I was reading the answer in the post Maximum angle between a vector $x$ and its linear transformation $A x$. I am not able to figure out how is minimizing over $x$ the following function \begin{align} \frac{a}{\sqrt{a^{2}+b^{2}}} \cos ^{2}x + \frac{b}{\sqrt{a^{2}+b^{2}}} \sin ^{2}x \end{align} equivalent to maximizing over $x$ the function \begin{align} \tan\left(\tan^{-1}\left(\frac{b}{a}\tan\:x\right)-x\right). \end{align} I tried using many trigonometric identities but could not figure out.
Minimization of $\frac{a}{\sqrt{a^{2}+b^{2}}} \cos ^{2}x + \frac{b}{\sqrt{a^{2}+b^{2}}} \sin ^{2}x $
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Using the half angle formulas and assuming $\tan \theta=a/b$ should do it. – Jun 02 '22 at 12:35
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I think there are some typos there. I think it should be $q_1 = [\cos x, \sin x]^\mathsf{T}$, $q_2 = [d_1\cos x, d_2\sin x]^\mathsf{T}$, then $\frac{q_1^\top q_2}{| q_2|} = \frac{d_1\cos^2 x + d_2\sin^2 x}{\sqrt{d_1^2\cos^2 x + d_2^2\sin^2 x}}$, it is not difficult to prove that $\frac{d_1\cos^2 x + d_2\sin^2 x}{\sqrt{d_1^2\cos^2 x + d_2^2\sin^2 x}} \ge \frac{2\sqrt{\frac{d_2}{d_1}}}{1 + \frac{d_2}{d_1}}$. – River Li Jun 02 '22 at 13:35