En trigonometría , el cosenu (abreviáu cos ) defínese como la razón ente'l catetu axacente y la hipotenusa . O tamién como l'abcisa correspondiente a un puntu que pertenez a una circunferencia unitaria centrada nel orixe.
Cosenu
función trigonométrica y función par (es)
En matemátiques , el cosenu ye la función algamada al facer variar la razón mentada, siendo una de les funciones trescendentes.
editar
θ
,
ϕ
∈
R
{\displaystyle \theta ,\phi \in \ \mathbb {R} }
cos
(
ϕ
+
θ
)
=
c
o
s
(
ϕ
)
c
o
s
(
θ
)
−
s
i
n
(
ϕ
)
s
i
n
(
θ
)
{\displaystyle \cos \left(\phi +\theta \right)=cos(\phi )cos(\theta )-sin(\phi )sin(\theta )}
Si facemos
cos
(
ϕ
+
(
−
θ
)
)
=
c
o
s
(
ϕ
)
c
o
s
(
−
θ
)
−
s
i
n
(
ϕ
)
s
i
n
(
−
θ
)
{\displaystyle \cos \left(\phi +(-\theta \right))=cos(\phi )cos(-\theta )-sin(\phi )sin(-\theta )}
algamamos la resta. Como'l cosenu ye par , el signu nun importa y como'l senu ye impar , el signu sal:
cos
(
ϕ
−
θ
)
=
c
o
s
(
ϕ
)
c
o
s
(
θ
)
+
s
i
n
(
ϕ
)
s
i
n
(
θ
)
{\displaystyle \cos \left(\phi -\theta \right)=cos(\phi )cos(\theta )+sin(\phi )sin(\theta )}
editar
Tenemos que
cos
(
ϕ
+
θ
)
=
c
o
s
(
ϕ
)
c
o
s
(
θ
)
−
s
i
n
(
ϕ
)
s
i
n
(
θ
)
{\displaystyle \cos \left(\phi +\theta \right)=cos(\phi )cos(\theta )-sin(\phi )sin(\theta )}
Faciendo
θ
=
ϕ
{\displaystyle \theta =\phi }
c
o
s
(
2
ϕ
)
=
c
o
s
2
(
ϕ
)
−
s
i
n
2
(
ϕ
)
{\displaystyle cos\left(2\phi \right)=cos^{2}(\phi )-sin^{2}(\phi )}
editar
Hai de decatase que namái con un cenciellu remanéu alxebraicu podemos algamar la fórmula del cosenu del ángulu mediu. Seya
α
,
ϕ
∈
R
{\displaystyle \alpha ,\phi \in \mathbb {R} }
Como
c
o
s
(
2
ϕ
)
=
c
o
s
2
(
ϕ
)
−
s
i
n
2
(
ϕ
)
{\displaystyle cos\left(2\phi \right)=cos^{2}(\phi )-sin^{2}(\phi )}
podemos escribila como
c
o
s
(
2
ϕ
)
=
1
−
2
s
i
n
2
(
ϕ
)
{\displaystyle cos\left(2\phi \right)=1-2sin^{2}(\phi )}
Seya
ϕ
=
α
2
{\displaystyle \phi ={\frac {\alpha }{2}}}
Entós algamamos
|
c
o
s
(
α
2
)
|
=
c
o
s
(
α
)
+
1
2
{\displaystyle |cos({\frac {\alpha }{2}})|={\sqrt {\frac {cos(\alpha )+1}{2}}}}
y analizando los signos de la espresión pa cada cuadrante, finamos con que:
c
o
s
(
α
2
)
=
c
o
s
(
α
)
+
1
2
{\displaystyle cos({\frac {\alpha }{2}})={\sqrt {\frac {cos(\alpha )+1}{2}}}}
C
o
s
(
ϕ
)
+
C
o
s
(
θ
)
=
2
c
o
s
(
ϕ
+
θ
2
)
c
o
s
(
ϕ
−
θ
2
)
{\displaystyle Cos(\phi )+Cos(\theta )=2cos({\frac {\phi +\theta }{2}})cos({\frac {\phi -\theta }{2}})}
Amosamientu
Tomamos
α
β
θ
,
ϕ
∈
R
