<s>[INST] Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ It is known that$$\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$$
for exactly one set of positive integers $(a,b,c)$ where $0<c<90$
find the value 
$$\dfrac{b+c}{a}$$
my idea,$ \sin 50^\circ >\sin 45^\circ >\frac{_5}{^8} $
so$\sqrt{9-8\sin 50^{\circ}}<2$,then $a=1$
then $$\dfrac{b^2}{16}(1-\cos{(2c)})+\dfrac{b}{4}\sin{c}=1-\sin{50^{0}}$$
so $b=4$
then we have $\sin{c}-\cos{(2c)}=-\sin{50^{0}}$
then my question: How can prove this $c$ must equality 10?
Thank you everyone: yesterday,when I go to bed, I have consider this:let $f(c)=\sin{c}-\cos{(2c)}$.then we have $f(10)=\sin{10}-\cos{20}=\cos{80}-\cos{20}=2\sin{\dfrac{80-20}{2}}\sin{\dfrac{80+20}{2}}=\sin{50}$, by other hand, we have $f'(c)=\cos{c}+2\sin{2c}>0,0<c<\dfrac{\pi}{2}$,so if we $f(c)=f(10)$,we must $c=10$
 [/INST] We have the following (working in degrees):
$$\cos 20 - \cos 80 = \cos(50-30) - \cos(50+30) = 2 \sin 50 \sin 30 = \sin 50$$
Thus we have that
$$1 - 2\sin^2 10 - \sin 10 = \sin 50$$
(using $\cos 20 = 1 - 2 \sin^2 10$ and $\cos 80 = \sin (90 - 80) = \sin 10$)
And so
$$9 - 8 \sin 50 = 9 - 8(1 - 2\sin^2 10 - \sin 10) = 1 + 8\sin 10 + 16\sin^2 10 = (1 + 4 \sin 10)^2$$
Thus $a=1, b=4, c=10$.
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