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Anónimo preguntado en Ciencias y matemáticasMatemáticas · hace 2 meses

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Calificación
  • hace 2 meses
    Respuesta preferida

    Do you know these identities?

    cos(a + b) = cos(a).cos(b) - sin(a).sin(b) ← memorize this as (1)

    cos(a - b) = cos(a).cos(b) + sin(a).sin(b) ← memorize this as (2)

    sin(a + b) = sin(a).cos(b) + cos(a).sin(b) ← memorize this as (3)

    sin(a - b) = sin(a).cos(b) - cos(a).sin(b) ← memorize this as (4)

    = [cos(a + b) - cos(a - b)] / [sin(a + b) + sin(a - b)] → recall (1)

    = [{cos(a).cos(b) - sin(a).sin(b)} - cos(a - b)] / [sin(a + b) + sin(a - b)] → recall (2)

    = [{cos(a).cos(b) - sin(a).sin(b)} - {cos(a).cos(b) + sin(a).sin(b)}] / [sin(a + b) + sin(a - b)] → you simplify

    = [cos(a).cos(b) - sin(a).sin(b) - cos(a).cos(b) - sin(a).sin(b)] / [sin(a + b) + sin(a - b)] → you simplify

    = [- 2.sin(a).sin(b)] / [sin(a + b) + sin(a - b)] → recall (3)

    = [- 2.sin(a).sin(b)] / [{sin(a).cos(b) + cos(a).sin(b)} + sin(a - b)] → recall (4)

    = [- 2.sin(a).sin(b)] / [{sin(a).cos(b) + cos(a).sin(b)} + {sin(a).cos(b) - cos(a).sin(b)}] → you simplify

    = [- 2.sin(a).sin(b)] / [sin(a).cos(b) + cos(a).sin(b) + sin(a).cos(b) - cos(a).sin(b)] → you simplify

    = [- 2.sin(a).sin(b)] / [2.sin(a).cos(b)] → you simplify

    = - [sin(a).sin(b)] / [sin(a).cos(b)] → you simplify

    = - [sin(b)] / [cos(b)]

    = - tan(b)

    ≠ tan(a + b)

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