MathLabs

Problem 4

Let f(θ)=1−acos⁡θ−bsin⁡θ−Acos⁡2θ−Bsin⁡2θf(\theta)=1-a\cos\theta-b\sin\theta-A\cos2\theta-B\sin2\theta, where a,b,A,Ba,b,A,B are real. Prove that if f(θ)≥0f(\theta)\ge0 for all real θ\theta, then a2+b2≤2a^2+b^2\le2 and A2+B2≤1A^2+B^2\le1.
Step 2 of 3: Isolate the first harmonic
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f(θ)+f(θ+π2)=2−(a+b)cos⁡θ+(a−b)sin⁡θ≥0f(\theta)+f(\theta+\frac{\pi}{2})=2-(a+b)\cos\theta+(a-b)\sin\theta\ge0
Detailed analysis

A shift by π2\frac{\pi}{2} changes the signs of both double-angle terms, so they cancel in the sum. The remaining sinusoid has amplitude (a+b)2+(a−b)2=2a2+2b2\sqrt{(a+b)^2+(a-b)^2}=\sqrt{2a^2+2b^2}. Nonnegativity gives 2a2+2b2≤2\sqrt{2a^2+2b^2}\le2, hence a2+b2≤2a^2+b^2\le2.