<s>[INST] Maximizing sin(a-b) given a trig relation Suppose $a$, $b$ are acute angle measures such that $\tan a = 5\tan b$. Find the maximum value of $\sin(a-b)$.
$\sin(a-b)=4\sin b \cos a$, but I don't know what to do from here.
 [/INST] Since $a,b\in(0,\frac\pi2)$ and $\tan$ is positive and strictly increasing there, the assumption $\tan a=5\tan b$ implies $a>b$.  Thus $a-b\in(0,\frac\pi2)$ also, so $\sin(a-b)>0$, and we might as well maximize $\sin^2(a-b)$ instead.  And since
$$ \sin^2(a-b) = 1 - \frac1{1+\tan^2(a-b)} $$
we might as well maximize $\tan^2(a-b)$ instead.
$$ \tan^2(a-b)
= \left(\frac{\tan a-\tan b}{1+\tan a\tan b}\right)^2
= \left(\frac{4\tan b}{1+5\tan^2 b}\right)^2
= \left(\frac{2}{\frac12\bigl(\frac1{\tan b}+5\tan b\bigr)}\right)^2
\le \left(\frac{2}{\sqrt5}\right)^2 = \frac45 $$
by AM/GM, with equality iff $b=\arctan\frac1{\sqrt5}$ (and so $a=\arctan\sqrt 5$).  Thus
$$\sin(a-b) \le \sqrt{1-\frac1{1+\frac45}} = \frac23 $$
with the same equality case.
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