Mixing
Largest possible value of trigonometric functions
Clash Royale CLAN TAG#URR8PPP
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7
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Find the largest possible value of
$$sin(a_1)cos(a_2) + sin(a_2)cos(a_3) + cdots + sin(a_2014)cos(a_1)$$
Since the range of the $sin$ and $cos$ function is between $1$ and $-1$, shouldn't the answer be $2014$?
trigonometry inequality optimization summation a.m.-g.m.-inequality
edited Aug 8 at 8:45
Michael Rozenberg
88.7k1579179
asked Aug 8 at 8:17
SuperMage1
707210
add a comment |Â
up vote
7
down vote
favorite
Find the largest possible value of
$$sin(a_1)cos(a_2) + sin(a_2)cos(a_3) + cdots + sin(a_2014)cos(a_1)$$
Since the range of the $sin$ and $cos$ function is between $1$ and $-1$, shouldn't the answer be $2014$?
trigonometry inequality optimization summation a.m.-g.m.-inequality
edited Aug 8 at 8:45
Michael Rozenberg
88.7k1579179
asked Aug 8 at 8:17
SuperMage1
707210
1
You cannot have $sin (a_1)=cos (a_1)=1$ so the maximum is not 2014.
– Kavi Rama Murthy
Aug 8 at 8:24
1
Fairly obviously it won't exceed $2014$ but it does not necessarily achieve it either.
– badjohn
Aug 8 at 8:24
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Find the largest possible value of
$$sin(a_1)cos(a_2) + sin(a_2)cos(a_3) + cdots + sin(a_2014)cos(a_1)$$
Since the range of the $sin$ and $cos$ function is between $1$ and $-1$, shouldn't the answer be $2014$?
trigonometry inequality optimization summation a.m.-g.m.-inequality
edited Aug 8 at 8:45
Michael Rozenberg
88.7k1579179
asked Aug 8 at 8:17
SuperMage1
707210
Find the largest possible value of
$$sin(a_1)cos(a_2) + sin(a_2)cos(a_3) + cdots + sin(a_2014)cos(a_1)$$
Since the range of the $sin$ and $cos$ function is between $1$ and $-1$, shouldn't the answer be $2014$?
trigonometry inequality optimization summation a.m.-g.m.-inequality
edited Aug 8 at 8:45
Michael Rozenberg
88.7k1579179
asked Aug 8 at 8:17
SuperMage1
707210
edited Aug 8 at 8:45
Michael Rozenberg
88.7k1579179
edited Aug 8 at 8:45
Michael Rozenberg
88.7k1579179
edited Aug 8 at 8:45
Michael Rozenberg
88.7k1579179
88.7k1579179
asked Aug 8 at 8:17
SuperMage1
707210
asked Aug 8 at 8:17
SuperMage1
707210
asked Aug 8 at 8:17
SuperMage1
707210
707210
1
You cannot have $sin (a_1)=cos (a_1)=1$ so the maximum is not 2014.
– Kavi Rama Murthy
Aug 8 at 8:24
1
Fairly obviously it won't exceed $2014$ but it does not necessarily achieve it either.
– badjohn
Aug 8 at 8:24
add a comment |Â
1
You cannot have $sin (a_1)=cos (a_1)=1$ so the maximum is not 2014.
– Kavi Rama Murthy
Aug 8 at 8:24
1
Fairly obviously it won't exceed $2014$ but it does not necessarily achieve it either.
– badjohn
Aug 8 at 8:24
1
1
You cannot have $sin (a_1)=cos (a_1)=1$ so the maximum is not 2014.
– Kavi Rama Murthy
Aug 8 at 8:24
You cannot have $sin (a_1)=cos (a_1)=1$ so the maximum is not 2014.
– Kavi Rama Murthy
Aug 8 at 8:24
1
1
Fairly obviously it won't exceed $2014$ but it does not necessarily achieve it either.
– badjohn
Aug 8 at 8:24
Fairly obviously it won't exceed $2014$ but it does not necessarily achieve it either.
