Question

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

Answer 1

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