos {\frac {x}{2}}\left(\cos {\frac {x}{2}}\pm \sin {\frac {x}{2}}\right)}
exercice 4-4 )
1
−
sin
x
+
cos
x
1
+
sin
x
+
cos
x
=
tan
(
π
4
−
x
2
)
{\displaystyle {\frac {1-\sin x+\cos x}{1+\sin x+\cos x}}=\tan \left({\frac {\pi }{4}}-{\frac {x}{2}}\right)}
Simplifier les expressions :
1°
cos
2
a
−
cos
2
b
sin
(
a
+
b
)
{\displaystyle {\frac {\cos ^{2}a-\cos ^{2}b}{\sin(a+b)}}}
2°
sin
(
a
−
b
)
sin
a
±
sin
b
{\displaystyle {\frac {\sin(a-b)}{\sin a\pm \sin b}}}
3°
sin
2
a
−
sin
2
b
(
cos
a
+
cos
b
)
2
{\displaystyle {\frac {\sin ^{2}a-\sin ^{2}b}{(\cos a+\cos b)^{2}}}}
4°
sin
a
−
sin
b
tan
a
−
tan
b
{\displaystyle {\frac {\sin a-\sin b}{\tan a-\tan b}}}
Solution
1°
cos
2
a
−
cos
2
b
sin
(
a
+
b
)
=
(
2
cos
a
+
b
2
cos
a
−
b
2
)
(
−
2
sin
a
+
b
2
sin
a
−
b
2
)
2
sin
a
+
b
2
cos
a
+
b
2
=
sin
(
b
−
a
)
{\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°
sin
(
a
−
b
)
sin
a
+
sin
b
=
2
sin
a
−
b
2
cos
a
−
b
2
2
sin
a
+
b
2
cos
a
−
b
2
=
sin
a
−
b
2
sin
a
+
b
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}}}}}
sin
(
a
−
b
)
sin
a
−
sin
b
=
2
sin
a
−
b
2
cos
a
−
b
2
2
sin
a
−
b
2
cos
a
+
b
2
=
cos
a
−
b
2
cos
a
+
b
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 {\cos {\frac {a-b}{2}}}{\cos {\frac {a+b}{2}}}}}
3°
sin
2
a
−
sin
2
b
(
cos
a
+
cos
b
)
2
=
tan
a
+
b
2
tan
a
−
b
2
{\displaystyle {\frac {\sin ^{2}a-\sin ^{2}b}{(\cos a+\cos b)^{2}}}=\tan {\frac {a+b}{2}}\tan {\frac {a-b}{2}}}
4°
sin
a
−
sin
b
tan
a
−
tan
b
=
2
sin
a
−
b
2
cos
a
+
b
2
2
sin
a
−
b
2
cos
a
−
b
2
cos
a
cos
b
=
cos
a
+
b
2
cos
a
−
b
2
cos
a
cos
b
{\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}
Simplifier les expressions :
1°
sin
a
+
sin
4
a
+
sin
7
a
cos
a
+
cos
4
a
+
cos
7
a
{\displaystyle {\frac {\sin a+\sin 4a+\sin 7a}{\cos a+\cos 4a+\cos 7a}}}
2°
sin
2
a
+
2
sin
3
a
+
sin
4
a
sin
3
a
+
2
sin
4
a
+
sin
5
a
{\displaystyle {\frac {\sin 2a+2\sin 3a+\sin 4a}{\sin 3a+2\sin 4a+\sin 5a}}}
Solution
1°
sin
a
+
sin
4
a
+
sin
7
a
cos
a
+
cos
4
a
+
cos
7
a
=
sin
4
a
+
2
sin
4
a
cos
3
a
cos
4
a
+
2
cos
4
a
cos
3
a
=
tan
4
a
{\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}
2°
sin
2
a
+
2
sin
3
a
+
sin
4
a
sin
3
a
+
2
sin
4
a
+
sin
5
a
=
2
sin
3
a
+
2
sin
3
a
cos
a
2
sin
4
a
+
2
sin
4
a
cos
a
=
sin
3
a
sin
4
a
{\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° On considère les expressions :
S
=
sin
x
+
sin
(
x
+
r
)
+
sin
(
x
+
2
r
)
+
⋯
+
sin
[
x
+
(
n
−
1
)
r
]
{\displaystyle S=\sin x+\sin(x+r)+\sin(x+2r)+\cdots +\sin[x+(n-1)r]}
S
′
=
cos
x
+
cos
(
x
+
r
)
+
cos
(
x
+
2
r
)
+
⋯
+
cos
[
x
+
(
n
−
1
)
r
]
{\displaystyle S'=\cos x+\cos(x+r)+\cos(x+2r)+\cdots +\cos[x+(n-1)r]}
que l'on se propose de simplifier.
