# Compare two fractions [closed]

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

• Hint: module 360 – John L Winters Dec 12 '18 at 19:27
• Where is this problem from? – Arthur Dec 12 '18 at 19:30
• @Arthur an olimpiad – Mark Tiukov Dec 12 '18 at 19:32
• And has it finished? Or is it ongoing? – Arthur Dec 12 '18 at 19:33
• @Arthur finished several years ago – Mark Tiukov Dec 12 '18 at 19:34

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 $$\begin{eqnarray*} &&\frac{\sin{2016°}}{\sin{2017°}}-\frac{\sin{2018°}}{\sin{2019°}}\\ &=&\frac{\sin{144°}}{\sin{143°}}-\frac{\sin{142°}}{\sin{141°}}\\ &=&\frac{\sin{36°}}{\sin{37°}}-\frac{\sin{38°}}{\sin{39°}}\\ &=&\frac{\sin{36°}\sin{39°}-\sin{37°}\sin{38°}}{\sin{37°}\sin{39°}}\\ &=&\frac12\frac{(\cos{3°}-\cos{75°})-(\cos{1°}-\cos{75°})}{\sin{37°}\sin{39°}}\\ &=&\frac12\frac{\cos{3°}-\cos{1°}}{\sin{37°}\sin{39°}}\\ &<&0. \end{eqnarray*}$$

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.

• Sorry, I typed the wrong problem. Can you solve it now? – Mark Tiukov Dec 12 '18 at 19:36

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 $$2019$$.

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: $$\frac{\sin(2018)}{\sin(2019)}=\frac{\sin(2018)}{\sin(2018)\cos(1)+\cos(2018)\sin(1)}$$

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