一、构造函数例1设α、m为常数,θ是任意实数,求证:眼cos(θ+α)+mcosθ演2≤1+2mcosα+m2.证明构造函数y=f(θ)=1+2mcosα+m2-眼cos(θ+α)+mcosθ演2,则只需证明y≥0即可.f(θ)=sin2(θ+α)+2m眼cosα-cosθcos(θ+α)演+m2sin2θ.令sin(θ+α)=x,则得二次函数y=x2+2msinθ· 1. Constructor Example 1 Let α and m be constants and θ be an arbitrary real number. Prove that: the eye cos(θ+α)+mcosθ plays 2≤1+2mcosα+m2. Prove the constructor y=f(θ)=1+ 2mcosα+m2-eye cos(θ+α)+mcosθ2, then only need to prove that y≥0. F(θ)=sin2(θ+α)+2m eye cosα-cosθcos(θ+α)+ M2sin2θ. Let sin(θ+α)=x, then get the quadratic function y=x2+2msinθ·