{\displaystyle \alpha \,\beta \,\theta ,\phi \in \ \mathbb {R} }
Entós
C
o
s
(
α
+
β
)
+
C
o
s
(
α
−
β
)
=
c
o
s
(
α
)
c
o
s
(
β
)
−
s
i
n
(
α
)
s
i
n
(
β
)
+
c
o
s
(
α
)
c
o
s
(
β
)
+
s
i
n
(
α
)
s
i
n
(
β
)
{\displaystyle Cos\left(\alpha +\beta \right)+Cos(\alpha -\beta )=cos(\alpha )cos(\beta )-sin(\alpha )sin(\beta )+cos(\alpha )cos(\beta )+sin(\alpha )sin(\beta )}
C
o
s
(
α
+
β
)
+
C
o
s
(
α
−
β
)
=
2
c
o
s
(
α
)
c
o
s
(
β
)
{\displaystyle Cos\left(\alpha +\beta \right)+Cos(\alpha -\beta )=2cos(\alpha )cos(\beta )}
Faciendo
θ
=
α
+
β
{\displaystyle \theta =\alpha +\beta }
ϕ
=
α
−
β
{\displaystyle \phi =\alpha -\beta }
Entós, resolviendo el sistema, tiense que
α
=
θ
+
ϕ
2
{\displaystyle \alpha \ ={\frac {\theta +\phi }{2}}}
β
=
θ
−
ϕ
2
{\displaystyle \beta \ ={\frac {\theta -\phi }{2}}}
Reemplazando algámase
C
o
s
(
ϕ
)
+
C
o
s
(
θ
)
=
2
c
o
s
(
θ
+
ϕ
2
)
c
o
s
(
θ
−
ϕ
2
)
{\displaystyle Cos\left(\phi \right)+Cos(\theta )=2cos({\frac {\theta +\phi }{2}})cos({\frac {\theta -\phi }{2}})}
Análogamente amuésase que
C
o
s
(
ϕ
)
−
C
o
s
(
θ
)
=
−
2
S
i
n
(
ϕ
+
θ
2
)
S
i
n
(
θ
−
ϕ
2
)
{\displaystyle Cos(\phi )-Cos(\theta )=-2Sin({\frac {\phi +\theta }{2}})Sin({\frac {\theta -\phi }{2}})}
Según la definición de derivada:
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
{\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}
lo que ye
cos
′
(
x
)
=
lim
h
→
0
cos
(
x
+
h
)
−
cos
(
x
)
h
{\displaystyle \cos '(x)=\lim _{h\rightarrow 0}{\frac {\cos(x+h)-\cos(x)}{h}}}
Entós, usando les fórmules anteriormente consiñaes, tiense que
cos
′
(
x
)
=
lim
h
→
0
cos
(
x
)
⋅
cos
(
h
)
−
sin
(
x
)
⋅
sin
(
h
)
−
cos
(
x
)
h
{\displaystyle \cos '(x)=\lim _{h\rightarrow 0}{\frac {\cos(x)\cdot \cos(h)-\sin(x)\cdot \sin(h)-\cos(x)}{h}}}
Fautorizando
cos
′
(
x
)
=
lim
h
→
0
cos
(
x
)
⋅
(
(
cos
(
h
)
−
1
)
)
−
sin
(
x
)
⋅
sin
(
h
)
h
{\displaystyle \cos '(x)=\lim _{h\rightarrow 0}{\frac {\cos(x)\cdot ((\cos(h)-1))-\sin(x)\cdot \sin(h)}{h}}}
Separtando tenemos
cos
′
(
x
)
=
lim
h
→
0
cos
(
x
)
⋅
(
(
cos
(
h
)
−
1
)
)
h
−
lim
h
→
0
sin
(
x
)
⋅
sin
(
h
)
h
{\displaystyle \cos '(x)=\lim _{h\rightarrow 0}{\frac {\cos(x)\cdot ((\cos(h)-1))}{h}}-\lim _{h\rightarrow 0}{\frac {\sin(x)\cdot \sin(h)}{h}}}
Sabiendo que
lim
h
→
0
sin
(
h
)
h
=
1
{\displaystyle \lim _{h\rightarrow 0}{\frac {\sin(h)}{h}}=1}
cos
′
(
x
)
=
−
sin
(
x
)
{\displaystyle \cos '\left(x\right)=-\sin(x)}
Datos: Q1256164
Multimedia: Cosine function
