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What is the least value of $tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta$? [duplicate]
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Find the minimum value of following trigonometric expression.
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What is the least value of this expression?
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta$$
Will putting $theta=45^circ$ give right answer?
trigonometry
edited Jul 19 at 13:53
prog_SAHIL
850217
asked Jul 19 at 13:30
Dani
91
marked as duplicate by Winther, Blue trigonometry
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Jul 19 at 14:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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down vote
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This question already has an answer here:
Find the minimum value of following trigonometric expression.
6 answers
What is the least value of this expression?
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta$$
Will putting $theta=45^circ$ give right answer?
trigonometry
edited Jul 19 at 13:53
prog_SAHIL
850217
asked Jul 19 at 13:30
Dani
91
marked as duplicate by Winther, Blue trigonometry
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Jul 19 at 14:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Might help to simplify the expression (using $sin^2theta +cos^2 theta =1$ and so on).
– lulu
Jul 19 at 13:33
To address the question of $theta$ being 45 degrees: the symmetry of replacing $theta$ by $90-theta$ in your expression implies that 45 degrees is a critical point. But if you follow down this direction you still need to find out whether the critical point is in fact the global minimum. So while 45 degrees does give you the correct answer, you will not know that for sure unless you do more work.
– Willie Wong
Jul 19 at 13:59
add a comment |Â
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1
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up vote
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down vote
favorite
This question already has an answer here:
Find the minimum value of following trigonometric expression.
6 answers
What is the least value of this expression?
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta$$
Will putting $theta=45^circ$ give right answer?
trigonometry
edited Jul 19 at 13:53
prog_SAHIL
850217
asked Jul 19 at 13:30
Dani
91
This question already has an answer here:
Find the minimum value of following trigonometric expression.
6 answers
What is the least value of this expression?
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta$$
Will putting $theta=45^circ$ give right answer?
This question already has an answer here:
Find the minimum value of following trigonometric expression.
6 answers
trigonometry
edited Jul 19 at 13:53
prog_SAHIL
850217
asked Jul 19 at 13:30
Dani
91
edited Jul 19 at 13:53
prog_SAHIL
850217
edited Jul 19 at 13:53
prog_SAHIL
850217
edited Jul 19 at 13:53
prog_SAHIL
850217
850217
asked Jul 19 at 13:30
Dani
91
asked Jul 19 at 13:30
Dani
91
asked Jul 19 at 13:30
Dani
91
91
marked as duplicate by Winther, Blue trigonometry
Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.
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Jul 19 at 14:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Might help to simplify the expression (using $sin^2theta +cos^2 theta =1$ and so on).
– lulu
Jul 19 at 13:33
To address the question of $theta$ being 45 degrees: the symmetry of replacing $theta$ by $90-theta$ in your expression implies that 45 degrees is a critical point. But if you follow down this direction you still need to find out whether the critical point is in fact the global minimum. So while 45 degrees does give you the correct answer, you will not know that for sure unless you do more work.
– Willie Wong
Jul 19 at 13:59
add a comment |Â
1
Might help to simplify the expression (using $sin^2theta +cos^2 theta =1$ and so on).
– lulu
Jul 19 at 13:33
To address the question of $theta$ being 45 degrees: the symmetry of replacing $theta$ by $90-theta$ in your expression implies that 45 degrees is a critical point. But if you follow down this direction you still need to find out whether the critical point is in fact the global minimum. So while 45 degrees does give you the correct answer, you will not know that for sure unless you do more work.
– Willie Wong
Jul 19 at 13:59
1
1
Might help to simplify the expression (using $sin^2theta +cos^2 theta =1$ and so on).
– lulu
Jul 19 at 13:33
Might help to simplify the expression (using $sin^2theta +cos^2 theta =1$ and so on).
– lulu
Jul 19 at 13:33
To address the question of $theta$ being 45 degrees: the symmetry of replacing $theta$ by $90-theta$ in your expression implies that 45 degrees is a critical point. But if you follow down this direction you still need to find out whether the critical point is in fact the global minimum. So while 45 degrees does give you the correct answer, you will not know that for sure unless you do more work.
– Willie Wong
Jul 19 at 13:59
To address the question of $theta$ being 45 degrees: the symmetry of replacing $theta$ by $90-theta$ in your expression implies that 45 degrees is a critical point. But if you follow down this direction you still need to find out whether the critical point is in fact the global minimum. So while 45 degrees does give you the correct answer, you will not know that for sure unless you do more work.
