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Exercice : Relations trigonométriques 1Trigonométrie/Exercices/Relations trigonométriques 1 », n'a pu être restituée correctement ci-dessus.
En supposant
a
+
b
+
c
=
π
{\displaystyle a+b+c=\pi }
1°
sin
a
+
sin
b
+
sin
c
sin
a
+
sin
b
−
sin
c
{\displaystyle {\frac {\sin a+\sin b+\sin c}{\sin a+\sin b-\sin c}}}
2°
1
+
cos
a
+
cos
b
−
cos
c
{\displaystyle 1+\cos a+\cos b-\cos c}
Solution
1°
sin
a
+
sin
b
+
sin
c
sin
a
+
sin
b
−
sin
c
=
sin
a
+
sin
b
+
sin
(
a
+
b
)
sin
a
+
sin
b
−
sin
(
a
+
b
)
=
2
sin
(
(
a
+
b
)
/
2
)
)
(
cos
(
(
a
−
b
)
/
2
)
)
+
cos
(
(
a
+
b
)
/
2
)
)
)
2
sin
(
(
a
+
b
)
/
2
)
)
(
cos
(
(
a
−
b
)
/
2
)
)
−
cos
(
(
a
+
b
)
/
2
)
)
)
=
cos
(
a
/
2
)
cos
(
b
/
2
)
sin
(
a
/
2
)
sin
(
b
/
2
)
=
cot
(
a
/
2
)
cot
(
b
/
2
)
{\displaystyle {\begin{aligned}{\frac {\sin a+\sin b+\sin c}{\sin a+\sin b-\sin c}}&={\frac {\sin a+\sin b+\sin(a+b)}{\sin a+\sin b-\sin(a+b)}}\\&={\frac {2\sin((a+b)/2))\left(\cos((a-b)/2))+\cos((a+b)/2))\right)}{2\sin((a+b)/2))\left(\cos((a-b)/2))-\cos((a+b)/2))\right)}}\\&={\frac {\cos(a/2)\cos(b/2)}{\sin(a/2)\sin(b/2)}}\\&=\cot(a/2)\cot(b/2)\end{aligned}}}
2°
1
+
cos
a
+
cos
b
−
cos
c
=
cos
a
+
cos
b
+
1
+
cos
(
a
+
b
)
=
2
cos
(
(
a
+
b
)
/
2
)
(
cos
(
(
a
−
b
)
/
2
)
+
cos
(
(
a
+
b
)
/
2
)
)
=
2
cos
(
(
π
−
c
)
/
2
)
×
2
cos
a
cos
b
=
4
sin
c
2
cos
a
cos
b
{\displaystyle {\begin{aligned}1+\cos a+\cos b-\cos c&=\cos a+\cos b+1+\cos(a+b)\\&=2\cos((a+b)/2)\left(\cos((a-b)/2)+\cos((a+b)/2)\right)\\&=2\cos((\pi -c)/2)\times 2\cos a\cos b\\&=4\sin {\frac {c}{2}}\cos a\cos b\end{aligned}}}
Calculer la somme :
S
=
sin
4
π
8
+
sin
4
3
π
8
+
sin
4
5
π
8
+
sin
4
7
π
8
{\displaystyle S=\sin ^{4}{\frac {\pi }{8}}+\sin ^{4}{\frac {3\pi }{8}}+\sin ^{4}{\frac {5\pi }{8}}+\sin ^{4}{\frac {7\pi }{8}}}
Solution
S
=
2
(
sin
4
π
8
+
cos
4
π
8
)
=
2
(
(
sin
2
π
8
+
cos
2
π
8
)
2
−
2
sin
2
π
8
cos
2
π
8
)
=
2
−
sin
2
π
4
=
3
2
.
