Nesta página aprenderemos a efetuar operações trigonométricas que envolvam a adição, subtração ou multiplicação de números reais.
Adição de arcos
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Cosseno da soma
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Considere a figura ao lado. Sejam três pontos A , {\displaystyle A\;\!,} B {\displaystyle B\;\!} e C {\displaystyle C\;\!} pertencentes à circunferência , cujas coordenadas são 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)\;\!} e C ( cos b , − s e n b ) . {\displaystyle C\left(\cos b,-\mathrm {sen} \,b\right)\;\!.}
Os arcos P B ^ {\displaystyle {\widehat {PB}}} e C A ^ {\displaystyle {\widehat {CA}}}
têm medidas iguais, logo as cordas P B ¯ {\displaystyle {\overline {PB}}} e C A ¯ {\displaystyle {\overline {CA}}} também têm a mesma medida. Após aplicarmos a fórmula da distância entre dois pontos da Geometria analítica, temos:
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\;\!}
Seno da soma
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Sabemos que s e n x = cos ( π 2 − x ) . {\displaystyle \mathrm {sen} \,x=\cos \left({\frac {\pi }{2}}-x\right).} A partir disto e sendo x = a + b , {\displaystyle x=a+b\;\!,} obtemos:
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} e s e n ( π 2 − a ) = cos a {\displaystyle \mathrm {sen} \,\left({\frac {\pi }{2}}-a\right)=\cos a} nesta expressão, então:
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\;\!}
Tangente da soma
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Sabendo que tan x = s e n x cos x {\displaystyle \tan x={\frac {\mathrm {sen} \,x}{\cos x}}} e utilizando as fórmulas anteriores para soma de senos e cossenos, podemos facilmente conseguir uma expressão para 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 } e a + b ≠ π 2 + k π , k ∈ Z , {\displaystyle a+b\neq {\frac {\pi }{2}}+k\pi ,k\in \mathbb {Z} ,} porque a relação tan x = s e n x cos x {\displaystyle \tan x={\frac {\mathrm {sen} \,x}{\cos x}}} só é válida se e somente se x ≠ π 2 , 3 π 2 . {\displaystyle x\neq {\frac {\pi }{2}},{\frac {3\pi }{2}}.}
Cotangente da soma
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Como cot x = cos x s e n x , {\displaystyle \cot x={\frac {\cos x}{\mathrm {sen} \,x}},} podemos obter, de maneira semelhante à formula da tangente da soma, uma expressão para 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}}} é válida se e somente se x ≠ 0 , π , 2 π , {\displaystyle x\neq 0,\pi ,2\pi \;\!,} a identidade que demonstramos acima só pode ser usada se a ≠ k π , b ≠ k π {\displaystyle a\neq k\pi ,b\neq k\pi \;\!} e 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}}}
Subtração de arcos
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Cosseno da diferença
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Para calcular cos ( a − b ) , {\displaystyle \cos \left(a-b\right)\;\!,} fazemos uso da igualdade a − b = a + ( − b ) {\displaystyle a-b=a+\left(-b\right)\;\!} na fórmula do cosseno da soma, conforme a seguir:
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\;\!}
Seno da diferença
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Podemos fazer a mesma substituição da igualdade a − b = a + ( − b ) {\displaystyle a-b=a+\left(-b\right)\;\!} para encontrar as outras relações de diferença de arcos. Para o seno, usaremos a fórmula do seno da soma e a igualdade citada acima, conforme a seguir:
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\;\!}
Tangente da diferença
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Usando novamente a igualdade a − b = a + ( − b ) {\displaystyle a-b=a+\left(-b\right)\;\!} e, desta vez, a fórmula da tangente da soma:
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 } e a − b ≠ π 2 + k π , k ∈ Z . {\displaystyle a-b\neq {\frac {\pi }{2}}+k\pi ,k\in \mathbb {Z} .}
Cotangente da diferença
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Mais uma vez, usaremos a igualdade a − b = a + ( − b ) {\displaystyle a-b=a+\left(-b\right)\;\!} e, desta vez, a fórmula da cotangente da soma:
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 \;\!} e 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\;\!} e tan β = 1 2 , {\displaystyle \tan \beta ={\frac {1}{2}}\;\!,} calcule 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}}}
Multiplicação de arcos
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É possível deduzir fórmulas para calcular as funções trigonométricas de 2 a , 3 a , . . . , {\displaystyle 2a,3a,...\;\!,} utilizando as fórmulas obtidas para a soma de arcos e fazendo 2 a = a + a , 3 a = 2 a + a , . . . , {\displaystyle 2a=a+a,3a=2a+a,...\;\!,} conforme será mostrado adiante.
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\;\!}
ou
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,...\;\!} são obtidas por processos semelhantes.
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,...\;\!} são obtidas por processos semelhantes.
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\;\!} e simplificarmos, obtemos como fórmula final:
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,...\;\!} são obtidas por processos semelhantes.
Se cot x = 5 3 {\displaystyle \cot x={\frac {5}{3}}\;\!} e 0 < x < π 2 , {\displaystyle 0<x<{\frac {\pi }{2}}\;\!,} calcule cos 2 x . {\displaystyle \cos 2x\;\!.} Precisamos encontrar s e n x {\displaystyle \mathrm {sen} \,x\;\!} para aplicarmos a fórmula. Para tanto, utilizaremos a identidade csc 2 x = 1 + cot 2 x , {\displaystyle \csc ^{2}x=1+\cot ^{2}x\;\!,} que relaciona as funções cotangente e cossecante. A partir da cossecante obtida, podemos encontrar o valor do seno, uma vez que csc x = 1 s e n x . {\displaystyle \csc x={\frac {1}{\mathrm {sen} \,x}}\;\!.}
Como 0 < x < π 2 , {\displaystyle 0<x<{\frac {\pi }{2}}\;\!,} o valor da cossecante é positivo.
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}}.}
Bissecção de arcos
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Vamos utilizar as duas fórmulas que encontramos para cos 2 a {\displaystyle \cos 2a\;\!} a fim de que, dado o cosseno de uma arco x {\displaystyle x\;\!} qualquer, possamos obter cos x 2 , s e n x 2 {\displaystyle \cos {\frac {x}{2}},\mathrm {sen} \,{\frac {x}{2}}\;\!} ou tan x 2 . {\displaystyle \tan {\frac {x}{2}}\;\!.} Para isto, consideraremos 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\;\!,} temos:
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}},} temos:
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\;\!,} sabendo que cos x = ± 1 − s e n 2 x , {\displaystyle \cos x=\pm {\sqrt {1-\mathrm {sen} \,^{2}x}},} calculamos cos x {\displaystyle \cos x\;\!} e usamos as fórmulas dadas logo acima para o cosseno.
Precisamos agora encontrar fórmulas que permitam calcular s e n x , {\displaystyle \mathrm {sen} \,x\;\!,} cos x {\displaystyle \cos x\;\!} e tan x , {\displaystyle \tan x\;\!,} conhecida a tan x 2 . {\displaystyle \tan {\frac {x}{2}}.} Para tanto, tomaremos as fórmulas de multiplicação
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\;\!,} de modo que:
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}},} com 0 < x < π 2 , {\displaystyle 0<x<{\frac {\pi }{2}},} calcule as funções circulares de 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}},} determine s e n x . {\displaystyle \mathrm {sen} \,x\;\!.}
Podemos aplicar diretamente a fórmula, de modo que:
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}}}