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 $c=10$
 A: 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 $a=1, b=4, c=10$.
A: $$\sin c-\cos(2c)$$
$$\implies \cos(90-c)-\cos(2c)$$
$$\implies -2\sin\Bigg(\dfrac{90+c}2\Bigg) \sin \Bigg(\dfrac{90-3c}2\Bigg)=-\sin(50^{\circ})$$
$$2\sin\theta \sin\phi=\sin 50^{\circ}$$
So, one of the solution comes when one of $\sin \theta^{\circ}$ or $\sin\phi$
is equal to $1/2$ and other is $\sin 50^{\circ}$
So, $$1)\dfrac{90+c}2=50 ;\sin\Bigg(\dfrac{90-3c}2\Bigg)=1/2$$
Or $$2)\dfrac{90-3c}2=50 ; \sin\Bigg(\dfrac{90+c}2\Bigg)=1/2$$
Case $2$ does not hold true , So, from $1$st case $c=10^{\circ}$
And as the solution is unique, it's the one we need.
A: Well the most forward way I think of is using the fact that $cos(2c)=1-2sin^2(c)$. This gives you some equation like:
$2X^2+X-(1+sin(50°))=0$ where $X=sin(c)$
Once you have $sin(c)$ you can get $c$ quite easily given your condition over $c$.
