Question

Demostrar las siguientes igualdades:
sen(a+b) · cos(a-b)= sen a · cos a + sen b · cos b

Answer 1

sen(a + b) cos(a - b) = (sen a cos b + sen b cos a)(cos a cos b +

sen a sen b)

sen(a + b) cos(a - b) = sen a cos a cos²b + sen²a sen b cos b +

sen b cos b cos²a + sen²b sen a cos a

sen(a + b) cos(a - b) = sen a cos a (sen²b + cos²b) +

sen b cos b (sen²a + cos²a)

sen(a + b) cos(a - b) = sen a cos a (1) + sen b cos b (1)

sen(a + b) cos(a - b) = sen a cos a + sen b cos b

Answer 2

Hola,

Demostrar la siguiente Identidad Trigonométrica:

$\sin(a + b) \times \cos(a - b) = \sin(a) \times \cos(a) + \sin(b) \times \cos(b) \\ ( \sin(a) \times \cos(b) + \cos(a) \times \sin(b) )( \cos(a) \times \cos(b) + \sin(a) \times \sin(b) ) = \sin(a) \times \cos(a) + \sin(b) \times \cos(b) \\ \sin(a) \times \cos(a) \times \cos {}^{2} (b) + \sin {}^{2} (a) \times \sin(b) \times \cos(b) + \sin(b) \times \cos {}^{2} (a) \times \cos(b) + \sin(a) \times \sin {}^{2} (b) \times \cos(a) =\sin(a) \times \cos(a) + \sin(b) \times \cos(b) \\ \sin(a) \times \cos(a) ( \cos {}^{2} (b) + \sin {}^{2} (b)) + \sin(b) \times \cos(b) ( \cos {}^{2} (a) + \sin {}^{2} (a) ) = \sin(a) \times \cos(a) + \sin(b) \times \cos(b) \\ \sin(a) \times \cos(a)(1) + \sin(b) \times \cos(b) (1) = \sin(a) \times \cos(a) + \sin(b) \times \cos(b) \\ \sin(a) \times \cos(a) + \sin(b) \times \cos(b) = \sin(a) \times \cos(a) + \sin(b) \times \cos(b)$
Espero que te sirva, Saludos.