Anónimo
Anónimo preguntado en Ciencias y matemáticasMatemáticas · hace 2 meses

¿Identidades Trigonometricas?

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2 respuestas

Calificación
  • hace 2 meses

    = sin(a).cot(a) → you know that: cot(a) = cos(a)/sin(a)

    = sin(a).[cos(a)/sin(a)] → you can simplify by sin(a)

    = cos(a)

    = csc(a).tan(a) → you know that: csc(a) = 1/sin(a)

    = [1/sin(a)].tan(a) → you know that: tan(a) = sin(a)/cos(a)

    = [1/sin(a)].[sin(a)/cos(a)]

    = sin(a)/[cos(a).sin(a)] → you can simplify by sin(a)

    = 1/cos(a) → you know that: 1/cos(a) = sec(a)

    = sec(a)

    = csc(a).[1 - cos²(a)] → recall: csc(a) = 1/sin(a)

    = [1/sin(a)].[1 - cos²(a)] → you know that: cos²(a) + sin²(a) = 1

    = [1/sin(a)].[cos²(a) + sin²(a) - cos²(a)]

    = [1/sin(a)].[sin²(a)]

    = sin²(a)/sin(a)

    = sin(a)

    = sin(a).[1 + cot²(a)] → recall: cot(a) = cos(a)/sin(a)

    = sin(a).[1 + {cos(a)/sin(a)}²]

    = sin(a).[1 + {cos²(a)/sin²(a)}]

    = sin(a).[{sin²(a)/sin²(a)} + {cos²(a)/sin²(a)}]

    = sin(a).[{sin²(a) + cos²(a)}/sin²(a)] → recall: cos²(a) + sin²(a) = 1

    = sin(a).[1/sin²(a)]

    = sin(a)/sin²(a)

    = 1/sin(a) → recall: 1/sin(a) = csc(a)

    = csc(a)

    = [tan²(a) + 1].cos²(a) → recall: tan(a) = sin(a)/cos(a)

    = [{sin(a)/cos(a)}² + 1].cos²(a)

    = [{sin²(a)/cos²(a)} + 1].cos²(a)

    = [{sin²(a)/cos²(a)} + {cos²(a)/cos²(a)}].cos²(a)

    = [{sin²(a) + cos²(a)}/cos²(a)].cos²(a) → recall: cos²(a) + sin²(a) = 1

    = [1/cos²(a)].cos²(a)

    = cos²(a)/cos²(a)

    = 1

    = sec²(a).[1 - sin²(a)] → recall: sec(a) = 1/cos(a)

    = [1/cos(a)]².[1 - sin²(a)]

    = [1/cos²(a)].[1 - sin²(a)]

    = [1 - sin²(a)]/cos²(a) → recall: cos²(a) + sin²(a) = 1

    = [cos²(a) + sin²(a) - sin²(a)]/cos²(a)

    = [cos²(a)]/cos²(a)

    = 1

    = sec(a).[1 - sin²(a)] → recall: sec(a) = 1/cos(a)

    = [1/cos(a)].[1 - sin²(a)]

    = [1 - sin²(a)]/cos(a) → recall: cos²(a) + sin²(a) = 1

    = [cos²(a) + sin²(a) - sin²(a)]/cos(a)

    = [cos²(a)]/cos(a)

    = cos(a)

    = [sin²(a) + cos²(a)]/tan²(a) → recall: cos²(a) + sin²(a) = 1

    = 1/tan²(a) → recall: tan(a) = sin(a)/cos(a)

    = 1/[sin(a)/cos(a)]²

    = 1/[sin²(a)/cos²(a)]

    = cos²(a)/sin²(a)

    = [cos(a)/sin(a)]² → recall: cos(a)/sin(a) = cot(a)

    = cot²(a)

    = [sin²(a) + cos²(a)]/[tan²(a) + 1] → recall: cos²(a) + sin²(a) = 1

    = 1/[tan²(a) + 1] → recall: tan(a) = sin(a)/cos(a)

    = 1/[{sin(a)/cos(a)}² + 1]

    = 1/[{sin²(a)/cos²(a)} + 1]

    = 1/[{sin²(a)/cos²(a)} + {cos²(a)/cos²(a)}]

    = 1/[{sin²(a) + cos²(a)}/cos²(a)] → recall: cos²(a) + sin²(a) = 1

    = 1/[1/cos²(a)]

    = cos²(a)

  • hace 2 meses

    senA cot A=cos A

    sen A cos A/ sen A=cos A

    cos A=cos A

    cscA tanA= secA

    1/senA senA/cosA=secA

    1/cosA=secA

    secA=secA

    cscA(1-cos²A)=senA

    1/senA sen²A= senA

    senA=senA

    senA(1+cot²A)=cscA

    senA(1+cos²A/sen²A)=cscA

    senA((sen²A+cos²A)/sen²A)=cscA

    1/senA=cscA

    cscA=cscA

    (tan²A+1)cos²A=1

    (sen²A/cos²A +1)cos²A=1

    ((sen²A+cos²A)/cos²A)cos²A=1

    (1/cos²A)cos²A=1

    1=1

    sec²A(1-sen²A)=1

    (1/cos²A)cos²A=1

    1=1

    secA(1-sen²A)=cosA

    (1/cosA)cos²A=cosA

    cosA=cosA

    (sen²A+cos²A)/tan²A=cot²A

    1/(sen²A/cos²A)=cot²A

    cos²A/sen²A=cot²A

    cot²A=cot²A

    (sen²A+cos²A)/tan²A+1=cos²A

    1/((sen²A/cos²A)+1=cos²A

    1/((sen²A+cos²A)/(cos²A))=cos²A

    1/1/cos²A=cos²A

    cos²A=cos²A

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