已知x+y=√2sin(θ+π/4),x-y=√2sin(θ-π/4),求证x²+y²=1
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2x=√2（sin(a+π/4)+sin(a-π/4))=√2(sin(a+π/4)-cos(a+π/4))
2y=√2(sin(a+π/4)+cos(a+π/4))
x^2+y^2=1/2(sin(a+π/4)-cos(a+π/4))^2+1/2(sin(a+π/4)+cos(a+π/4))^2
=1/2(1-sin2a)+1/2(1+sin2a)=1
2y=√2(sin(a+π/4)+cos(a+π/4))
x^2+y^2=1/2(sin(a+π/4)-cos(a+π/4))^2+1/2(sin(a+π/4)+cos(a+π/4))^2
=1/2(1-sin2a)+1/2(1+sin2a)=1
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