若实数x,y满足x^2+y^2-12x+4y+40=0,求xsin(-25π/3)+ysin(-15π/4)的值求详解 急需!

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若实数x,y满足x^2+y^2-12x+4y+40=0,求xsin(-25π/3)+ysin(-15π/4)的值求详解  急需!

若实数x,y满足x^2+y^2-12x+4y+40=0,求xsin(-25π/3)+ysin(-15π/4)的值求详解 急需!
若实数x,y满足x^2+y^2-12x+4y+40=0,求xsin(-25π/3)+ysin(-15π/4)的值
求详解 急需!

若实数x,y满足x^2+y^2-12x+4y+40=0,求xsin(-25π/3)+ysin(-15π/4)的值求详解 急需!
答:
实数x,y满足x^2+y^2-12x+4y+40=0
(x^2-12x+36)+(y^2+4y+4)=0
(x-6)^2+(y+2)^2=0
完全平方数具有非负性质,同时为0其和为0
所以:
x-6=0
y+2=0
解得:x=6,y=-2
所以:
xsin(-25π/3)+ysin(-15π/4)
=6sin(-π/3)-2sin(π/4)
=-6*(√3/2)-2*(√2/2)
=-3√3-√2