已知asinθ+bcosθ=c,bsinθ+acosθ=d,求证:(ac-bd)²+(ad-bc)²=(a²-b²)²
问题描述:
已知asinθ+bcosθ=c,bsinθ+acosθ=d,求证:(ac-bd)²+(ad-bc)²=(a²-b²)²
答
很容易的把sinθ和cosθ看成未知数,解原方程组,得cosθ=(ad-bc)/(a²-b²)sinθ=(ac-bd)/(a²-b²)把这两个代入sin²θ+cos²θ=1即可得(ac-bd)²+(ad-bc)²=(a²-b²)...