Respuesta :
Ok, f(z) is everywhere continuous and differentiable, hence MVT is applicable.
MVT says for every x,y,
f'(z) = cos(z) = (sin(x)-sin(y)) / (x-y)
But you know -1 ≤ cos(z) ≤ 1. i.e. 0 ≤ |cos(z)| ≤ 1
=> -1 ≤ (sin(x)-sin(y)) / (x-y) ≤ 1
Multiply across by (x-y), and note the inequality remains valid for both the +ve and -ve cases due to the symmetry (i.e. -1 swaps with 1 and v.v.):
=> -(x-y) ≤ (sin(x)-sin(y)) ≤ (x-y)
Take || (abs) of both sides.
=> 0 ≤ | sin(x)-sin(y) | ≤ | x-y |
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
MVT says for every x,y,
f'(z) = cos(z) = (sin(x)-sin(y)) / (x-y)
But you know -1 ≤ cos(z) ≤ 1. i.e. 0 ≤ |cos(z)| ≤ 1
=> -1 ≤ (sin(x)-sin(y)) / (x-y) ≤ 1
Multiply across by (x-y), and note the inequality remains valid for both the +ve and -ve cases due to the symmetry (i.e. -1 swaps with 1 and v.v.):
=> -(x-y) ≤ (sin(x)-sin(y)) ≤ (x-y)
Take || (abs) of both sides.
=> 0 ≤ | sin(x)-sin(y) | ≤ | x-y |
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
