Considere a figura ao lado. Sejam três pontos
A
,
{\displaystyle A\;\!,}
B
{\displaystyle B\;\!}
C
{\displaystyle C\;\!}
A
(
cos
a
,
s
e
n
a
)
,
{\displaystyle A\left(\cos a,\mathrm {sen} \,a\right)\;\!,}
B
(
cos
(
a
+
b
)
,
s
e
n
(
a
+
b
)
)
{\displaystyle B\left(\cos \left(a+b\right),\mathrm {sen} \,\left(a+b\right)\right)\;\!}
C
(
cos
b
,
−
s
e
n
b
)
.
{\displaystyle C\left(\cos b,-\mathrm {sen} \,b\right)\;\!.}
P
B
^
{\displaystyle {\widehat {PB}}}
C
A
^
{\displaystyle {\widehat {CA}}}
P
B
¯
{\displaystyle {\overline {PB}}}
C
A
¯
{\displaystyle {\overline {CA}}}
d
P
B
2
=
2
−
2
⋅
cos
(
a
+
b
)
{\displaystyle d_{PB}^{2}=2-2\cdot \cos \left(a+b\right)\;\!}
d
C
A
2
=
2
−
2
⋅
cos
a
⋅
cos
b
+
2
⋅
s
e
n
a
⋅
s
e
n
b
{\displaystyle d_{CA}^{2}=2-2\cdot \cos a\cdot \cos b+2\cdot \mathrm {sen} \,a\cdot \mathrm {sen} \,b\;\!}
Ao igualarmos as duas expressões, temos a fórmula:
cos
(
a
+
b
)
=
cos
a
⋅
cos
b
−
s
e
n
a
⋅
s
e
n
b
{\displaystyle \cos \left(a+b\right)=\cos a\cdot \cos b-\mathrm {sen} \,a\cdot \mathrm {sen} \,b\;\!}
Sabemos que
s
e
n
x
=
cos
(
π
2
−
x
)
.
{\displaystyle \mathrm {sen} \,x=\cos \left({\frac {\pi }{2}}-x\right).}
x
=
a
+
b
,
{\displaystyle x=a+b\;\!,}
s
e
n
(
a
+
b
)
=
cos
[
π
2
−
(
a
+
b
)
]
=
cos
[
(
π
2
−
a
)
−
b
]
{\displaystyle \mathrm {sen} \,\left(a+b\right)=\cos \left[{\frac {\pi }{2}}-\left(a+b\right)\right]=\cos \left[\left({\frac {\pi }{2}}-a\right)-b\right]}
Utilizando a fórmula do cosseno da diferença de dois arcos nessa última expressão:
cos
(
π
2
−
a
)
⋅
cos
b
+
s
e
n
(
π
2
−
a
)
⋅
s
e
n
b
{\displaystyle \cos \left({\frac {\pi }{2}}-a\right)\cdot \cos b+\mathrm {sen} \,\left({\frac {\pi }{2}}-a\right)\cdot \mathrm {sen} \,b}
Substituindo
cos
(
π
2
−
a
)
=
s
e
n
a
{\displaystyle \cos \left({\frac {\pi }{2}}-a\right)=\mathrm {sen} \,a}
s
e
n
(
π
2
−
a
)
=
cos
a
{\displaystyle \mathrm {sen} \,\left({\frac {\pi }{2}}-a\right)=\cos a}
s
e
n
(
a
+
b
)
=
s
e
n
a
⋅
cos
b
+
s
e
n
b
⋅
cos
a
{\displaystyle \mathrm {sen} \,\left(a+b\right)=\mathrm {sen} \,a\cdot \cos b+\mathrm {sen} \,b\cdot \cos a\;\!}
Sabendo que
tan
x
=
s
e
n
x
cos
x
{\displaystyle \tan x={\frac {\mathrm {sen} \,x}{\cos x}}}
tan
(
a
+
b
)
:
{\displaystyle \tan \left(a+b\right)\;\!:}
tan
(
a
+
b
)
=
s
e
n
(
a
+
b
)
cos
(
a
+
b
)
=
s
e
n
a
⋅
cos
b
+
s
e
n
b
⋅
cos
a
cos
a
⋅
cos
b
−
s
e
n
a
⋅
s
e
n
b
{\displaystyle \tan \left(a+b\right)={\frac {\mathrm {sen} \,\left(a+b\right)}{\cos \left(a+b\right)}}={\frac {\mathrm {sen} \,a\cdot \cos b+\mathrm {sen} \,b\cdot \cos a}{\cos a\cdot \cos b-\mathrm {sen} \,a\cdot \mathrm {sen} \,b}}}
=
s
e
n
a
⋅
cos
b
+
s
e
n
b
⋅
cos
a
cos
a
⋅
cos
b
cos
a
⋅
cos
b
−
s
e
n
a
⋅
s
e
n
b
cos
a
⋅
cos
b
{\displaystyle ={\frac {\frac {\mathrm {sen} \,a\cdot \cos b+\mathrm {sen} \,b\cdot \cos a}{\cos a\cdot \cos b}}{\frac {\cos a\cdot \cos b-\mathrm {sen} \,a\cdot \mathrm {sen} \,b}{\cos a\cdot \cos b}}}}
Então:
tan
(
a
+
b
)
=
tan
a
+
tan
b
1
−
tan
a
⋅
tan
b
{\displaystyle \tan \left(a+b\right)={\frac {\tan a+\tan b}{1-\tan a\cdot \tan b}}}
Vale lembrar que essa fórmula só pode ser usada se
a
≠
π
2
+
k
π
,
b
≠
π
2
+
k
π
{\displaystyle a\neq {\frac {\pi }{2}}+k\pi ,b\neq {\frac {\pi }{2}}+k\pi }
a
+
b
≠
π
2
+
k
π
,
k
∈
Z
,
{\displaystyle a+b\neq {\frac {\pi }{2}}+k\pi ,k\in \mathbb {Z} ,}
tan
x
=
s
e
n
x
cos
x
{\displaystyle \tan x={\frac {\mathrm {sen} \,x}{\cos x}}}
x
≠
π
2
,
3
π
2
.
