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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 $c=10$

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1  
Why is $b=4$? What can't it be $8$ or $32$ or ... –  Aryabhata Apr 12 '13 at 17:19
    
@Aryabhata For that the LHS will for sure be $>$ RHS –  Mr.ØØ7 Apr 12 '13 at 17:28
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@exploringnet: It may be true, but I don't see it without more work. The way it was stated was as if it is very obvious. For instance, as $c \to 0$, we have that LHS $\to 0$ for any $b$ (even large ones), so why can't $c=1$ and $b$ be larger than $4$? (And I don't see why it has to be a multiple of $4$ too, which I presume was part of the reason $4$ was chosen...). –  Aryabhata Apr 12 '13 at 17:45

3 Answers 3

up vote 5 down vote accepted

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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+1, Very nice. You have proved that there is one set of solutions. How does one show that this set is unique? –  Américo Tavares Apr 12 '13 at 18:25
    
@AméricoTavares: It is given that it is unique, as part of the problem statement. One just needs to find them. I agree, the question how that was determined is probably more interesting. –  Aryabhata Apr 12 '13 at 18:27
    
You are right. ${}{}{}{}$ –  Américo Tavares Apr 12 '13 at 18:29
    
very nice , 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 $c=10$ –  math110 Apr 13 '13 at 0:20
    
@math110: Your argument that $b=4$ is not convincing though... –  Aryabhata Apr 13 '13 at 2:46

$$sinc-cos(2c)$$ $$=>cos(90-c)-cos(2c)$$ $$=>-2.sin\Bigg(\dfrac{(90+c)}2\Bigg).sin\Bigg(\dfrac{(90-3c)}2\Bigg)=-sin(50)$$ $$2sin\theta sin\phi=sin50$$

So, one of the solution comes when one of $sin\theta^0$ or $sin\phi$ is equal to 1/2 and other is $sin50$

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 1st case $c=10^0$

And as the solution is unique, it's the one we need.

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Thank you exploringnet,But I think you methods don't strict,Because why $\dfrac{90+c}{2}=50$,and$ \sin{\dfrac{90-3c}{2}}=\dfrac{1}{2}$,you mean if $ab=cd$,then $a=c,b=d$,That's some mistake. –  math110 Apr 12 '13 at 17:05
    
@math110 Then try quadratic solving. –  Mr.ØØ7 Apr 12 '13 at 17:36
    
@math110 Now I see after the other answer that the solution set is unique and thus if we find even one , then it's more than enough. –  Mr.ØØ7 Apr 13 '13 at 4:23

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 $c$.

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Thank you ,But I think this methods can't solve my question,because $X=\dfrac{-1+\sqrt{9-8\sin{50^{0}}}}{4}$, So this return it. –  math110 Apr 12 '13 at 16:53

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