a) À cet effet, on calculera
S
sin
r
2
{\displaystyle S\sin {\frac {r}{2}}}
S
′
sin
r
2
{\displaystyle S'\sin {\frac {r}{2}}}
b) Étudier le cas où
r
=
2
π
n
{\displaystyle r={\frac {2\pi }{n}}}
2° Pour
a
{\displaystyle a}
π
{\displaystyle \pi }
a)
sin
a
+
sin
2
a
+
⋯
+
sin
n
a
{\displaystyle \sin a+\sin 2a+\cdots +\sin na}
b)
1
+
cos
a
+
cos
2
a
+
⋯
+
cos
n
a
{\displaystyle 1+\cos a+\cos 2a+\cdots +\cos na}
c)
sin
2
a
+
sin
2
2
a
+
⋯
+
sin
2
n
a
{\displaystyle \sin ^{2}a+\sin ^{2}2a+\cdots +\sin ^{2}na}
d)
1
+
cos
2
a
+
cos
2
2
a
+
⋯
+
cos
2
n
a
{\displaystyle 1+\cos ^{2}a+\cos ^{2}2a+\cdots +\cos ^{2}na}
3° Pour
n
a
{\displaystyle na}
π
2
{\displaystyle {\frac {\pi }{2}}}
sin
a
+
sin
3
a
+
sin
5
a
+
⋯
+
sin
(
2
n
−
1
)
a
cos
a
+
cos
3
a
+
cos
5
a
+
⋯
+
cos
(
2
n
−
1
)
a
{\displaystyle {\frac {\sin a+\sin 3a+\sin 5a+\cdots +\sin(2n-1)a}{\cos a+\cos 3a+\cos 5a+\cdots +\cos(2n-1)a}}}
Solution
1° a)
La formule
sin
a
sin
b
=
cos
(
a
−
b
)
−
cos
(
a
+
b
)
2
{\displaystyle \sin a\sin b={\frac {\cos(a-b)-\cos(a+b)}{2}}}
sin
(
x
+
k
r
)
sin
r
2
=
cos
(
x
+
(
k
−
1
2
)
r
)
−
cos
(
x
+
(
k
+
1
2
)
r
)
2
{\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
k
=
0
{\displaystyle k=0}
k
=
n
−
1
{\displaystyle k=n-1}
les termes intermédiaires s'éliminent 2 par 2 ) :
S
sin
r
2
=
cos
(
x
−
r
2
)
−
cos
(
x
+
(
n
−
1
2
)
r
)
2
{\displaystyle S\sin {\frac {r}{2}}={\frac {\cos(x-{\frac {r}{2}})-\cos(x+(n-{\frac {1}{2}})r)}{2}}}
puis, la formule
cos
a
−
cos
b
=
2
sin
a
+
b
2
sin
b
−
a
2
{\displaystyle \cos a-\cos b=2\sin {\frac {a+b}{2}}\sin {\frac {b-a}{2}}}
S
sin
r
2
=
sin
2
x
+
(
n
−
1
)
r
2
sin
n
r
2
{\displaystyle S\sin {\frac {r}{2}}=\sin {\frac {2x+(n-1)r}{2}}\sin {\frac {nr}{2}}}
De même, la formule
cos
a
sin
b
=
sin
(
a
+
b
)
−
sin
(
a
−
b
)
2
{\displaystyle \cos a\sin b={\frac {\sin(a+b)-\sin(a-b)}{2}}}
cos
(
x
+
k
r
)
sin
r
2
=
−
sin
(
x
+
(
k
−
1
2
)
r
)
+
sin
(
x
+
(
k
+
1
2
)
r
)
2
{\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
S
′
sin
r
2
=
sin
(
x
+
(
n
−
1
2
)
r
)
−
sin
(
x
−
r
2
)
2
{\displaystyle S'\sin {\frac {r}{2}}={\frac {\sin(x+(n-{\frac {1}{2}})r)-\sin(x-{\frac {r}{2}})}{2}}}
puis, la formule
sin
a
−
sin
b
=
2
cos
(
a
+
b
2