{\displaystyle x\neq {\frac {\pi }{2}},{\frac {3\pi }{2}}.}
Como
cot
x
=
cos
x
s
e
n
x
,
{\displaystyle \cot x={\frac {\cos x}{\mathrm {sen} \,x}},}
cot
(
a
+
b
)
:
{\displaystyle \cot \left(a+b\right)\;\!:}
cot
(
a
+
b
)
=
c
o
s
(
a
+
b
)
s
e
n
(
a
+
b
)
=
cos
a
⋅
cos
b
−
s
e
n
a
⋅
s
e
n
b
s
e
n
a
⋅
cos
b
+
s
e
n
b
⋅
cos
a
{\displaystyle \cot \left(a+b\right)={\frac {\ cos\left(a+b\right)}{\mathrm {sen} \,\left(a+b\right)}}={\frac {\cos a\cdot \cos b-\mathrm {sen} \,a\cdot \mathrm {sen} \,b}{\mathrm {sen} \,a\cdot \cos b+\mathrm {sen} \,b\cdot \cos a}}}
=
cos
a
⋅
cos
b
−
s
e
n
a
⋅
s
e
n
b
s
e
n
a
⋅
s
e
n
b
s
e
n
a
⋅
cos
b
+
s
e
n
b
⋅
cos
a
s
e
n
a
⋅
s
e
n
b
{\displaystyle ={\frac {\frac {\cos a\cdot \cos b-\mathrm {sen} \,a\cdot \mathrm {sen} \,b}{\mathrm {sen} \,a\cdot \mathrm {sen} \,b}}{\frac {\mathrm {sen} \,a\cdot \cos b+\mathrm {sen} \,b\cdot \cos a}{\mathrm {sen} \,a\cdot \mathrm {sen} \,b}}}}
Simplificando, temos:
cot
(
a
+
b
)
=
cot
a
⋅
cot
b
−
1
cot
a
+
cot
b
{\displaystyle \cot \left(a+b\right)={\frac {\cot a\cdot \cot b-1}{\cot a+\cot b}}}
Como
cot
x
=
cos
x
s
e
n
x
{\displaystyle \cot x={\frac {\cos x}{\mathrm {sen} \,x}}}
x
≠
0
,
π
,
2
π
,
{\displaystyle x\neq 0,\pi ,2\pi \;\!,}
a
≠
k
π
,
b
≠
k
π
{\displaystyle a\neq k\pi ,b\neq k\pi \;\!}
a
+
b
≠
k
π
,
k
∈
Z
.
