19. If $\sqrt{9-8 \sin 50^{\circ}}=a+b \csc 50^{\circ}$ where $a, b$ are integers, find $a b$.
(2 marks)若 $\sqrt{9-8 \sin 50^{\circ}}=a+b \csc 50^{\circ}$, 其中 $a 、 b$ 爲整數, 求 $a b$ 。
19. -3
19. Set $x=\sin 50^{\circ}$. Then the given equation becomes $\sqrt{9-8 x}=a+\frac{b}{x}$. Squaring both sides and rearranging terms, we get the cubic equation $8 x^{3}+\left(a^{2}-9\right) x^{2}+2 a b x+b^{2}=0$.
This cubic equation reminds us of the triple angle formula $\sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta$, which gives $\frac{1}{2}=\sin 150^{\circ}=3 x-4 x^{3}$ and hence $8 x^{3}-6 x+1=0$. Reversing the steps in the previous paragraph, we get $(9-8 x) x^{2}=(3 x-1)^{2}$ and hence $\sqrt{9-8 x}=3-\frac{1}{x}$, noting that $x>\frac{1}{3}$. It follows that we may take $(a, b)=(3,-1)$ which gives the answer $a b=-3$.
Finally, we show that the choices for $a, b$ are unique. Suppose on the contrary that $a+b \csc 50^{\circ}=a^{\prime}+b^{\prime} \csc 50^{\circ}$ where $a \neq a^{\prime}$. Then rearranging terms gives $x=\sin 50^{\circ}=\frac{b^{\prime}-b}{a-a^{\prime}}$, which is rational. However, it is easy to check that $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}$ and $\pm \frac{1}{8}$ are not roots of $8 x^{3}-6 x+1=0$, so the equation has no rational root, a contradiction.