Question

已知cos(α+β)=-1/3,cos2α=-5/13,α,β均为钝角,求cos(α-β)的值

Answer 1

解：∵α,β均为钝角，即π/2<α<π，π/2<β<π
==>π<α+β<2π，π<2α<2π
∴sin(α+β)<0,sin(2α)<0
∵cos(α+β)=-1/3,cos(2α)=-5/13
∴sin(α+β)=-√[1-cos²(α+β)]=-2√2/3
sin(2α)=-√[1-cos²(2α)]=-12/13
故 cos(α-β)=cos[2α-(α+β)]
=cos(2α)cos(α+β)+sin(2α)sin(α+β) (应用余弦差角公式)
=(-5/13)(-1/3)+(-12/13)(-2√2/3)
=(5+24√2)/39。