– badjohn
Aug 8 at 8:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
20
down vote
accepted
Let $a_2015=a_1.$
Thus, by AM-GM
$$sum_k=1^2014sina_kcosa_k+1 leqsum_k=1^2014|sina_k||cosa_k+1| le$$
$$leqsum_k=1^2014fracsin^2a_k+cos^2a_k+12=sum_k=1^2014fracsin^2a_k+cos^2a_k2=frac20142=1007.$$
The equality occurs for $a_i=45^circ,$ which says that $1007$ is a maximal value.
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more?
– Daniel Wagner
Aug 8 at 13:19
@Daniel Wagner I used $ableqfraca^2+b^22.$
– Michael Rozenberg
Aug 8 at 13:32
Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification.
– Daniel Wagner
Aug 8 at 13:35
You are welcome!
– Michael Rozenberg
Aug 8 at 13:37
add a comment |Â
up vote
4
down vote
From
$$
f_n(a) = sum_k=1^n sin a_k cos a_k+1
$$
with $a_n+1 = a_1$
the stationary points are located at the solutions for
$$
fracpartial partial a_kf_n(a) = -sin a_k-1sin a_k + cos a_k cos a_k+1 = 0
$$
and then
$$
tan a_ntan a_n-1cdotstan a_2 = cot a_1
$$
or
$$
tan a_ntan a_n-1cdotstan a_2tan a_1 = 1
$$
or
$$
prod_ksin a_k = prod_kcos a_k
$$
which is obtained for $a_k = fracpi4$ when
$$
f_n(a) = frac n2
$$
answered Aug 8 at 13:31
Cesareo
5,8782412
Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=fracpi4$ is the absolute maximum? The $tan a_ntan a_n-1cdotstan a_2tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum...
– Taladris
Aug 8 at 13:49
@Taladris There are many relative extrema. For instance for $n = 3$ we have: $left-frac32,frac32,-sqrt2,sqrt2right$ but with some additional considerations like symmetry...
– Cesareo
Aug 8 at 14:44
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
20
down vote
accepted
Let $a_2015=a_1.$
Thus, by AM-GM
$$sum_k=1^2014sina_kcosa_k+1 leqsum_k=1^2014|sina_k||cosa_k+1| le$$
$$leqsum_k=1^2014fracsin^2a_k+cos^2a_k+12=sum_k=1^2014fracsin^2a_k+cos^2a_k2=frac20142=1007.$$
The equality occurs for $a_i=45^circ,$ which says that $1007$ is a maximal value.
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more?
– Daniel Wagner
Aug 8 at 13:19
@Daniel Wagner I used $ableqfraca^2+b^22.$
– Michael Rozenberg
Aug 8 at 13:32
Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification.
– Daniel Wagner
Aug 8 at 13:35
You are welcome!
– Michael Rozenberg
Aug 8 at 13:37
add a comment |Â
up vote
20
down vote
accepted
Let $a_2015=a_1.$
Thus, by AM-GM
$$sum_k=1^2014sina_kcosa_k+1 leqsum_k=1^2014|sina_k||cosa_k+1| le$$
$$leqsum_k=1^2014fracsin^2a_k+cos^2a_k+12=sum_k=1^2014fracsin^2a_k+cos^2a_k2=frac20142=1007.$$
The equality occurs for $a_i=45^circ,$ which says that $1007$ is a maximal value.
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more?
– Daniel Wagner
Aug 8 at 13:19
@Daniel Wagner I used $ableqfraca^2+b^22.$
– Michael Rozenberg
Aug 8 at 13:32
Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification.
– Daniel Wagner
Aug 8 at 13:35
You are welcome!
– Michael Rozenberg
Aug 8 at 13:37
add a comment |Â
up vote
20
down vote
accepted
up vote
20
down vote
accepted
Let $a_2015=a_1.$
Thus, by AM-GM
$$sum_k=1^2014sina_kcosa_k+1 leqsum_k=1^2014|sina_k||cosa_k+1| le$$
$$leqsum_k=1^2014fracsin^2a_k+cos^2a_k+12=sum_k=1^2014fracsin^2a_k+cos^2a_k2=frac20142=1007.$$
The equality occurs for $a_i=45^circ,$ which says that $1007$ is a maximal value.
