Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$? 
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$

$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left side, $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = \sqrt{9\sin^250^\circ-8\sin^350^{\circ}}$$
Now How can i solve it after that , Help required, Thanks

If you use the formula for the triple angle:
$-8\sin^3(50)=2\sin(150)-6\sin(50)=1-6\sin(50)$
so your last square root becomes
$\sqrt{9\sin^2(50)-6\sin(50)+1}=\sqrt{(3\sin(50)-1)^2}=3\sin(50)-1$
as the last expression is positive. So:
$3\sin(50)-1=a\sin(50)+b$
Now if $a$ and $b$ are rational/integer, we have $a=3$, $b=-1$ so $ab=-3$ but in the case of real numbers there is no unique solution. For example if $a=0,b=3\sin(50)-1$ we have $ab=0$
In any case, the last equation is much simpler to work out than the first one :)
