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How to compare $\frac{\sin{2016°}}{\sin{2017°}}$ and $\frac{\sin{2018°}}{\sin{2019°}}$?

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  • $\begingroup$ Hint: module 360 $\endgroup$ – John L Winters Dec 12 '18 at 19:27
  • $\begingroup$ Where is this problem from? $\endgroup$ – Arthur Dec 12 '18 at 19:30
  • $\begingroup$ @Arthur an olimpiad $\endgroup$ – Mark Tiukov Dec 12 '18 at 19:32
  • $\begingroup$ And has it finished? Or is it ongoing? $\endgroup$ – Arthur Dec 12 '18 at 19:33
  • $\begingroup$ @Arthur finished several years ago $\endgroup$ – Mark Tiukov Dec 12 '18 at 19:34
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Since $$ 20176^\circ=6\cdot360^\circ-144^\circ, 2017^\circ=6\cdot360^\circ-143^\circ, 2018^\circ=6\cdot360^\circ-142^\circ, 2019^\circ=6\cdot360^\circ-141^\circ $$ one has \begin{eqnarray*} &&\frac{\sin{2016°}}{\sin{2017°}}-\frac{\sin{2018°}}{\sin{2019°}}\\ &=&\frac{\sin{144°}}{\sin{143°}}-\frac{\sin{142°}}{\sin{141°}}\\ &=&\frac{\sin{36°}}{\sin{37°}}-\frac{\sin{38°}}{\sin{39°}}\\ &=&\frac{\sin{36°}\sin{39°}-\sin{37°}\sin{38°}}{\sin{37°}\sin{39°}}\\ &=&\frac12\frac{(\cos{3°}-\cos{75°})-(\cos{1°}-\cos{75°})}{\sin{37°}\sin{39°}}\\ &=&\frac12\frac{\cos{3°}-\cos{1°}}{\sin{37°}\sin{39°}}\\ &<&0. \end{eqnarray*}

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First, check that all these sines are positive, then we have: $2\sin(2017^{\circ})\sin(2019^{\circ})= \cos(2^{\circ})-\cos(4036^{\circ}), 2\sin^2(2018^{\circ})= 1-\cos(4036^{\circ})$. Since $1 > \cos(2^{\circ})$, it follows that $\sin^2(2018^{\circ})> \sin(2017^{\circ})\sin(2019^{\circ})\implies \dfrac{\sin(2017^{\circ})}{\sin(2018^{\circ})} < \dfrac{\sin(2018^{\circ})}{\sin(2019^{\circ})}$.

Note 1: For your question, as suggested above, you should use mod $180^{\circ}$ to reduce it to an angle between $0^{\circ}$ and $180^{\circ}$ .

Note 2: For the edited problem, use the formula $\cos(a-b) - \cos(a+b) = 2\sin(a)\sin(b)$ to convert from a sine to a cosine, and it is easier to handle.

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  • $\begingroup$ Sorry, I typed the wrong problem. Can you solve it now? $\endgroup$ – Mark Tiukov Dec 12 '18 at 19:36
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It's easier to compare ratios by taking logs:

$$ \log \frac{\sin 2016^\circ}{\sin 2017^\circ} = \log \frac{-\sin 2016^\circ}{-\sin 2017^\circ} = \log(-\sin 2016^\circ) - \log(-\sin 2017^\circ). $$ So we want to compare the change in $f(x)=\log(-\sin x)$ when we go from $x=2016$ to $x=2017$, versus when we go from $x=2018$ to $x=2019$.

We have $f'(x) = \cot x = \frac{\cos x}{\sin x}$, and $f''(x) = -\frac{1}{\sin^2 x}$, making $f$ strictly concave everywhere it is defined. (It is only defined when $\sin x$ is negative, but this holds for $1980^\circ < x < 2160^\circ$.) For concave functions, slopes are always decreasing, so we have $$ \log(-\sin 2017^\circ) - \log(-\sin 2016^\circ) > \log(-\sin 2019^\circ) - \log(-\sin 2018^\circ) $$ which is equivalent to $\frac{\sin 2016^\circ}{\sin 2017^\circ} < \frac{\sin 2018^\circ}{\sin 2019^\circ}$.

You might complain that for concave functions tangent slopes are always negative, and that's not what we're using. To get the statement above, we could use the mean value theorem: the change $f(2017) - f(2016)$ is equal to $f'(x)$ for some $x$ between $2016$ and $2017$, and the change $f(2019) - f(2018)$ is equal to $f'(x)$ for some $x$ between $2018$ and $2019$.

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you could try to use the compound angle formula: $$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$ so: $$\frac{\sin(2018)}{\sin(2017)}=\frac{\sin(2017)\cos(1)+\cos(2017)\sin(1)}{\sin(2017)}=\cos(1)+\cot(2017)\sin(1)$$ now: $$\frac{\sin(2018)}{\sin(2019)}=\frac{\sin(2018)}{\sin(2018)\cos(1)+\cos(2018)\sin(1)}$$

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  • $\begingroup$ Sorry, I typed the wrong problem, can you solve it now (it's fixed) $\endgroup$ – Mark Tiukov Dec 12 '18 at 19:36

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