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Home » \[ T'^\mu_\mu = T^\mu_\mu = \sigma \] - Londonproperty

\[ T'^\mu_\mu = T^\mu_\mu = \sigma \] - Londonproperty

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \fra...
But it’s simpler to minimize directly:
Let $ s = \sin x + \cos x $, $ p = \sin x \cos x $. Then:
s^2 = 1 + 2p \Rightarrow p = \frac{s^2 - 1}{2}
We minimize this expression for $ s \in (\sqrt{2}, \sqrt{2}] $, since...
f(s) = \left( \frac{2s}{s^2 - 1} \right)^2
We minimize $ f(s) $ on $ (1, \sqrt{2}) $.
Let’s test $ s = \sqrt{2} $:
f(\sqrt{2}) = \left( \frac{2\sqrt{2}}{2 - 1} \right)^2 = (2\sqrt{2})^2...
Now check derivative or behavior: as $ s \to 1^+ $, $ f(s) \to \infty...

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