Anónimo
Anónimo preguntado en Ciencias y matemáticasMatemáticas · hace 1 mes

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Calificación
• hace 1 mes
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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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