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
Let $a_2015=a_1.$
Thus, by AM-GM
$$sum_k=1^2014sina_kcosa_k+1 leqsum_k=1^2014|sina_k||cosa_k+1| le$$
$$leqsum_k=1^2014fracsin^2a_k+cos^2a_k+12=sum_k=1^2014fracsin^2a_k+cos^2a_k2=frac20142=1007.$$
The equality occurs for $a_i=45^circ,$ which says that $1007$ is a maximal value.
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
edited Aug 8 at 8:52
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
answered Aug 8 at 8:41
Michael Rozenberg
88.7k1579179
88.7k1579179
You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more?
– Daniel Wagner
Aug 8 at 13:19
@Daniel Wagner I used $ableqfraca^2+b^22.$
– Michael Rozenberg
Aug 8 at 13:32
Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification.
– Daniel Wagner
Aug 8 at 13:35
You are welcome!
– Michael Rozenberg
Aug 8 at 13:37
add a comment |Â
You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more?
– Daniel Wagner
Aug 8 at 13:19
@Daniel Wagner I used $ableqfraca^2+b^22.$
– Michael Rozenberg
Aug 8 at 13:32
Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification.
– Daniel Wagner
Aug 8 at 13:35
You are welcome!
– Michael Rozenberg
Aug 8 at 13:37
You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more?
– Daniel Wagner
Aug 8 at 13:19
You mention AM-GM, which makes me expect to see a big sum turn into a big product at some point. But it never happens. Something clever is clearly going on in the third inequality -- but what? Could you expand this just a bit more?
– Daniel Wagner
Aug 8 at 13:19
@Daniel Wagner I used $ableqfraca^2+b^22.$
– Michael Rozenberg
Aug 8 at 13:32
@Daniel Wagner I used $ableqfraca^2+b^22.$
– Michael Rozenberg
Aug 8 at 13:32
Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification.
– Daniel Wagner
Aug 8 at 13:35
Ah! Very simple -- I kept trying to figure out where the calculation of a 2014th root went. Thanks for the clarification.
– Daniel Wagner
Aug 8 at 13:35
You are welcome!
– Michael Rozenberg
Aug 8 at 13:37
You are welcome!
– Michael Rozenberg
Aug 8 at 13:37
add a comment |Â
up vote
4
down vote
From
$$
f_n(a) = sum_k=1^n sin a_k cos a_k+1
$$
with $a_n+1 = a_1$
the stationary points are located at the solutions for
$$
fracpartial partial a_kf_n(a) = -sin a_k-1sin a_k + cos a_k cos a_k+1 = 0
$$
and then
$$
tan a_ntan a_n-1cdotstan a_2 = cot a_1
$$
or
$$
tan a_ntan a_n-1cdotstan a_2tan a_1 = 1
$$
or
$$
prod_ksin a_k = prod_kcos a_k
$$
which is obtained for $a_k = fracpi4$ when
$$
f_n(a) = frac n2
$$
answered Aug 8 at 13:31
Cesareo
5,8782412
Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=fracpi4$ is the absolute maximum? The $tan a_ntan a_n-1cdotstan a_2tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum...
– Taladris
Aug 8 at 13:49
@Taladris There are many relative extrema. For instance for $n = 3$ we have: $left-frac32,frac32,-sqrt2,sqrt2right$ but with some additional considerations like symmetry...
– Cesareo
Aug 8 at 14:44
add a comment |Â
up vote
4
down vote
From
$$
f_n(a) = sum_k=1^n sin a_k cos a_k+1
$$
with $a_n+1 = a_1$
the stationary points are located at the solutions for
$$
fracpartial partial a_kf_n(a) = -sin a_k-1sin a_k + cos a_k cos a_k+1 = 0
$$
and then
$$
tan a_ntan a_n-1cdotstan a_2 = cot a_1
$$
or
$$
tan a_ntan a_n-1cdotstan a_2tan a_1 = 1
$$
or
$$
prod_ksin a_k = prod_kcos a_k
$$
which is obtained for $a_k = fracpi4$ when
$$
f_n(a) = frac n2
$$
answered Aug 8 at 13:31
Cesareo
5,8782412
Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=fracpi4$ is the absolute maximum? The $tan a_ntan a_n-1cdotstan a_2tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum...