{\displaystyle {\begin{aligned}S&=2\left(\sin ^{4}{\frac {\pi }{8}}+\cos ^{4}{\frac {\pi }{8}}\right)\\&=2\left(\left(\sin ^{2}{\frac {\pi }{8}}+\cos ^{2}{\frac {\pi }{8}}\right)^{2}-2\sin ^{2}{\frac {\pi }{8}}\cos ^{2}{\frac {\pi }{8}}\right)\\&=2-\sin ^{2}{\frac {\pi }{4}}\\&={\frac {3}{2}}.\end{aligned}}}
De la relation
cos
2
x
=
2
cos
2
x
−
1
{\displaystyle \cos 2x=2\cos ^{2}x-1}
cos
4
x
=
3
8
+
cos
2
x
2
+
cos
4
x
8
{\displaystyle \cos ^{4}x={\frac {3}{8}}+{\frac {\cos 2x}{2}}+{\frac {\cos 4x}{8}}}
Calculer :
S
=
cos
4
π
8
+
cos
4
3
π
8
+
cos
4
5
π
8
+
cos
4
7
π
8
{\displaystyle S=\cos ^{4}{\frac {\pi }{8}}+\cos ^{4}{\frac {3\pi }{8}}+\cos ^{4}{\frac {5\pi }{8}}+\cos ^{4}{\frac {7\pi }{8}}}
Solution
cos
4
x
=
(
1
+
cos
2
x
2
)
2
=
1
+
2
cos
2
x
+
cos
2
2
x
4
=
1
+
2
cos
2
x
+
1
+
cos
4
x
2
4
=
3
8
+
cos
2
x
2
+
cos
4
x
8
{\displaystyle \cos ^{4}x=\left({\frac {1+\cos 2x}{2}}\right)^{2}={\frac {1+2\cos 2x+\cos ^{2}2x}{4}}={\frac {1+2\cos 2x+{\frac {1+\cos 4x}{2}}}{4}}={\frac {3}{8}}+{\frac {\cos 2x}{2}}+{\frac {\cos 4x}{8}}}
S
2
=
cos
4
π
8
+
cos
4
3
π
8
=
3
4
+
cos
π
4
+
cos
3
π
4
2
+
cos
π
2
+
cos
3
π
2
8
=
3
4
{\displaystyle {\frac {S}{2}}=\cos ^{4}{\frac {\pi }{8}}+\cos ^{4}{\frac {3\pi }{8}}={\frac {3}{4}}+{\frac {\cos {\frac {\pi }{4}}+\cos {\frac {3\pi }{4}}}{2}}+{\frac {\cos {\frac {\pi }{2}}+\cos {\frac {3\pi }{2}}}{8}}={\frac {3}{4}}}
S
=
3
2
{\displaystyle S={\frac {3}{2}}}
Mettre sous forme de produit ou de quotient les expressions :
1°
sin
a
+
2
sin
2
a
+
sin
3
a
{\displaystyle \sin a+2\sin 2a+\sin 3a}
2°
cos
a
+
2
cos
2
a
+
cos
3
a
{\displaystyle \cos a+2\cos 2a+\cos 3a}
3°
sin
2
a
+
sin
2
b
−
sin
2
(
a
+
b
)
{\displaystyle \sin ^{2}a+\sin ^{2}b-\sin ^{2}(a+b)}
4°
1
+
sin
x
+
cos
x
+
sin
x
cos
x
{\displaystyle 1+\sin x+\cos x+\sin x\cos x}
5°
sin
2
b
−
sin
2
a
=
cos
2
a
−
cos
2
b
{\displaystyle \sin ^{2}b-\sin ^{2}a=\cos ^{2}a-\cos ^{2}b}
Solution
1°
sin
a
+
2
sin
2
a
+
sin
3
a
=
2
sin
2
a
(
1
+
cos
a
)
=
4
sin
2
a
cos
2
a
2
{\displaystyle \sin a+2\sin 2a+\sin 3a=2\sin 2a\left(1+\cos a\right)=4\sin 2a\cos ^{2}{\frac {a}{2}}}
2°
cos
a
+
2
cos
2
a
+
cos
3
a
=
2
cos
2
a
(
1
+
cos
a
)
=
4
cos
2
a
cos
2
a
2
{\displaystyle \cos a+2\cos 2a+\cos 3a=2\cos 2a\left(1+\cos a\right)=4\cos 2a\cos ^{2}{\frac {a}{2}}}
3°
sin
2
a
+
sin
2
b
−
sin
2
(
a
+
b
)
=
sin
2
a
+
sin
2
b
−
(
sin
a
cos
b
+
sin
b
cos
a
)
2
{\displaystyle \sin ^{2}a+\sin ^{2}b-\sin ^{2}(a+b)=\sin ^{2}a+\sin ^{2}b-\left(\sin a\cos b+\sin b\cos a\right)^{2}}
=
sin
2
a
(
