inequalitytrigonometry
If $A>0$, $B>0$ and $A+B=30^\circ$, find the minimum value of $\tan A+\tan B$.
A similar question has been asked here. Many good answers are posted there. I understood that.
One particular answer by @Bill Kleinhans intrigued me as it was short and straightforward. But I didn't fully understand that answer. It uses Jensen's inequality, which I know, but don't know how to use that to get the answer here.
Not posting this as a comment there because Bill has been away close to three years now.
You can use the same intuition when $\theta$ lies in other quadrants, drawing the same line.
$\sin(20^\circ)$ and $\cos(20^\circ)$ can both be found using the triple-angle formulas. (I'm assuming your "20" was in degrees, not radians.)
Unfortunately, both of those will end up with you having to solve a cubic; you can use Cardano's formula to do that.
You can't solve it with just quadratics and algebra, for if you could, it'd be possible to trisect a 60-degree angle; but that is in fact exactly the example used generally to show that trisection is impossible, because $\cos(60^{\circ})$ is not a surd.
Best Answer
Bill's idea may be this..
consider $f(x)=\tan x$ $f''(x)=2\sec^2 x \tan x>0$ for $x\in (0,\pi/3)$
.By jensen $$f(x)+f(y)\le \frac{f(x)+f(y}{2}\Rightarrow \frac{\tan A+\tan B}{2}\ge {\tan(\frac{A+B}{2})}=\tan 15^o$$