Compare two fractions [closed]












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






trigonometry contest-math







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closed as off-topic by user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf Dec 13 '18 at 9:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    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
    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
    Dec 12 '18 at 19:34
















0












$begingroup$


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






trigonometry contest-math







share|cite|improve this question











$endgroup$



closed as off-topic by user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf Dec 13 '18 at 9:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    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
    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
    Dec 12 '18 at 19:34














0












0








0





$begingroup$


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






trigonometry contest-math







share|cite|improve this question











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






trigonometry contest-math




trigonometry contest-math






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share|cite|improve this question













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edited Dec 12 '18 at 19:35









user10354138

7,4422925




7,4422925










asked Dec 12 '18 at 19:24









MarkMark

278




278




closed as off-topic by user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf Dec 13 '18 at 9:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf Dec 13 '18 at 9:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user10354138, Mike Pierce, Cesareo, José Carlos Santos, supinf

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    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
    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
    Dec 12 '18 at 19:34


















  • $begingroup$
    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
    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
    Dec 12 '18 at 19:34
















$begingroup$
Hint: module 360
$endgroup$
– John L Winters
Dec 12 '18 at 19:27




$begingroup$
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




$begingroup$
Where is this problem from?
$endgroup$
– Arthur
Dec 12 '18 at 19:30












$begingroup$
@Arthur an olimpiad
$endgroup$
– Mark
Dec 12 '18 at 19:32




$begingroup$
@Arthur an olimpiad
$endgroup$
– Mark
Dec 12 '18 at 19:32












$begingroup$
And has it finished? Or is it ongoing?
$endgroup$
– Arthur
Dec 12 '18 at 19:33






$begingroup$
And has it finished? Or is it ongoing?
$endgroup$
– Arthur
Dec 12 '18 at 19:33














$begingroup$
@Arthur finished several years ago
$endgroup$
– Mark
Dec 12 '18 at 19:34




$begingroup$
@Arthur finished several years ago
$endgroup$
– Mark
Dec 12 '18 at 19:34










4 Answers
4






active

oldest

votes


















2












$begingroup$

Since
$$ 20176^circ=6cdot360^circ-144^circ, 2017^circ=6cdot360^circ-143^circ, 2018^circ=6cdot360^circ-142^circ, 2019^circ=6cdot360^circ-141^circ $$
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°}}\
&=&frac12frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
&=&frac12frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
&<&0.
end{eqnarray*}






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    1












    $begingroup$

    First, check that all these sines are positive, then we have: $2sin(2017^{circ})sin(2019^{circ})= cos(2^{circ})-cos(4036^{circ}), 2sin^2(2018^{circ})= 1-cos(4036^{circ})$. 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 $cos(a-b) - cos(a+b) = 2sin(a)sin(b)$ to convert from a sine to a cosine, and it is easier to handle.






    share|cite|improve this answer











    $endgroup$













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





















    1












    $begingroup$

    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 $2019$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

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






      share|cite|improve this answer









      $endgroup$













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


















      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Since
      $$ 20176^circ=6cdot360^circ-144^circ, 2017^circ=6cdot360^circ-143^circ, 2018^circ=6cdot360^circ-142^circ, 2019^circ=6cdot360^circ-141^circ $$
      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°}}\
      &=&frac12frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
      &=&frac12frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
      &<&0.
      end{eqnarray*}






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Since
        $$ 20176^circ=6cdot360^circ-144^circ, 2017^circ=6cdot360^circ-143^circ, 2018^circ=6cdot360^circ-142^circ, 2019^circ=6cdot360^circ-141^circ $$
        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°}}\
        &=&frac12frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
        &=&frac12frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
        &<&0.
        end{eqnarray*}






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Since
          $$ 20176^circ=6cdot360^circ-144^circ, 2017^circ=6cdot360^circ-143^circ, 2018^circ=6cdot360^circ-142^circ, 2019^circ=6cdot360^circ-141^circ $$
          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°}}\
          &=&frac12frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
          &=&frac12frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
          &<&0.
          end{eqnarray*}






          share|cite|improve this answer









          $endgroup$



          Since
          $$ 20176^circ=6cdot360^circ-144^circ, 2017^circ=6cdot360^circ-143^circ, 2018^circ=6cdot360^circ-142^circ, 2019^circ=6cdot360^circ-141^circ $$
          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°}}\
          &=&frac12frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
          &=&frac12frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
          &<&0.
          end{eqnarray*}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 19:42









          xpaulxpaul

          22.7k24455




          22.7k24455























              1












              $begingroup$

              First, check that all these sines are positive, then we have: $2sin(2017^{circ})sin(2019^{circ})= cos(2^{circ})-cos(4036^{circ}), 2sin^2(2018^{circ})= 1-cos(4036^{circ})$. 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 $cos(a-b) - cos(a+b) = 2sin(a)sin(b)$ to convert from a sine to a cosine, and it is easier to handle.






              share|cite|improve this answer











              $endgroup$













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


















              1












              $begingroup$

              First, check that all these sines are positive, then we have: $2sin(2017^{circ})sin(2019^{circ})= cos(2^{circ})-cos(4036^{circ}), 2sin^2(2018^{circ})= 1-cos(4036^{circ})$. 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 $cos(a-b) - cos(a+b) = 2sin(a)sin(b)$ to convert from a sine to a cosine, and it is easier to handle.






              share|cite|improve this answer











              $endgroup$













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
















              1












              1








              1





              $begingroup$

              First, check that all these sines are positive, then we have: $2sin(2017^{circ})sin(2019^{circ})= cos(2^{circ})-cos(4036^{circ}), 2sin^2(2018^{circ})= 1-cos(4036^{circ})$. 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 $cos(a-b) - cos(a+b) = 2sin(a)sin(b)$ to convert from a sine to a cosine, and it is easier to handle.






              share|cite|improve this answer











              $endgroup$



              First, check that all these sines are positive, then we have: $2sin(2017^{circ})sin(2019^{circ})= cos(2^{circ})-cos(4036^{circ}), 2sin^2(2018^{circ})= 1-cos(4036^{circ})$. 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 $cos(a-b) - cos(a+b) = 2sin(a)sin(b)$ to convert from a sine to a cosine, and it is easier to handle.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 12 '18 at 19:38

























              answered Dec 12 '18 at 19:31









              DeepSeaDeepSea

              71.2k54487




              71.2k54487












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




















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


















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






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













              1












              $begingroup$

              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 $2019$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                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 $2019$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 $2019$.






                  share|cite|improve this answer









                  $endgroup$



                  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 $2019$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 20:01









                  Misha LavrovMisha Lavrov

                  46.2k656107




                  46.2k656107























                      0












                      $begingroup$

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






                      share|cite|improve this answer









                      $endgroup$













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
















                      0












                      $begingroup$

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






                      share|cite|improve this answer









                      $endgroup$













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














                      0












                      0








                      0





                      $begingroup$

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






                      share|cite|improve this answer









                      $endgroup$



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







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 12 '18 at 19:32









                      Henry LeeHenry Lee

                      1,984219




                      1,984219












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


















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
















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




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



                      -

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