Margie Marx

2022-01-22

Derive $\mathrm{log}\left(a+b\right)=\mathrm{log}\left(a\right)-2\mathrm{log}\left(\mathrm{cos}\left(\mathrm{arctan}\left(\sqrt{\frac{b}{a}}\right)\right)\right)$

Thomas Lynn

Expert

Nothing too magical is happening - its

lovagwb

Expert

$\mathrm{log}\left(a+b\right)=\mathrm{log}\left(a\right)-2\mathrm{log}\left(\mathrm{cos}\left(\mathrm{arctan}\left(\sqrt{\frac{b}{a}}\right)\right)\right)$
Start from RHS:
Let $\mathrm{arctan}\left(\sqrt{\frac{b}{a}}\right)=y⇒\mathrm{tan}y=\frac{b}{a}$
Using the theorem according to pythagoras:
$\mathrm{cos}y=\frac{\sqrt{a}}{\sqrt{a+b}}$
The equation then reduces to:
$\mathrm{log}\left(a\right)-2\mathrm{log}\left(\mathrm{cos}y\right)=\mathrm{log}\left(a\right)-2\mathrm{log}\left(\frac{\sqrt{a}}{\sqrt{a+b}}\right)=\mathrm{log}\left(a\right)-\mathrm{log}\left(\frac{a}{a+b}\right)$
$\mathrm{log}\left(a\right)-\mathrm{log}\left(\frac{a}{a+b}\right)=\mathrm{log}\left(a\right)-\mathrm{log}\left(a\right)-\left(-\mathrm{log}\left(a+b\right)\right)=\mathrm{log}\left(a+b\right)$
=LHS

RizerMix

Expert

By subtracting $\mathrm{log}a$, $-\frac{1}{2}\mathrm{log}\left(1+\frac{b}{a}\right)=\mathrm{log}\mathrm{cos}\mathrm{arctan}\sqrt{\frac{b}{a}}$ Using the identity $\mathrm{cos}\mathrm{arctan}x=\frac{1}{\sqrt{{x}^{2}+1}}$ $\mathrm{log}\mathrm{cos}\mathrm{arctan}\sqrt{\frac{b}{a}}=\mathrm{log}\frac{1}{\sqrt{\frac{b}{a}+1}}=-\frac{1}{2}\mathrm{log}\left(1+\frac{b}{a}\right)$ as desired.