12
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Is the assumption that $F(\theta) \geq 0$ for all values of $\theta$, or just for one value of $\theta$? $\endgroup$ – user49640 May 22 '17 at 6:13
  • $\begingroup$ It is valid for all values of $\theta$ $\endgroup$ – Harsh Sharma May 22 '17 at 6:14
  • 1
    $\begingroup$ To prove $A^2 + B^2 \leq 1$, write the inequality $F(\theta) + F(\theta + \pi) \geq 0$. $\endgroup$ – user49640 May 22 '17 at 6:17
5
$\begingroup$

We have $$0 \leq F(\theta) + F(\theta + \pi) = 2 - 2(A \cos 2\theta + B \sin 2\theta).$$

Let $u$ be the vector $(A,B)$. Pick $\theta$ so that $v = (\cos 2\theta, \sin 2\theta)$ is a unit vector in the same direction as $u$. Then

$$\sqrt{A^2 + B^2} = |u| = u \cdot v = A \cos 2\theta + B \sin 2\theta \leq 1.$$

For the second part, write $$0 \leq F(\theta) + F(\theta + \pi/2) = 2 - (b + a)\cos \theta - (b- a)\sin \theta.$$

Now pick $\theta$ so that $v = (\cos \theta,\sin \theta)$ is a unit vector in the same direction as $u = (b + a,b - a)$. Then $$\sqrt{2}\sqrt{a^2 + b^2} = \sqrt{(b + a)^2 + (b - a)^2} = |u| = u \cdot v \leq 2.$$

$\endgroup$
1
$\begingroup$

Here is another (?) approach: define $c=a-bi$, $d=A-Bi$. Then we are given that $$ \operatorname{Re}(cz+dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1.\tag{1} $$ Replacing $z$ with $-z$ and with $iz$ in (1) we get two more inequalities \begin{eqnarray} z\mapsto -z:\quad\operatorname{Re}(-cz+dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1,\tag{2}\\ z\mapsto iz:\quad\ \operatorname{Re}(icz-dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1.\tag{3} \end{eqnarray} Adding (1) and (2) gives $$ \operatorname{Re}(dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1\quad\Leftrightarrow\quad |d|\le 1. $$ Similarly adding (1) and (3) gives $$ \operatorname{Re}((1+i)cz)\le 2,\quad\forall z\in\Bbb C\colon |z|=1\quad\Leftrightarrow\quad |(1+i)c|\le 2. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.