已知x+y=根号2sin(α+π/4),x-y=根号2sin(α-π/4),求证x²+y²=1
3个回答

联立方程

x+y=根号2sin(α+π/4)

x-y=根号2sin(α-π/4)

解得:x=(根号2/2)*(sin(α+π/4)+sin(α-π/4)),y=(根号2/2)*(sin(α+π/4)-sin(α-π/4));

由三角和差化积公式得:sin(α+π/4)=sinα*cos(π/4)+cosα*sin(π/4),sin(α-π/4)=sinα*cos(π/4)-cosα*sin(π/4);

故sin(α+π/4)+sin(α-π/4)=2*sinα*cos(π/4)=根号2*sinα;

sin(α+π/4)-sin(α-π/4)=2*cosα*sin(π/4)=根号2*cosα;

代入x,y可得:x=sinα,y=cosα;

所以,x²+y²=sin²α+cos²α=1.