)
sin
(
a
−
b
2
)
{\displaystyle \sin a-\sin b=2\cos \left({\frac {a+b}{2}}\right)\sin \left({\frac {a-b}{2}}\right)}
S
′
sin
r
2
=
cos
2
x
+
(
n
−
1
)
r
2
sin
n
r
2
{\displaystyle S'\sin {\frac {r}{2}}=\cos {\frac {2x+(n-1)r}{2}}\sin {\frac {nr}{2}}}
Finalement, on a trouvé
S
=
sin
2
x
+
(
n
−
1
)
r
2
sin
n
r
2
sin
r
2
et
S
′
=
cos
2
x
+
(
n
−
1
)
r
2
sin
n
r
2
sin
r
2
{\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
sin
r
2
=
0
{\displaystyle \sin {\frac {r}{2}}=0}
r
{\displaystyle r}
2
π
{\displaystyle 2\pi }
S
=
n
sin
x
{\displaystyle S=n\sin x}
S
′
=
n
cos
x
{\displaystyle S'=n\cos x}
b) Lorsque
r
=
2
π
n
{\displaystyle r={\frac {2\pi }{n}}}
sin
n
r
2
=
0
{\displaystyle \sin {\frac {nr}{2}}=0}
S
=
S
′
=
0
{\displaystyle S=S'=0}
2° a) Dans le cas
x
=
r
=
a
{\displaystyle x=r=a}
sin
a
+
sin
2
a
+
⋯
+
sin
n
a
=
S
=
sin
(
n
+
1
)
a
2
sin
n
a
2
sin
a
2
{\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
x
=
0
{\displaystyle x=0}
r
=
a
{\displaystyle r=a}
n
{\displaystyle n}
n
+
1
{\displaystyle n+1}
S
′
{\displaystyle S'}
1
+
cos
a
+
cos
2
a
+
⋯
+
cos
n
a
=
cos
n
a
2
sin
(
n
+
1
)
a
2
sin
a
2
{\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)
sin
2
a
+
sin
2
2
a
+
⋯
+
sin
2
n
a
=
1
−
cos
2
a
2
+
1
−
cos
4
a
2
+
⋯
+
1
−
cos
2
n
a
2
=
n
+
1
2
−
cos
n
a
sin
(
n
+
1
)
a
2
sin
a
{\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)
1
+
cos
2
a
+
cos
2
2
a
+
⋯
+
cos
2
n
a
=
1
+
1
+
cos
2
a
2
+
1
+
cos
4
a
2
+
⋯
+
1
+
cos
2
n
a
2
=
n
+
1
2
+
cos
n
a
sin
(
n
+
1
)
a
2
sin
a
{\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}}}
c) + d) = n + 1).3° La question 1, appliquée à
x
=
a
{\displaystyle x=a}
r
=
2
a
{\displaystyle r=2a}
sin
a
+
sin
3
a
+
sin
5
a
+
⋯
+
sin
(
2
n
−
1
)
a
cos
a
+
cos
3
a
+
cos
5
a
+
⋯
+
cos
(
2
n
−
1
)
a
=
S
S
′
=
tan
(
n
a
)
{\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)}
![{\displaystyle S=\sin x+\sin(x+r)+\sin(x+2r)+\cdots +\sin[x+(n-1)r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e68b8a884242ebd3cda6fe4d74c949e867dd5c20)
![{\displaystyle S'=\cos x+\cos(x+r)+\cos(x+2r)+\cdots +\cos[x+(n-1)r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47de1f4e31a3ac6d445e38edc18067e4c3794550)