– Willie Wong
Jul 19 at 13:59
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
Let $s=sin theta$ and $c=cos theta$, then
beginalign
E &= tan^2 theta+cot^2 theta+sin^2 theta+
cos^2 theta+sec^2 theta+csc^2 theta \
&= fracs^2c^2+fracc^2s^2+1+frac1c^2+frac1s^2 \
&= fracs^4+c^4+s^2c^2+s^2+c^2s^2c^2 \
&= frac(s^2+c^2)^2-s^2c^2+1s^2c^2 \
&= frac2-s^2c^2s^2c^2 \
&= frac2sin^2 theta cos^2 theta-1 \
&= frac8sin^2 2theta-1 \
& ge 7
endalign
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
1
As per Dani's conjecture, $theta=fracpi4$ minimises $E$.
– J.G.
Jul 19 at 13:56
@J.G. its kind of a flunk.
– prog_SAHIL
Jul 19 at 14:02
@prog_SAHIL I assume that means it's not surprising?
– J.G.
Jul 19 at 14:34
add a comment |Â
up vote
2
down vote
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta=3+2(tan^2theta+cot^2theta)$$
Now $tan^2theta+cot^2theta=(cottheta-tantheta)^2+2ge?$
the equality occurs if $cottheta-tantheta=0ifftan^2theta=1iffcos2theta=0iff2theta=(2m+1)dfracpi2$ where $m$ is any integer
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
Thanks , this is very easy explaination.
– Dani
Jul 19 at 14:33
add a comment |Â
up vote
1
down vote
One can also answer using the "guess" that the value is attained at $theta= pi/4$.
Guessing $theta$
The expression you care about is
- periodic with period $pi/2$; and
- symmetric when you replace $theta$ by $pi/2 - theta$.
So you just need to look at the range $theta in (0,pi/2)$; the endpoints are ruled out since your function is infinite there. The symmetry implies that $theta = pi/4$ is a critical point.
$pi/4$ is the only critical point
Observe that $sin^2theta + cos^2theta = 1$. Observe further that $sec, csc, tan,cot$ are all
- non-negative on $(0,pi/2)$
- convex on $(0,pi/2)$
these imply that $tan^2$ etc. are all convex functions on $(0,pi/2)$. Hence their sum is convex. A convex function can only have at most one critical point, and the critical point, if exists, is the global minimum.
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Let $s=sin theta$ and $c=cos theta$, then
beginalign
E &= tan^2 theta+cot^2 theta+sin^2 theta+
cos^2 theta+sec^2 theta+csc^2 theta \
&= fracs^2c^2+fracc^2s^2+1+frac1c^2+frac1s^2 \
&= fracs^4+c^4+s^2c^2+s^2+c^2s^2c^2 \
&= frac(s^2+c^2)^2-s^2c^2+1s^2c^2 \
&= frac2-s^2c^2s^2c^2 \
&= frac2sin^2 theta cos^2 theta-1 \
&= frac8sin^2 2theta-1 \
& ge 7
endalign
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
1
As per Dani's conjecture, $theta=fracpi4$ minimises $E$.
– J.G.
Jul 19 at 13:56
@J.G. its kind of a flunk.
– prog_SAHIL
Jul 19 at 14:02
@prog_SAHIL I assume that means it's not surprising?
– J.G.
Jul 19 at 14:34
add a comment |Â
up vote
4
down vote
Let $s=sin theta$ and $c=cos theta$, then
beginalign
E &= tan^2 theta+cot^2 theta+sin^2 theta+
cos^2 theta+sec^2 theta+csc^2 theta \
&= fracs^2c^2+fracc^2s^2+1+frac1c^2+frac1s^2 \
&= fracs^4+c^4+s^2c^2+s^2+c^2s^2c^2 \
&= frac(s^2+c^2)^2-s^2c^2+1s^2c^2 \
&= frac2-s^2c^2s^2c^2 \
&= frac2sin^2 theta cos^2 theta-1 \
&= frac8sin^2 2theta-1 \
& ge 7
endalign
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
1
As per Dani's conjecture, $theta=fracpi4$ minimises $E$.
– J.G.
Jul 19 at 13:56
@J.G. its kind of a flunk.
– prog_SAHIL
Jul 19 at 14:02
@prog_SAHIL I assume that means it's not surprising?
– J.G.
