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.
 A: 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.
A: We have $$\frac{\sin A\cos B}{\cos A\sin B}=\frac51$$
Applying Componendo and dividendo,  $$\frac{\sin(A-B)}{\sin(A+B)}=\frac{5-1}{5+1}$$
$$\implies\sin(A-B)=\frac23\sin(A+B)$$
Now for real $A+B,-1\le\sin(A+B)\le1$
Hope you can take it from here
A: you can use the Lagrange Multiplier Method, defining the function
$f(a,b,\lambda)=\sin(a-b)+\lambda(\tan(a)-5\tan(b))$
now you must solve the system
$f_a=0;$
$f_b=0;$
$f_{\lambda}=0$
this means
$$\cos(a-b)+\lambda(1+\tan(a)^2)=0$$
$$-\cos(a-b)+\lambda(-5-5\tan(b)^2)=0$$
$$\tan(a)-5\tan(b)=0$$