{\displaystyle a+b\neq k\pi ,k\in \mathbb {Z} \;\!.}
(
1
)
{\displaystyle \left(1\right)\;\!}
cos
75
∘
:
{\displaystyle \cos 75^{\circ }\;\!:}
(
2
)
{\displaystyle \left(2\right)\;\!}
s
e
n
105
∘
:
{\displaystyle \mathrm {sen} \,105^{\circ }\;\!:}
(
3
)
{\displaystyle \left(3\right)\;\!}
tan
105
∘
:
{\displaystyle \tan 105^{\circ }\;\!:}
(
4
)
{\displaystyle \left(4\right)\;\!}
cot
75
∘
{\displaystyle \cot 75^{\circ }\;\!}
(
1
)
{\displaystyle \left(1\right)\;\!}
cos
75
∘
=
cos
(
30
∘
+
45
∘
)
=
cos
30
∘
⋅
cos
45
∘
−
s
e
n
30
∘
⋅
s
e
n
45
∘
{\displaystyle \cos 75^{\circ }=\cos \left(30^{\circ }+45^{\circ }\right)=\cos 30^{\circ }\cdot \cos 45^{\circ }-\mathrm {sen} \,30^{\circ }\cdot \mathrm {sen} \,45^{\circ }}
=
3
2
⋅
2
2
−
1
2
⋅
2
2
=
6
−
2
4
:
{\displaystyle ={\frac {\sqrt {3}}{2}}\cdot {\frac {\sqrt {2}}{2}}-{\frac {1}{2}}\cdot {\frac {\sqrt {2}}{2}}={\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}:}
(
2
)
{\displaystyle \left(2\right)\;\!}
s
e
n
105
∘
=
s
e
n
(
45
∘
+
60
∘
)
=
s
e
n
45
∘
⋅
cos
60
∘
+
s
e
n
60
∘
⋅
cos
45
∘
{\displaystyle \mathrm {sen} \,105^{\circ }=\mathrm {sen} \,\left(45^{\circ }+60^{\circ }\right)=\mathrm {sen} \,45^{\circ }\cdot \cos 60^{\circ }+\mathrm {sen} \,60^{\circ }\cdot \cos 45^{\circ }}
=
2
2
⋅
1
2
+
3
2
⋅
2
2
=
2
+
6
4
:
{\displaystyle ={\frac {\sqrt {2}}{2}}\cdot {\frac {1}{2}}+{\frac {\sqrt {3}}{2}}\cdot {\frac {\sqrt {2}}{2}}={\frac {{\sqrt {2}}+{\sqrt {6}}}{4}}:}
(
3
)
{\displaystyle \left(3\right)\;\!}
tan
105
∘
=
tan
(
45
∘
+
60
∘
)
=
tan
45
∘
+
tan
60
∘
1
−
tan
45
∘
⋅
tan
60
∘
{\displaystyle \tan 105^{\circ }=\tan \left(45^{\circ }+60^{\circ }\right)={\frac {\tan 45^{\circ }+\tan 60^{\circ }}{1-\tan 45^{\circ }\cdot \tan 60^{\circ }}}}
=
1
+
3
1
−
1
⋅
3
=
1
+
3
1
−
3
:
{\displaystyle ={\frac {1+{\sqrt {3}}}{1-1\cdot {\sqrt {3}}}}={\frac {1+{\sqrt {3}}}{1-{\sqrt {3}}}}:}
(
4
)
{\displaystyle \left(4\right)\;\!}
cot
75
∘
=
cot
(
30
∘
+
45
∘
)
=
cot
30
∘
⋅
cot
45
∘
−
1
cot
30
∘
+
cot
45
∘
{\displaystyle \cot 75^{\circ }=\cot \left(30^{\circ }+45^{\circ }\right)={\frac {\cot 30^{\circ }\cdot \cot 45^{\circ }-1}{\cot 30^{\circ }+\cot 45^{\circ }}}}
=
3
⋅
1
−
1
3
+
1
=
3
−
1
3
+
1
{\displaystyle ={\frac {{\sqrt {3}}\cdot 1-1}{{\sqrt {3}}+1}}={\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}}
Para calcular
cos
(
a
−
b
)
,
{\displaystyle \cos \left(a-b\right)\;\!,}
a
−
b
=
a
+
(
−
b
)
{\displaystyle a-b=a+\left(-b\right)\;\!}
cos
(
a
−
b
)
=
cos
[
a
+
(
−
b
)
]
{\displaystyle \cos \left(a-b\right)=\cos \left[a+\left(-b\right)\right]\;\!}
=
cos
a
⋅
cos
(
−
b
)
−
s
e
n
a
⋅
s
e
n
(
−
b
)
{\displaystyle =\cos a\cdot \cos \left(-b\right)-\mathrm {sen} \,a\cdot \mathrm {sen} \,\left(-b\right)\;\!}
=
cos
a
⋅
cos
b
−
s
e
n
a
⋅
(
−
s
e
n
b
)
{\displaystyle =\cos a\cdot \cos b-\mathrm {sen} \,a\cdot \left(-\mathrm {sen} \,b\right)\;\!}
Então:
cos
(
a
−
b
)
=
cos
a
⋅
cos
b
+
s
e
n
a
⋅
s
e
n
b
