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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Étude de fonctions 4Fonctions trigonométriques/Exercices/Étude de fonctions 4 », n'a pu être restituée correctement ci-dessus.
Étudier et tracer la fonction
f
{\displaystyle f}
f
(
x
)
=
cos
3
x
1
+
cos
2
x
{\displaystyle f(x)={\frac {\cos 3x}{1+\cos 2x}}}
Solution
f
{\displaystyle f}
f
(
x
+
π
)
=
−
f
(
x
)
{\displaystyle f(x+\pi )=-f(x)}
[
0
,
π
2
[
{\displaystyle \left[0,{\frac {\pi }{2}}\right[}
Sur cet intervalle,
cos
{\displaystyle \cos }
f
=
2
cos
−
3
2
cos
{\displaystyle f=2\cos -{\frac {3}{2\cos }}}
f
(
0
)
=
1
2
{\displaystyle f(0)={\frac {1}{2}}}
lim
π
2
−
f
=
−
∞
{\displaystyle \lim _{{\frac {\pi }{2}}^{-}}f=-\infty }
Tracé sur Google
Étudier et tracer la fonction
f
{\displaystyle f}
f
(
x
)
=
6
sin
x
−
3
cos
2
x
+
2
sin
3
x
{\displaystyle f(x)=6\sin x-3\cos 2x+2\sin 3x}
Solution
f
{\displaystyle f}
2
π
{\displaystyle 2\pi }
f
(
π
−
x
)
=
f
(
x
)
{\displaystyle f(\pi -x)=f(x)}
[
−
π
2
,
π
2
]
{\displaystyle \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right]}
f
(
x
)
=
g
(
sin
x
)
{\displaystyle f(x)=g(\sin x)}
g
(
s
)
=
6
s
−
3
(
1
−
2
s
2
)
+
2
(
3
s
−
4
s
3
)
=
−
8
s
3
+
6
s
2
+
12
s
−
3
{\displaystyle g(s)=6s-3(1-2s^{2})+2(3s-4s^{3})=-8s^{3}+6s^{2}+12s-3}
g
′
(
s
)
=
−
12
(
s
−
1
)
(
s
+
1
2
)
{\displaystyle g'(s)=-12\left(s-1\right)\left(s+{\frac {1}{2}}\right)}
sin
′
=
cos
{\displaystyle \sin '=\cos }
f
′
{\displaystyle f'}
sin
x
+
1
2
{\displaystyle \sin x+{\frac {1}{2}}}
f
{\displaystyle f}
[
−
π
2
,
−
π
6
]
{\displaystyle \left[-{\frac {\pi }{2}},-{\frac {\pi }{6}}\right]}
[
−
π
6
,
π
2
]
{\displaystyle \left[-{\frac {\pi }{6}},{\frac {\pi }{2}}\right]}
f
(
−
π
2
)
=
g
(
−
1
)
=
−
1
{\displaystyle f\left(-{\frac {\pi }{2}}\right)=g(-1)=-1}
f
(
−
π
6
)
=
g
(
−
1
2
)
=
−
6
,
5
{\displaystyle f\left(-{\frac {\pi }{6}}\right)=g\left(-{\frac {1}{2}}\right)=-6{,}5}
f
(
π
2
)
=
g
(
1
)
=
7
{\displaystyle f\left({\frac {\pi }{2}}\right)=g(1)=7}
Tracé sur Google
Étudier et tracer la fonction
f
{\displaystyle f}
f
(
x
)
=
sin
x
+
cos
x
sin
x
−
cos
x
{\displaystyle f(x)={\frac {\sin x+\cos x}{\sin x-\cos x}}}
Solution
Les variations et la courbe de
f
(
x
)
=
−
tan
(
x
+
π
4
)
{\displaystyle f(x)=-\tan \left(x+{\frac {\pi }{4}}\right)}
tan
{\displaystyle \tan }
Tracé sur Google
Étudier et tracer la fonction
f
{\displaystyle f}
f
(
x
)
=
1
1
+
cos
2
x
−
tan
x
{\displaystyle f(x)={\frac {1}{1+\cos 2x}}-\tan x}
Étudier et tracer la fonction
f
{\displaystyle f}
f
(
x
)
=
ln
cos
x
{\displaystyle f(x)=\ln \cos x}
![{\displaystyle \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54381f086ac9ffe8306d413f813abcb616e95dee)
![{\displaystyle \left[-{\frac {\pi }{2}},-{\frac {\pi }{6}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb16da9a8b8ed53ad86b3e4735656b6ad9f4b9b2)
![{\displaystyle \left[-{\frac {\pi }{6}},{\frac {\pi }{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9167f7e22dad93682121fdd3986016e3a6017b41)
![{\displaystyle \left]-{\frac {\pi }{2}},{\frac {\pi }{2}}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6b123273d134ae0a355186bca3947bf36996ceb)
![{\displaystyle \left]-{\frac {\pi }{2}},{\frac {\pi }{4}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc1ccdf31f688f65238b9676439d2743acd332f8)
![{\displaystyle \left]-\pi /2,\pi /2\right[+2\pi \mathbb {Z} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f101269c32830a8e986cf2fac48953ec5ff08e78)
