简单三角恒等已知x+y=(根号2)sin(a+π/4),x-y=(根号2)sin(a-π/4),求证x^2+y^2=1

Question

简单三角恒等
已知x+y=(根号2)sin(a+π/4),x-y=(根号2)sin(a-π/4),求证x^2+y^2=1

Answer 1

由x+y=(根号2)sin(a+π/4)得x²+y²+2xy=2sin²（a+π/4)
由x-y=(根号2)sin(a-π/4)得x²+y²-2xy=2sin²（a-π/4)
∵sin（a-π/4)=cos(a+π/4)
∴两式相加得
2（x²+y²）=2
∴x²+y²=1