Compare two fractions [closed]
$begingroup$
How to compare $frac{sin{2016°}}{sin{2017°}}$ and $frac{sin{2018°}}{sin{2019°}}$?
trigonometry contest-math
edited Dec 12 '18 at 19:35
user10354138
7,4422925
asked Dec 12 '18 at 19:24
MarkMark
278
$endgroup$
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.
|
show 2 more comments
$begingroup$
How to compare $frac{sin{2016°}}{sin{2017°}}$ and $frac{sin{2018°}}{sin{2019°}}$?
trigonometry contest-math
edited Dec 12 '18 at 19:35
user10354138
7,4422925
asked Dec 12 '18 at 19:24
MarkMark
278
$endgroup$
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
|
show 2 more comments
$begingroup$
How to compare $frac{sin{2016°}}{sin{2017°}}$ and $frac{sin{2018°}}{sin{2019°}}$?
trigonometry contest-math
edited Dec 12 '18 at 19:35
user10354138
7,4422925
asked Dec 12 '18 at 19:24
MarkMark
278
$endgroup$
How to compare $frac{sin{2016°}}{sin{2017°}}$ and $frac{sin{2018°}}{sin{2019°}}$?
trigonometry contest-math
trigonometry contest-math
edited Dec 12 '18 at 19:35
user10354138
7,4422925
asked Dec 12 '18 at 19:24
MarkMark
278
edited Dec 12 '18 at 19:35
user10354138
7,4422925
asked Dec 12 '18 at 19:24
MarkMark
278
edited Dec 12 '18 at 19:35
user10354138
7,4422925
edited Dec 12 '18 at 19:35
user10354138
7,4422925
edited Dec 12 '18 at 19:35
user10354138
7,4422925
7,4422925
asked Dec 12 '18 at 19:24
MarkMark
278
asked Dec 12 '18 at 19:24
MarkMark
278
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
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°}}\
&=½frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
&=½frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
&<&0.
end{eqnarray*}
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 19:31
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
Sorry, I typed the wrong problem. Can you solve it now?
$endgroup$
– Mark
Dec 12 '18 at 19:36
add a comment |
$begingroup$
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 $2019$.
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
$endgroup$
add a comment |
$begingroup$
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:
$$frac{sin(2018)}{sin(2019)}=frac{sin(2018)}{sin(2018)cos(1)+cos(2018)sin(1)}$$
answered Dec 12 '18 at 19:32
Henry LeeHenry Lee
1,984219
$endgroup$
$begingroup$
Sorry, I typed the wrong problem, can you solve it now (it's fixed)
$endgroup$
– Mark
Dec 12 '18 at 19:36
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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°}}\
&=½frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
&=½frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
&<&0.
end{eqnarray*}
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
$endgroup$
add a comment |
$begingroup$
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°}}\
&=½frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
&=½frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
&<&0.
end{eqnarray*}
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
$endgroup$
add a comment |
$begingroup$
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°}}\
&=½frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
&=½frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
&<&0.
end{eqnarray*}
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
$endgroup$
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°}}\
&=½frac{(cos{3°}-cos{75°})-(cos{1°}-cos{75°})}{sin{37°}sin{39°}}\
&=½frac{cos{3°}-cos{1°}}{sin{37°}sin{39°}}\
&<&0.
end{eqnarray*}
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
answered Dec 12 '18 at 19:42
xpaulxpaul
22.7k24455
22.7k24455
add a comment |
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 19:31
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
Sorry, I typed the wrong problem. Can you solve it now?
$endgroup$
– Mark
Dec 12 '18 at 19:36
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 19:31
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
Sorry, I typed the wrong problem. Can you solve it now?
$endgroup$
– Mark
Dec 12 '18 at 19:36
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 19:31
DeepSeaDeepSea
71.2k54487
$endgroup$
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.
answered Dec 12 '18 at 19:31
DeepSeaDeepSea
71.2k54487
edited Dec 12 '18 at 19:38
answered Dec 12 '18 at 19:31
DeepSeaDeepSea
71.2k54487
answered Dec 12 '18 at 19:31
DeepSeaDeepSea
71.2k54487
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $2019$.
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
$endgroup$
add a comment |
$begingroup$
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 $2019$.
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
$endgroup$
add a comment |
$begingroup$
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 $2019$.
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
$endgroup$
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 $2019$.
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
answered Dec 12 '18 at 20:01
Misha LavrovMisha Lavrov
46.2k656107
46.2k656107
add a comment |
add a comment |
$begingroup$
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:
$$frac{sin(2018)}{sin(2019)}=frac{sin(2018)}{sin(2018)cos(1)+cos(2018)sin(1)}$$
answered Dec 12 '18 at 19:32
Henry LeeHenry Lee
1,984219
$endgroup$
$begingroup$
Sorry, I typed the wrong problem, can you solve it now (it's fixed)
$endgroup$
– Mark
Dec 12 '18 at 19:36
add a comment |
$begingroup$
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:
$$frac{sin(2018)}{sin(2019)}=frac{sin(2018)}{sin(2018)cos(1)+cos(2018)sin(1)}$$
answered Dec 12 '18 at 19:32
Henry LeeHenry Lee
1,984219
$endgroup$
$begingroup$
Sorry, I typed the wrong problem, can you solve it now (it's fixed)
$endgroup$
– Mark
Dec 12 '18 at 19:36
add a comment |
$begingroup$
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:
$$frac{sin(2018)}{sin(2019)}=frac{sin(2018)}{sin(2018)cos(1)+cos(2018)sin(1)}$$
answered Dec 12 '18 at 19:32
Henry LeeHenry Lee
1,984219
$endgroup$
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:
$$frac{sin(2018)}{sin(2019)}=frac{sin(2018)}{sin(2018)cos(1)+cos(2018)sin(1)}$$
answered Dec 12 '18 at 19:32
Henry LeeHenry Lee
1,984219
answered Dec 12 '18 at 19:32
Henry LeeHenry Lee
1,984219
answered Dec 12 '18 at 19:32
Henry LeeHenry Lee
1,984219
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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