Ensino Médio ⇒ Trigonometria Tópico resolvido

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Código: Selecionar tudo

[tex3]arc \cos\frac{3}{5}+arc \cos\frac{12}{13}=arc \cos\frac{16}{65}[/tex3]

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[alevini98]

[tex3]\arccos{\frac{3}{5}}+\arccos{\frac{12}{13}}=\arccos{\frac{16}{65}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\arccos{\frac{3}{5}}+\arccos{\frac{12}{13}}=\arccos{\frac{16}{65}}[/tex3]

[tex3]\arccos{\frac{3}{5}}=a\\\\\boxed{\cos{a}=\frac{3}{5}}\\\\\sen^2{a}+\cos^2{a}=1\\\\\sen^2{a}+\frac{9}{25}=1\\\\\boxed{\sen{a}=\frac{4}{5}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\arccos{\frac{3}{5}}=a\\\\\boxed{\cos{a}=\frac{3}{5}}\\\\\sen^2{a}+\cos^2{a}=1\\\\\sen^2{a}+\frac{9}{25}=1\\\\\boxed{\sen{a}=\frac{4}{5}}[/tex3]

[tex3]\arccos{\frac{12}{13}}=b\\\\\boxed{\cos{b}=\frac{12}{13}}\\\\\sen^2{b}+\cos^2{b}=1\\\\\sen^2{b}+\frac{144}{169}=1\\\\\boxed{\sen{b}=\frac{5}{13}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\arccos{\frac{12}{13}}=b\\\\\boxed{\cos{b}=\frac{12}{13}}\\\\\sen^2{b}+\cos^2{b}=1\\\\\sen^2{b}+\frac{144}{169}=1\\\\\boxed{\sen{b}=\frac{5}{13}}[/tex3]

[tex3]\arccos{\frac{3}{5}}+\arccos{\frac{12}{13}}=\arccos{\frac{16}{65}}\\\\a+b=\arccos{\frac{16}{5}}\\\\\cos{(a+b)}=\frac{16}{65}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\arccos{\frac{3}{5}}+\arccos{\frac{12}{13}}=\arccos{\frac{16}{65}}\\\\a+b=\arccos{\frac{16}{5}}\\\\\cos{(a+b)}=\frac{16}{65}[/tex3]

Rearranjando o primeiro membro da equação:

[tex3]\cos{a}\cdot\cos{b}-\sen{a}\cdot\sen{b}=\frac{16}{65}\\\\\frac{3}{5}\cdot\frac{12}{13}-\frac{4}{5}\cdot\frac{5}{13}=\frac{16}{65}\\\\\frac{36}{65}-\frac{20}{65}=\frac{16}{65}\\\\\boxed{\frac{16}{65}=\frac{16}{65}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\cos{a}\cdot\cos{b}-\sen{a}\cdot\sen{b}=\frac{16}{65}\\\\\frac{3}{5}\cdot\frac{12}{13}-\frac{4}{5}\cdot\frac{5}{13}=\frac{16}{65}\\\\\frac{36}{65}-\frac{20}{65}=\frac{16}{65}\\\\\boxed{\frac{16}{65}=\frac{16}{65}}[/tex3]