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
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![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](/uploads/image/z/14604165-45-5.jpg?t=%E7%AE%80%E5%8D%95%E4%B8%89%E8%A7%92%E6%81%92%E7%AD%89%E5%B7%B2%E7%9F%A5x%2By%3D%28%E6%A0%B9%E5%8F%B72%29sin%28a%2B%CF%80%2F4%29%2Cx-y%3D%28%E6%A0%B9%E5%8F%B72%29sin%28a-%CF%80%2F4%29%2C%E6%B1%82%E8%AF%81x%5E2%2By%5E2%3D1)
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
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
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
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