已知:a^2+b^2=c^2+d^2=1,求证:(ac-bd)^2+(ad+bc)^2=1

2025-12-18 00:53:12
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回答1:

证明:
a^2+b^2=c^2+d^2=1
a^2=1-b^2,c^2=1-d^2

(ac-bd)^2+(ad+bc)^2
=a^2c^2+b^2d^2-2abcd+a^2d^2+b^2c^2+2abcd
=a^2c^2+b^2d^2+a^2d^2+b^2c^2
=(1-b^2)(1-d^2)+b^2d^2+(1-b^2)d^2+b^2(1-d^2)
=1-b^2-d^2+b^2d^2+b^2d^2+d^2-b^2d^2+b2-b^2d^2
=1

∴(ac-bd)^2+(ad+bc)^2=1