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
 A: 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 :)
A: $$\sqrt{9-8\sin 50^\circ}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin^350^\circ}$$
$$(\text{using }\sin^2x=1-\cos^x)$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ(1-\cos^250^\circ)}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ+8\sin50^\circ\cos^250^\circ}$$
$$(\text{using }2\sin x\cos x=\sin2x)$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ+4\sin100^\circ\cos50^\circ}$$
$$(\text{using }2\sin x\cos y=\sin(x+y)-\sin(x-y))$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ+2(\sin150^\circ+\sin50^\circ)}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-6\sin50^\circ+2\sin150^\circ}$$
$$(\text{using }\sin150^\circ=\sin30^\circ=\frac{1}{2})$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-6\sin50^\circ+1}$$
$$=\csc50^\circ\sqrt{(3\sin50^\circ-1)^2}$$
$$\text{(Taking the positive root.)}$$
$$=\csc50^\circ(3\sin50^\circ-1)$$
$$=3-\csc50^\circ$$
$$\text{So }a=3\text{ and }b=-1$$
A: There is not a unique solution; we will be able to provide a solution as a function of $a$.  Note that if $a = 0$, then $ab=0$.  Henceforth, assume $a \neq 0$.
Let $x = \sin(50^\circ) \neq 0$ and $b = c/a$ so that we wish to solve for $c$.  Then the given equation is
$$  \sqrt{9 - 8 x} = a + \frac{c}{a x}  \text{.}  $$
Squaring, we get
$$  9 - 8 x = a^2 + \frac{2 c}{x} + \frac{c^2}{a^2 x^2}  \text{.}  $$
(And we will remember to check all solutions back in the original equation since this step may have introduced spurious solutions.)  Multiply through by $a^2 x^2$ and collect everything on the right (then swap sides):
$$  c^2 + 2a^2 x c + a^2 x^2( a^2 + 8 x - 9) = 0  \text{.}  $$
Applying the quadratic formula, \begin{align*}
c &= \frac{ -2 a^2 x \pm \sqrt{4 a^4 x^2 - 4a^2 x^2(a^2 + 8 x - 9)}}{2}  \\
  &= -a^2 x \pm a x \sqrt{a^2 - a^2 - (8 x - 9)}  \\
  &= -a^2 x \pm a x \sqrt{9 - 8 x}  \text{.}
\end{align*}
Plugging back in to the first display above, we quickly see that the $+$ choice works and the $-$ choice does not.  Converting back to the original symbols, the solution is 
$$  ab = -a \sin(50^\circ) \left(a - \sqrt{9 - 8 \sin(50^\circ)}\right)  \text{.}  $$
Happily, when $a=0$, this is also $0$, so we do not need to report the solution piecewise.