{\displaystyle \cos \left(a-b\right)=\cos a\cdot \cos b+\mathrm {sen} \,a\cdot \mathrm {sen} \,b\;\!}
Podemos fazer a mesma substituição da igualdade
a
−
b
=
a
+
(
−
b
)
{\displaystyle a-b=a+\left(-b\right)\;\!}
s
e
n
(
a
−
b
)
=
s
e
n
[
a
+
(
−
b
)
]
=
s
e
n
a
⋅
cos
(
−
b
)
+
s
e
n
(
−
b
)
⋅
cos
a
{\displaystyle \mathrm {sen} \,\left(a-b\right)=\mathrm {sen} \,\left[a+\left(-b\right)\right]=\mathrm {sen} \,a\cdot \cos \left(-b\right)+\mathrm {sen} \,\left(-b\right)\cdot \cos a\;\!}
Logo,
s
e
n
(
a
−
b
)
=
s
e
n
a
⋅
cos
b
−
s
e
n
b
⋅
cos
a
{\displaystyle \mathrm {sen} \,\left(a-b\right)=\mathrm {sen} \,a\cdot \cos b-\mathrm {sen} \,b\cdot \cos a\;\!}
Usando novamente a igualdade
a
−
b
=
a
+
(
−
b
)
{\displaystyle a-b=a+\left(-b\right)\;\!}
tan
(
a
−
b
)
=
tan
[
a
+
(
−
b
)
]
=
tan
a
+
tan
(
−
b
)
1
−
tan
a
⋅
tan
(
−
b
)
{\displaystyle \tan \left(a-b\right)=\tan \left[a+\left(-b\right)\right]={\frac {\tan a+\tan \left(-b\right)}{1-\tan a\cdot \tan \left(-b\right)}}}
Simplificando, temos:
tan
(
a
−
b
)
=
tan
a
−
tan
b
1
+
tan
a
⋅
tan
b
{\displaystyle \tan \left(a-b\right)={\frac {\tan a-\tan b}{1+\tan a\cdot \tan b}}}
Pelos motivos já citados anteriormente, esta fórmula só é válida se
a
≠
π
2
+
k
π
,
b
≠
π
2
+
k
π
{\displaystyle a\neq {\frac {\pi }{2}}+k\pi ,b\neq {\frac {\pi }{2}}+k\pi }
a
−
b
≠
π
2
+
k
π
,
k
∈
Z
.
{\displaystyle a-b\neq {\frac {\pi }{2}}+k\pi ,k\in \mathbb {Z} .}
Mais uma vez, usaremos a igualdade
a
−
b
=
a
+
(
−
b
)
{\displaystyle a-b=a+\left(-b\right)\;\!}
cot
(
a
−
b
)
=
cot
[
a
+
(
−
b
)
]
=
cot
a
⋅
cot
(
−
b
)
−
1
cot
a
+
cot
(
−
b
)
{\displaystyle \cot \left(a-b\right)=\cot \left[a+\left(-b\right)\right]={\frac {\cot a\cdot \cot \left(-b\right)-1}{\cot a+\cot \left(-b\right)}}}
Logo, obtemos a identidade:
cot
(
a
−
b
)
=
cot
a
⋅
cot
b
+
1
cot
b
−
cot
a
{\displaystyle \cot \left(a-b\right)={\frac {\cot a\cdot \cot b+1}{\cot b-\cot a}}}
Está fórmula só pode ser aplicada se
a
≠
k
π
,
b
≠
k
π
{\displaystyle a\neq k\pi ,b\neq k\pi \;\!}
a
−
b
≠
k
π
,
k
∈
Z
.
{\displaystyle a-b\neq k\pi ,k\in \mathbb {Z} \;\!.}
(
1
)
{\displaystyle \left(1\right)\;\!}
cos
15
∘
:
{\displaystyle \cos 15^{\circ }\;\!:}
(
2
)
{\displaystyle \left(2\right)\;\!}
s
e
n
15
∘
:
{\displaystyle \mathrm {sen} \,15^{\circ }\;\!:}
(
3
)
{\displaystyle \left(3\right)\;\!}
cot
15
∘
{\displaystyle \cot 15^{\circ }\;\!}
(
1
)
{\displaystyle \left(1\right)\;\!}
cos
15
∘
=
cos
15
∘
(
45
∘
−
30
∘
)
=
cos
45
∘
⋅
cos
30
∘
+
s
e
n
45
∘
⋅
s
e
n
30
∘
{\displaystyle \cos 15^{\circ }=\cos 15^{\circ }\left(45^{\circ }-30^{\circ }\right)=\cos 45^{\circ }\cdot \cos 30^{\circ }+\mathrm {sen} \,45^{\circ }\cdot \mathrm {sen} \,30^{\circ }}
=
2
2
⋅
3
2
+
2
2
⋅
1
2
=
6
+
2
4
{\displaystyle ={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {3}}{2}}+{\frac {\sqrt {2}}{2}}\cdot {\frac {1}{2}}={\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}}
(
2
)
{\displaystyle \left(2\right)\;\!}
s
e
n
15
∘
=
s
e
n
15