– Taladris
Aug 8 at 13:49
@Taladris There are many relative extrema. For instance for $n = 3$ we have: $left-frac32,frac32,-sqrt2,sqrt2right$ but with some additional considerations like symmetry...
– Cesareo
Aug 8 at 14:44
add a comment |Â
up vote
4
down vote
up vote
4
down vote
From
$$
f_n(a) = sum_k=1^n sin a_k cos a_k+1
$$
with $a_n+1 = a_1$
the stationary points are located at the solutions for
$$
fracpartial partial a_kf_n(a) = -sin a_k-1sin a_k + cos a_k cos a_k+1 = 0
$$
and then
$$
tan a_ntan a_n-1cdotstan a_2 = cot a_1
$$
or
$$
tan a_ntan a_n-1cdotstan a_2tan a_1 = 1
$$
or
$$
prod_ksin a_k = prod_kcos a_k
$$
which is obtained for $a_k = fracpi4$ when
$$
f_n(a) = frac n2
$$
answered Aug 8 at 13:31
Cesareo
5,8782412
From
$$
f_n(a) = sum_k=1^n sin a_k cos a_k+1
$$
with $a_n+1 = a_1$
the stationary points are located at the solutions for
$$
fracpartial partial a_kf_n(a) = -sin a_k-1sin a_k + cos a_k cos a_k+1 = 0
$$
and then
$$
tan a_ntan a_n-1cdotstan a_2 = cot a_1
$$
or
$$
tan a_ntan a_n-1cdotstan a_2tan a_1 = 1
$$
or
$$
prod_ksin a_k = prod_kcos a_k
$$
which is obtained for $a_k = fracpi4$ when
$$
f_n(a) = frac n2
$$
answered Aug 8 at 13:31
Cesareo
5,8782412
answered Aug 8 at 13:31
Cesareo
5,8782412
answered Aug 8 at 13:31
Cesareo
5,8782412
answered Aug 8 at 13:31
Cesareo
5,8782412
5,8782412
Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=fracpi4$ is the absolute maximum? The $tan a_ntan a_n-1cdotstan a_2tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum...
– Taladris
Aug 8 at 13:49
@Taladris There are many relative extrema. For instance for $n = 3$ we have: $left-frac32,frac32,-sqrt2,sqrt2right$ but with some additional considerations like symmetry...
– Cesareo
Aug 8 at 14:44
add a comment |Â
Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=fracpi4$ is the absolute maximum? The $tan a_ntan a_n-1cdotstan a_2tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum...
– Taladris
Aug 8 at 13:49
@Taladris There are many relative extrema. For instance for $n = 3$ we have: $left-frac32,frac32,-sqrt2,sqrt2right$ but with some additional considerations like symmetry...
– Cesareo
Aug 8 at 14:44
Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=fracpi4$ is the absolute maximum? The $tan a_ntan a_n-1cdotstan a_2tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum...
– Taladris
Aug 8 at 13:49
Nice answer. The calculus is easier than it looks at first. Is it obvious that $a_k=fracpi4$ is the absolute maximum? The $tan a_ntan a_n-1cdotstan a_2tan a_1 = 1$ may have many solutions but I guess they are not all absolute maximum...
– Taladris
Aug 8 at 13:49
@Taladris There are many relative extrema. For instance for $n = 3$ we have: $left-frac32,frac32,-sqrt2,sqrt2right$ but with some additional considerations like symmetry...
– Cesareo
Aug 8 at 14:44
@Taladris There are many relative extrema. For instance for $n = 3$ we have: $left-frac32,frac32,-sqrt2,sqrt2right$ but with some additional considerations like symmetry...
– Cesareo
Aug 8 at 14:44
add a comment |Â
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1
You cannot have $sin (a_1)=cos (a_1)=1$ so the maximum is not 2014.
– Kavi Rama Murthy
Aug 8 at 8:24
1
Fairly obviously it won't exceed $2014$ but it does not necessarily achieve it either.
– badjohn
Aug 8 at 8:24