1
−
cos
2
b
)
+
sin
2
b
(
1
−
cos
2
a
)
−
2
sin
a
cos
a
sin
b
cos
b
=
2
sin
a
sin
b
(
sin
a
sin
b
−
cos
a
cos
b
)
{\displaystyle =\sin ^{2}a\left(1-\cos ^{2}b\right)+\sin ^{2}b\left(1-\cos ^{2}a\right)-2\sin a\cos a\sin b\cos b=2\sin a\sin b\left(\sin a\sin b-\cos a\cos b\right)}
=
−
2
sin
a
sin
b
cos
(
a
+
b
)
{\displaystyle =-2\sin a\sin b\cos \left(a+b\right)}
4°
1
+
sin
x
+
cos
x
+
sin
x
cos
x
=
(
1
+
sin
x
)
(
1
+
cos
x
)
=
4
sin
π
+
2
x
4
cos
π
−
2
x
4
cos
2
x
2
{\displaystyle 1+\sin x+\cos x+\sin x\cos x=\left(1+\sin x\right)\left(1+\cos x\right)=4\sin {\frac {\pi +2x}{4}}\cos {\frac {\pi -2x}{4}}\cos ^{2}{\frac {x}{2}}}
5°
cos
2
a
−
cos
2
b
=
(
cos
a
+
cos
b
)
(
cos
a
−
cos
b
)
=
−
4
cos
a
+
b
2
cos
a
−
b
2
sin
a
+
b
2
sin
a
−
b
2
=
−
sin
(
a
+
b
)
sin
(
a
−
b
)
{\displaystyle \cos ^{2}a-\cos ^{2}b=\left(\cos a+\cos b\right)\left(\cos a-\cos b\right)=-4\cos {\frac {a+b}{2}}\cos {\frac {a-b}{2}}\sin {\frac {a+b}{2}}\sin {\frac {a-b}{2}}=-\sin \left(a+b\right)\sin \left(a-b\right)}
Mettre sous forme de produit ou de quotient les expressions :
1°
1
−
sin
2
x
−
sin
2
y
{\displaystyle 1-\sin ^{2}x-\sin ^{2}y}
2°
cos
2
x
−
cos
2
x
{\displaystyle \cos 2x-\cos ^{2}x}
3°
sin
x
+
sin
2
x
{\displaystyle \sin x+\sin 2x}
4°
1
+
tan
x
tan
2
x
{\displaystyle 1+\tan x\tan 2x}
5°
1
+
cos
2
x
+
2
cos
x
{\displaystyle 1+\cos 2x+2\cos x}
Solution
1°
1
−
sin
2
x
−
sin
2
y
=
cos
2
x
−
cos
2
(
π
2
−
y
)
=
−
sin
(
x
+
π
2
−
y
)
sin
(
x
−
π
2
+
y
)
{\displaystyle 1-\sin ^{2}x-\sin ^{2}y=\cos ^{2}x-\cos ^{2}\left({\frac {\pi }{2}}-y\right)=-\sin \left(x+{\frac {\pi }{2}}-y\right)\sin \left(x-{\frac {\pi }{2}}+y\right)}
2°
cos
2
x
−
cos
2
x
=
cos
2
x
−
1
=
−
sin
2
x
{\displaystyle \cos 2x-\cos ^{2}x=\cos ^{2}x-1=-\sin ^{2}x}
3°
sin
x
+
sin
2
x
=
2
sin
x
(
1
2
+
cos
x
)
=
2
sin
x
(
cos
π
3
+
cos
x
)
=
4
sin
x
cos
π
3
+
x
2
cos
π
3
−
x
2
{\displaystyle \sin x+\sin 2x=2\sin x\left({\frac {1}{2}}+\cos x\right)=2\sin x\left(\cos {\frac {\pi }{3}}+\cos x\right)=4\sin x\cos {\frac {{\frac {\pi }{3}}+x}{2}}\cos {\frac {{\frac {\pi }{3}}-x}{2}}}
4°
1
+
tan
x
tan
2
x
=
1
+
2
tan
x
cos
x
sin
x
cos
2
x
=
cos
2
x
+
2
sin
2
x
cos
2
x
=
1
cos
2
x
{\displaystyle 1+\tan x\tan 2x=1+{\frac {2\tan x\cos x\sin x}{\cos 2x}}={\frac {\cos 2x+2\sin ^{2}x}{\cos 2x}}={\frac {1}{\cos 2x}}}
5°
1
+
cos
2
x
+
2
cos
x
=
2
cos
x
(
cos
x
+
1
)
=
4
cos
x
cos
2
x
2
{\displaystyle 1+\cos 2x+2\cos x=2\cos x\left(\cos x+1\right)=4\cos x\cos ^{2}{\frac {x}{2}}}
Démontrer les relations :
1°
S
=
sin