∘
(
45
∘
−
30
∘
)
=
s
e
n
45
∘
⋅
cos
30
∘
−
s
e
n
30
∘
⋅
cos
45
∘
{\displaystyle \mathrm {sen} \,15^{\circ }=\mathrm {sen} \,15^{\circ }\left(45^{\circ }-30^{\circ }\right)=\mathrm {sen} \,45^{\circ }\cdot \cos 30^{\circ }-\mathrm {sen} \,30^{\circ }\cdot \cos 45^{\circ }}
=
2
2
⋅
3
2
−
1
2
⋅
2
2
=
6
−
2
4
{\displaystyle ={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {3}}{2}}-{\frac {1}{2}}\cdot {\frac {\sqrt {2}}{2}}={\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}}
(
3
)
{\displaystyle \left(3\right)\;\!}
cot
15
∘
=
cot
15
∘
(
60
∘
−
45
∘
)
=
cot
60
∘
⋅
cot
45
∘
+
1
cot
45
∘
−
cot
60
∘
{\displaystyle \cot 15^{\circ }=\cot 15^{\circ }\left(60^{\circ }-45^{\circ }\right)={\frac {\cot 60^{\circ }\cdot \cot 45^{\circ }+1}{\cot 45^{\circ }-\cot 60^{\circ }}}}
=
3
3
⋅
1
+
1
1
−
3
3
=
3
+
3
3
−
3
{\displaystyle ={\frac {{\frac {\sqrt {3}}{3}}\cdot 1+1}{1-{\frac {\sqrt {3}}{3}}}}={\frac {3+{\sqrt {3}}}{3-{\sqrt {3}}}}}
Dados
tan
α
=
1
{\displaystyle \tan \alpha =1\;\!}
tan
β
=
1
2
,
{\displaystyle \tan \beta ={\frac {1}{2}}\;\!,}
tan
(
α
−
β
)
.
{\displaystyle \tan \left(\alpha -\beta \right)\;\!.}
tan
(
α
−
β
)
=
tan
α
−
tan
β
1
+
tan
α
⋅
tan
β
{\displaystyle \tan \left(\alpha -\beta \right)={\frac {\tan \alpha -\tan \beta }{1+\tan \alpha \cdot \tan \beta }}}
=
1
−
1
2
1
+
1
⋅
1
2
=
1
2
3
2
=
1
2
⋅
2
3
=
1
3
{\displaystyle ={\frac {1-{\frac {1}{2}}}{1+1\cdot {\frac {1}{2}}}}={\frac {\frac {1}{2}}{\frac {3}{2}}}={\frac {1}{2}}\cdot {\frac {2}{3}}={\frac {1}{3}}}
É possível deduzir fórmulas para calcular as funções trigonométricas de
2
a
,
3
a
,
.
.
.
,
{\displaystyle 2a,3a,...\;\!,}
2
a
=
a
+
a
,
3
a
=
2
a
+
a
,
.
.
.
,
{\displaystyle 2a=a+a,3a=2a+a,...\;\!,}
Usando a fórmula do cosseno da soma, temos:
cos
2
a
=
cos
(
a
+
a
)
=
cos
a
⋅
cos
a
−
s
e
n
a
⋅
s
e
n
a
=
cos
2
a
−
s
e
n
2
a
{\displaystyle \cos 2a=\cos \left(a+a\right)=\cos a\cdot \cos a-\mathrm {sen} \,a\cdot \mathrm {sen} \,a=\cos ^{2}a-\mathrm {sen} \,^{2}a\;\!}
Logo, utilizando a Identidade relacional básica , podemos obter duas fórmulas finais:
cos
2
a
=
2
cos
2
a
−
1
{\displaystyle \cos 2a=2\cos ^{2}a-1\;\!}
cos
2
a
=
1
−
2
s
e
n
2
a
{\displaystyle \cos 2a=1-2\mathrm {sen} \,^{2}a\;\!}
cos
3
a
=
cos
(
2
a
+
a
)
=
cos
2
a
⋅
cos
a
−
s
e
n
2
a
⋅
s
e
n
a
{\displaystyle \cos 3a=\cos \left(2a+a\right)=\cos 2a\cdot \cos a-\mathrm {sen} \,2a\cdot \mathrm {sen} \,a\;\!}
=
(
2
cos
2
a
−
1
)
⋅
cos
a
−
(
2
⋅
s
e
n
a
cos
a
)
⋅
s
e
n
a
{\displaystyle =\left(2\cos ^{2}a-1\right)\cdot \cos a-\left(2\cdot \mathrm {sen} \,a\cos a\right)\cdot \mathrm {sen} \,a\;\!}
=
(
2
cos
2
a
−
1
)
⋅
cos
a
−
2
s
e
n
2
a
⋅
cos
a
{\displaystyle =\left(2\cos ^{2}a-1\right)\cdot \cos a-2\mathrm {sen} \,^{2}a\cdot \cos a\;\!}
Utilizando a Identidade relacional básica e trabalhando algebricamente, temos:
cos
3
a
=
4
cos
3
a
−
3
cos
a
{\displaystyle \cos 3a=4\cos ^{3}a-3\cos a\;\!}
Expressões para
cos
4
a
,
cos
5
a
,
.
