1) sin²α+cos²α=1
sin²α=1-cos²α
sinα=+-√1-cos²α
берем знак "-", т.к. α в 4-ой четверти
sinα= -√1-cos²α = - √1-(3/5)² = - √1-9/25 = - √16/25 = -4/5
аналогично для sinβ:
sinβ=+-√1-cos²β
берем знак "-", т.к. β в 3-ой четверти
sinβ=-√1-cos²β= - √1-(-7/25)² = - √1-49/625 = - √576/625 = -24/25
sin(α-β)=sinα*cosβ - sinβ*cosα = -4/5*(-7/25) - (-24/25)*3/5=(28+72)/(25*5)=100/(25*5)=4/5
2) cos(arcsin(15/17)+arccos(-12/13))=
={по формуле cos(α+β)= cosα*cosβ - sinα*sinβ}=
=cos(arcsin(15/17)*cos(arccos(-12/13)) - sin(arcsin(15/17))*sin(arccos(-12/13))=
cos(arcsin(15/17)*cos(arccos(-12/13)) - sin(arcsin(15/17))*sin(arccos(-12/13))=
=√1-(sin(arcsin(15/17))²*(-12/13) - 15/17*√1-(cos(arccos(-12/13))²=
=√1-(15/17)² * (-12/13) - 15/17 * √1-(-12/13)² =
=√64/289 * (-12/13) - 15/17 * √25/169 =
=8/17 * (-12/13) - 15/17 * 5/13 = (-96-75)/(17*13) =-171/221