I tried to solve this question but things got too much complicated and hence my efforts were completely futile.
Let $a,A,b,B \in \mathbb{R}$ and $$F(\theta)= 1- a\cos \theta - b\sin \theta- A\cos 2\theta- B\sin 2\theta$$ It is given that $$F(\theta) \ge 0 \;\forall\; \theta $$ and we have to prove that $\color{red}{a^2+b^2 \le 2}$ and $\color{green}{A^2+B^2 \le 1}$.
MY ATTEMPT
We need to prove that $$a\cos \theta + b\sin \theta+A\cos 2\theta+ B\sin 2\theta \le 1$$ $$\begin{align} & = a\cos \theta + b\sin \theta+A( \cos^2 \theta- \sin^2 \theta)+ B \sin \theta \cdot \cos \theta + B \sin \theta \cdot \cos \theta \le 1 \\ & =\ cos \theta (a+A \cos \theta+ B \sin \theta)+\sin \theta(b-A \sin \theta+B \ \cos \theta) \le1 \\ \end{align}$$
We know that $-\sqrt{x^2 + y^2} \le x \cos \theta + y \sin \theta \le \sqrt{x^2 + y^2}$ $$\Rightarrow (a+A \cos \theta+ B \sin \theta)^2 + (b-A \sin \theta+B \ \cos \theta)^2 \le 1$$ After solving this equation we get $$a^2 + b^2 + 2(A^2 +B^2)+ \cos \theta (2aA+2bB) + \sin \theta (2aB-2bA) \le1$$
Now If I again apply the same property, certainly the things are going to become more complicated and hence I think my approach is not at all right. Kindly Help me with this question.