.
.
{\displaystyle \cos 4a,\cos 5a,...\;\!}
Ultilizando a fórmula do seno da soma:
s
e
n
2
a
=
s
e
n
(
a
+
a
)
=
s
e
n
a
⋅
cos
a
+
s
e
n
a
⋅
cos
a
{\displaystyle \mathrm {sen} \,2a=\mathrm {sen} \,\left(a+a\right)=\mathrm {sen} \,a\cdot \cos a+\mathrm {sen} \,a\cdot \cos a\;\!}
Então, temos:
s
e
n
2
a
=
2
⋅
s
e
n
a
cos
a
{\displaystyle \mathrm {sen} \,2a=2\cdot \mathrm {sen} \,a\cos a\;\!}
s
e
n
3
a
=
s
e
n
(
2
a
+
a
)
=
s
e
n
2
a
⋅
cos
a
+
cos
2
a
⋅
s
e
n
a
{\displaystyle \mathrm {sen} \,3a=\mathrm {sen} \,\left(2a+a\right)=\mathrm {sen} \,2a\cdot \cos a+\cos 2a\cdot \mathrm {sen} \,a\;\!}
=
(
2
⋅
s
e
n
a
⋅
cos
a
)
⋅
cos
a
+
s
e
n
a
(
1
−
2
⋅
s
e
n
2
a
)
{\displaystyle =\left(2\cdot \mathrm {sen} \,a\cdot \cos a\right)\cdot \cos a+\mathrm {sen} \,a\left(1-2\cdot \ sen^{2}a\right)\;\!}
Utilizando a Identidade relacional básica :
=
2
⋅
s
e
n
a
⋅
(
1
−
s
e
n
2
a
)
+
s
e
n
a
⋅
(
1
−
2
⋅
s
e
n
2
a
)
{\displaystyle =2\cdot \mathrm {sen} \,a\cdot \left(1-\mathrm {sen} \,^{2}a\right)+\mathrm {sen} \,a\cdot \left(1-2\cdot \mathrm {sen} \,^{2}a\right)\;\!}
Logo:
s
e
n
3
a
=
3
⋅
s
e
n
a
−
4
s
e
n
3
a
{\displaystyle \mathrm {sen} \,3a=3\cdot \mathrm {sen} \,a-4\mathrm {sen} \,^{3}a\;\!}
Expressões para
s
e
n
4
a
,
s
e
n
5
a
,
.
.
.
{\displaystyle \mathrm {sen} \,4a,\mathrm {sen} \,5a,...\;\!}
A partir da fórmula da tangente da soma:
tan
2
a
=
tan
(
a
+
a
)
=
tan
a
+
tan
a
1
−
tan
a
⋅
tan
a
{\displaystyle \tan 2a=\tan \left(a+a\right)={\frac {\tan a+\tan a}{1-\tan a\cdot \tan a}}\;\!}
Logo:
tan
2
a
=
2
⋅
tan
a
1
−
tan
2
a
{\displaystyle \tan 2a={\frac {2\cdot \tan a}{1-\tan ^{2}a}}}
tan
3
a
=
tan
(
2
a
+
a
)
=
tan
2
a
+
tan
a
1
−
tan
2
a
⋅
tan
a
{\displaystyle \tan 3a=\tan \left(2a+a\right)={\frac {\tan 2a+\tan a}{1-\tan 2a\cdot \tan a}}\;\!}
Ao subtituimos a fórmula anterior para
tan
2
a
{\displaystyle \tan 2a\;\!}
tan
3
a
=
3
⋅
tan
a
−
tan
3
a
1
−
3
⋅
tan
2
a
{\displaystyle \tan 3a={\frac {3\cdot \tan a-\tan ^{3}a}{1-3\cdot \tan ^{2}a}}\;\!}
Expressões para
tan
4
a
,
tan
5
a
,
.
.
.
{\displaystyle \tan 4a,\tan 5a,...\;\!}
Se
cot
x
=
5
3
{\displaystyle \cot x={\frac {5}{3}}\;\!}
0
<
x
<
π
2
,
{\displaystyle 0<x<{\frac {\pi }{2}}\;\!,}
cos
2
x
.
{\displaystyle \cos 2x\;\!.}
Precisamos encontrar
s
e
n
x
{\displaystyle \mathrm {sen} \,x\;\!}
csc
2
x
=
1
+
cot
2
x
,
{\displaystyle \csc ^{2}x=1+\cot ^{2}x\;\!,}
csc
x
=
1
s
e
n
x
.
