Question

Найдите значение выражения sin(arccos 4/5-arccos 3/5)

Answer 1

$sin(arccos 4/5-arccos 3/5)=sin(arccos 4/5)cos(arccos 3/5)- \ -cos(arccos 4/5)sin(arccos 3/5)= >\ \ sin(arccos 4/5)= sqrt{1-cos^2(arccos 4/5)}= sqrt{1-16/25}= sqrt{9/25}=3/5 \ cos(arccos 3/5)=3/5 \ cos(arccos 4/5)= 4/5 \ sin(arccos 3/5)= sqrt{1-cos^2(arccos 3/5)}= sqrt{1-9/25}= sqrt{16/25}=4/5 \ \ => = 3/5*3/5-4/5*4/5=9/25-16/25=-7/25$

Answer 2

СпасибО!!!