若实数x,y满足x²+y²-12x+4y+40=0求xsin(-25π/3)+ysin(-15π/4)的值

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若实数x,y满足x²+y²-12x+4y+40=0求xsin(-25π/3)+ysin(-15π/4)的值若实数x,y满足x²+y²-12x+4y+40=0求x

若实数x,y满足x²+y²-12x+4y+40=0求xsin(-25π/3)+ysin(-15π/4)的值
若实数x,y满足x²+y²-12x+4y+40=0求xsin(-25π/3)+ysin(-15π/4)的值

若实数x,y满足x²+y²-12x+4y+40=0求xsin(-25π/3)+ysin(-15π/4)的值
x²+y²-12x+4y+40=0
(x²-12x+36)+(y²+4y+4)=0
(x-6)²+(y+2)²=0
解得x=6 ,y=-2
xsin(-25π/3)+ysin(-15π/4)
=xsin(-π/3)+ysin(π/4)
=6*(-√3/2)-2*(√2/2)
=-3√3-√2