{\displaystyle \csc x={\frac {1}{\mathrm {sen} \,x}}\;\!.}
0
<
x
<
π
2
,
{\displaystyle 0<x<{\frac {\pi }{2}}\;\!,}
csc
x
=
1
+
cot
2
x
=
1
+
25
9
=
34
9
=
34
3
{\displaystyle \csc x={\sqrt {1+\cot ^{2}x}}={\sqrt {1+{\frac {25}{9}}}}={\sqrt {\frac {34}{9}}}={\frac {\sqrt {34}}{3}}}
De onde vem
s
e
n
x
=
3
34
.
{\displaystyle \mathrm {sen} \,x={\frac {3}{\sqrt {34}}}.}
Podemos finalmente calcular:
cos
2
x
=
1
−
2
⋅
s
i
n
2
x
=
1
−
2
⋅
9
34
=
1
−
18
34
=
8
17
.
{\displaystyle \cos 2x=1-2\cdot \ sin^{2}x=1-2\cdot {\frac {9}{34}}=1-{\frac {18}{34}}={\frac {8}{17}}.}
Vamos utilizar as duas fórmulas que encontramos para
cos
2
a
{\displaystyle \cos 2a\;\!}
x
{\displaystyle x\;\!}
cos
x
2
,
s
e
n
x
2
{\displaystyle \cos {\frac {x}{2}},\mathrm {sen} \,{\frac {x}{2}}\;\!}
tan
x
2
.
{\displaystyle \tan {\frac {x}{2}}\;\!.}
2
a
=
x
.
{\displaystyle 2a=x\;\!.}
A partir de
cos
2
a
=
2
cos
2
a
−
1
:
{\displaystyle \cos 2a=2\cos ^{2}a-1\;\!:}
cos
x
=
2
⋅
cos
2
x
2
−
1
{\displaystyle \cos x=2\cdot \cos ^{2}{\frac {x}{2}}-1}
⇒
cos
x
2
=
±
1
+
cos
x
2
{\displaystyle \Rightarrow \cos {\frac {x}{2}}=\pm {\sqrt {\frac {1+\cos x}{2}}}}
A partir de
cos
2
a
=
1
−
2
s
e
n
2
a
,
{\displaystyle \cos 2a=1-2\mathrm {sen} \,^{2}a\;\!,}
cos
x
=
1
−
2
⋅
s
e
n
2
x
2
{\displaystyle \cos x=1-2\cdot \mathrm {sen} \,^{2}{\frac {x}{2}}\;\!}
⇒
s
e
n
x
2
=
±
1
−
cos
x
2
{\displaystyle \Rightarrow \mathrm {sen} \,{\frac {x}{2}}=\pm {\sqrt {\frac {1-\cos x}{2}}}}
Finalmente, sabendo que
tan
x
=
s
e
n
x
cos
x
,
{\displaystyle \tan x={\frac {\mathrm {sen} \,x}{\cos x}},}
tan
x
2
=
s
e
n
x
2
cos
x
2
{\displaystyle \tan {\frac {x}{2}}={\frac {\mathrm {sen} \,{\frac {x}{2}}}{\cos {\frac {x}{2}}}}}
⇒
tan
x
2
=
±
1
−
cos
x
1
+
cos
x
{\displaystyle \Rightarrow \tan {\frac {x}{2}}=\pm {\sqrt {\frac {1-\cos x}{1+\cos x}}}}
Caso nos seja dado o
s
e
n
x
,
{\displaystyle \mathrm {sen} \,x\;\!,}
cos
x
=
±
1
−
s
e
n
2
x
,
{\displaystyle \cos x=\pm {\sqrt {1-\mathrm {sen} \,^{2}x}},}
cos
x
{\displaystyle \cos x\;\!}
Precisamos agora encontrar fórmulas que permitam calcular
s
e
n
x
,
{\displaystyle \mathrm {sen} \,x\;\!,}
cos
x
{\displaystyle \cos x\;\!}
tan
x
,
{\displaystyle \tan x\;\!,}
tan
x
2
.
{\displaystyle \tan {\frac {x}{2}}.}
s
e
n
2
a
=
2
⋅
s
e
n
a
cos
a
=
2
⋅
s
e
n
a
⋅
cos
2
a
cos
a
=
2
⋅
s
e
n
a
cos
a
⋅
1
sec
2
a
=
2
⋅
tan
a
1
+
tan
2
a
{\displaystyle \mathrm {sen} \,2a=2\cdot \mathrm {sen} \,a\cos a=2\cdot \mathrm {sen} \,a\cdot {\frac {\cos ^{2}a}{\cos a}}=2\cdot {\frac {\mathrm {sen} \,a}{\cos a}}\cdot {\frac {1}{\sec ^{2}a}}={\frac {2\cdot \tan a}{1+\tan ^{2}a}}\;\!}
tan
2
a
=
2
⋅
tan
a
1
−
tan
2
a
{\displaystyle \tan 2a={\frac {2\cdot \tan a}{1-\tan ^{2}a}}}
e consideraremos
2
a
=
x
,
{\displaystyle 2a=x\;\!,}
s
e
n
x
=
2
⋅
tan
x
2
1
+
tan
2
x
2
{\displaystyle \mathrm {sen} \,x={\frac {2\cdot \tan {\frac {x}{2}}}{1+\tan ^{2}{\frac {x}{2}}}}}
tan
x
=
2
⋅
tan
x
2
1
−
tan
2
x
2
{\displaystyle \tan x={\frac {2\cdot \tan {\frac {x}{2}}}{1-\tan ^{2}{\frac {x}{2}}}}}
cos
x
=
1
−
tan
2
x
2
1
+
tan
2
x
2
{\displaystyle \cos x={\frac {1-\tan ^{2}{\frac {x}{2}}}{1+\tan ^{2}{\frac {x}{2}}}}}
Se
s
e
n
x
=
4
5
,
{\displaystyle \mathrm {sen} \,x={\frac {4}{5}},}
0
<
x
<
π
2
,
{\displaystyle 0<x<{\frac {\pi }{2}},}
x
2
.