π
9
sin
2
π
9
sin
3
π
9
sin
4
π
9
=
3
16
{\displaystyle S=\sin {\frac {\pi }{9}}\sin {\frac {2\pi }{9}}\sin {\frac {3\pi }{9}}\sin {\frac {4\pi }{9}}={\frac {3}{16}}}
2°
C
=
cos
π
9
cos
2
π
9
cos
3
π
9
cos
4
π
9
=
1
16
{\displaystyle C=\cos {\frac {\pi }{9}}\cos {\frac {2\pi }{9}}\cos {\frac {3\pi }{9}}\cos {\frac {4\pi }{9}}={\frac {1}{16}}}
Solution
1° La partie réelle de
1
−
(
cos
t
+
i
sin
t
)
18
{\displaystyle 1-\left(\cos t+\mathrm {i} \sin t\right)^{18}}
1
−
∑
k
=
0
9
(
18
2
k
)
(
−
1
)
k
sin
2
k
t
(
1
−
sin
2
t
)
9
−
k
=
sin
2
t
P
(
sin
t
)
{\displaystyle 1-\sum _{k=0}^{9}{\binom {18}{2k}}(-1)^{k}\sin ^{2k}t\left(1-\sin ^{2}t\right)^{9-k}=\sin ^{2}t\,P(\sin t)}
P
{\displaystyle P}
±
sin
π
9
,
±
sin
2
π
9
,
±
sin
3
π
9
,
±
sin
4
π
9
{\displaystyle \pm \sin {\frac {\pi }{9}},\pm \sin {\frac {2\pi }{9}},\pm \sin {\frac {3\pi }{9}},\pm \sin {\frac {4\pi }{9}}}
P
{\displaystyle P}
∑
k
=
0
9
(
18
2
k
)
=
2
17
{\displaystyle \sum _{k=0}^{9}{\binom {18}{2k}}=2^{17}}
−
(
18
0
)
(
−
1
)
0
(
−
9
)
−
(
18
2
)
(
−
1
)
1
=
9
+
18
×
17
2
=
9
×
18
{\displaystyle -{\binom {18}{0}}(-1)^{0}(-9)-{\binom {18}{2}}(-1)^{1}=9+{\frac {18\times 17}{2}}=9\times 18}
S
=
9
×
18
2
17
4
=
3
16
{\displaystyle S={\sqrt[{4}]{\frac {9\times 18}{2^{17}}}}={\frac {3}{16}}}
2° La partie réelle de
1
−
(
cos
t
+
i
sin
t
)
18
{\displaystyle 1-\left(\cos t+\mathrm {i} \sin t\right)^{18}}
1
−
∑
k
=
0
9
(
18
2
k
)
(
−
1
)
k
(
1
−
cos
2
t
)
k
cos
18
−
2
k
t
=
(
1
−
cos
2
t
)
Q
(
cos
t
)
{\displaystyle 1-\sum _{k=0}^{9}{\binom {18}{2k}}(-1)^{k}\left(1-\cos ^{2}t\right)^{k}\cos ^{18-2k}t=\left(1-\cos ^{2}t\right)\,Q(\cos t)}
Q
{\displaystyle Q}
±
cos
π
9
,
±
cos
2
π
9
,
±
cos
3
π
9
,
±
cos
4
π
9
{\displaystyle \pm \cos {\frac {\pi }{9}},\pm \cos {\frac {2\pi }{9}},\pm \cos {\frac {3\pi }{9}},\pm \cos {\frac {4\pi }{9}}}
Q
{\displaystyle Q}
2
17
{\displaystyle 2^{17}}
2
{\displaystyle 2}
C
=
2
2
17
4
=
1
16
{\displaystyle C={\sqrt[{4}]{\frac {2}{2^{17}}}}={\frac {1}{16}}}
Par conséquent,
∏
k
=
1
8
cos
k
π
9
=
C
2
=
1
4
4
{\displaystyle \prod _{k=1}^{8}\cos {\frac {k\pi }{9}}=C^{2}={\frac {1}{4^{4}}}}
Polynôme/Exercices/Racines de polynômes#Exercice 1-7 .
![{\displaystyle S={\sqrt[{4}]{\frac {9\times 18}{2^{17}}}}={\frac {3}{16}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef2aa5467b9488e43c139ebc39f0a7afea4af843)
![{\displaystyle C={\sqrt[{4}]{\frac {2}{2^{17}}}}={\frac {1}{16}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61d024c524717262c6f7292be3c52028fbbfefca)