{\displaystyle {\frac {x}{2}}.}
cos
x
=
1
−
s
e
n
2
x
=
1
−
16
25
=
9
25
=
3
5
{\displaystyle \cos x={\sqrt {1-\mathrm {sen} \,^{2}x}}={\sqrt {1-{\frac {16}{25}}}}={\sqrt {\frac {9}{25}}}={\frac {3}{5}}}
Logo, temos:
s
e
n
x
2
=
1
−
cos
x
2
=
1
−
3
5
2
=
1
5
:
{\displaystyle \mathrm {sen} \,{\frac {x}{2}}={\sqrt {\frac {1-\cos x}{2}}}={\sqrt {\frac {1-{\frac {3}{5}}}{2}}}={\sqrt {\frac {1}{5}}}:}
cos
x
2
=
1
+
cos
x
2
=
1
+
3
5
2
=
4
5
=
2
5
5
:
{\displaystyle \cos {\frac {x}{2}}={\sqrt {\frac {1+\cos x}{2}}}={\sqrt {\frac {1+{\frac {3}{5}}}{2}}}={\sqrt {\frac {4}{5}}}={\frac {2{\sqrt {5}}}{5}}:}
tan
x
2
=
1
−
cos
x
1
+
cos
x
=
1
−
3
5
1
+
3
5
=
1
4
=
1
2
{\displaystyle \tan {\frac {x}{2}}={\sqrt {\frac {1-\cos x}{1+\cos x}}}={\sqrt {\frac {1-{\frac {3}{5}}}{1+{\frac {3}{5}}}}}={\sqrt {\frac {1}{4}}}={\frac {1}{2}}}
Se
tan
x
2
=
1
4
,
{\displaystyle \tan {\frac {x}{2}}={\frac {1}{4}},}
s
e
n
x
.
{\displaystyle \mathrm {sen} \,x\;\!.}
s
e
n
x
=
2
⋅
tan
x
2
1
+
tan
2
x
2
=
2
⋅
1
4
1
+
1
16
=
1
2
17
16
=
8
17
{\displaystyle \mathrm {sen} \,x={\frac {2\cdot \tan {\frac {x}{2}}}{1+\tan ^{2}{\frac {x}{2}}}}={\frac {2\cdot {\frac {1}{4}}}{1+{\frac {1}{16}}}}={\frac {\frac {1}{2}}{\frac {17}{16}}}={\frac {8}{17}}}
![{\displaystyle \mathrm {sen} \,\left(a+b\right)=\cos \left[{\frac {\pi }{2}}-\left(a+b\right)\right]=\cos \left[\left({\frac {\pi }{2}}-a\right)-b\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0816e1dd2a297cfb5d28dab327714734a890da1)
![{\displaystyle \cos \left(a-b\right)=\cos \left[a+\left(-b\right)\right]\;\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/025272648f7390cf4e3d1adad5add5a6c558d1aa)
![{\displaystyle \mathrm {sen} \,\left(a-b\right)=\mathrm {sen} \,\left[a+\left(-b\right)\right]=\mathrm {sen} \,a\cdot \cos \left(-b\right)+\mathrm {sen} \,\left(-b\right)\cdot \cos a\;\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee9609d266adb1eb06767bb9f997d1ba70a46b4b)
![{\displaystyle \tan \left(a-b\right)=\tan \left[a+\left(-b\right)\right]={\frac {\tan a+\tan \left(-b\right)}{1-\tan a\cdot \tan \left(-b\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36b2c317c03ad279d7bc689e53c3c5d511474b98)
![{\displaystyle \cot \left(a-b\right)=\cot \left[a+\left(-b\right)\right]={\frac {\cot a\cdot \cot \left(-b\right)-1}{\cot a+\cot \left(-b\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/563cf30ca50270ab92ad566f1fa339e3513d3c38)