Jul 19 at 14:34
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let $s=sin theta$ and $c=cos theta$, then
beginalign
E &= tan^2 theta+cot^2 theta+sin^2 theta+
cos^2 theta+sec^2 theta+csc^2 theta \
&= fracs^2c^2+fracc^2s^2+1+frac1c^2+frac1s^2 \
&= fracs^4+c^4+s^2c^2+s^2+c^2s^2c^2 \
&= frac(s^2+c^2)^2-s^2c^2+1s^2c^2 \
&= frac2-s^2c^2s^2c^2 \
&= frac2sin^2 theta cos^2 theta-1 \
&= frac8sin^2 2theta-1 \
& ge 7
endalign
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
Let $s=sin theta$ and $c=cos theta$, then
beginalign
E &= tan^2 theta+cot^2 theta+sin^2 theta+
cos^2 theta+sec^2 theta+csc^2 theta \
&= fracs^2c^2+fracc^2s^2+1+frac1c^2+frac1s^2 \
&= fracs^4+c^4+s^2c^2+s^2+c^2s^2c^2 \
&= frac(s^2+c^2)^2-s^2c^2+1s^2c^2 \
&= frac2-s^2c^2s^2c^2 \
&= frac2sin^2 theta cos^2 theta-1 \
&= frac8sin^2 2theta-1 \
& ge 7
endalign
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
answered Jul 19 at 13:46
Ng Chung Tak
13k31130
13k31130
1
As per Dani's conjecture, $theta=fracpi4$ minimises $E$.
– J.G.
Jul 19 at 13:56
@J.G. its kind of a flunk.
– prog_SAHIL
Jul 19 at 14:02
@prog_SAHIL I assume that means it's not surprising?
– J.G.
Jul 19 at 14:34
add a comment |Â
1
As per Dani's conjecture, $theta=fracpi4$ minimises $E$.
– J.G.
Jul 19 at 13:56
@J.G. its kind of a flunk.
– prog_SAHIL
Jul 19 at 14:02
@prog_SAHIL I assume that means it's not surprising?
– J.G.
Jul 19 at 14:34
1
1
As per Dani's conjecture, $theta=fracpi4$ minimises $E$.
– J.G.
Jul 19 at 13:56
As per Dani's conjecture, $theta=fracpi4$ minimises $E$.
– J.G.
Jul 19 at 13:56
@J.G. its kind of a flunk.
– prog_SAHIL
Jul 19 at 14:02
@J.G. its kind of a flunk.
– prog_SAHIL
Jul 19 at 14:02
@prog_SAHIL I assume that means it's not surprising?
– J.G.
Jul 19 at 14:34
@prog_SAHIL I assume that means it's not surprising?
– J.G.
Jul 19 at 14:34
add a comment |Â
up vote
2
down vote
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta=3+2(tan^2theta+cot^2theta)$$
Now $tan^2theta+cot^2theta=(cottheta-tantheta)^2+2ge?$
the equality occurs if $cottheta-tantheta=0ifftan^2theta=1iffcos2theta=0iff2theta=(2m+1)dfracpi2$ where $m$ is any integer
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
Thanks , this is very easy explaination.
– Dani
Jul 19 at 14:33
add a comment |Â
up vote
2
down vote
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta=3+2(tan^2theta+cot^2theta)$$
Now $tan^2theta+cot^2theta=(cottheta-tantheta)^2+2ge?$
the equality occurs if $cottheta-tantheta=0ifftan^2theta=1iffcos2theta=0iff2theta=(2m+1)dfracpi2$ where $m$ is any integer
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
Thanks , this is very easy explaination.
– Dani
Jul 19 at 14:33
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta=3+2(tan^2theta+cot^2theta)$$
Now $tan^2theta+cot^2theta=(cottheta-tantheta)^2+2ge?$
the equality occurs if $cottheta-tantheta=0ifftan^2theta=1iffcos2theta=0iff2theta=(2m+1)dfracpi2$ where $m$ is any integer
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
$$tan^2 theta + cot^2 theta + sin^2 theta + cos^2 theta + sec^2 theta+ textrmcosec^2 theta=3+2(tan^2theta+cot^2theta)$$
Now $tan^2theta+cot^2theta=(cottheta-tantheta)^2+2ge?$
the equality occurs if $cottheta-tantheta=0ifftan^2theta=1iffcos2theta=0iff2theta=(2m+1)dfracpi2$ where $m$ is any integer
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
answered Jul 19 at 14:02
lab bhattacharjee
215k14152264
215k14152264
Thanks , this is very easy explaination.
– Dani
Jul 19 at 14:33
add a comment |Â
Thanks , this is very easy explaination.
– Dani
Jul 19 at 14:33
Thanks , this is very easy explaination.
– Dani
Jul 19 at 14:33
Thanks , this is very easy explaination.
– Dani
Jul 19 at 14:33
add a comment |Â
up vote
1
down vote
One can also answer using the "guess" that the value is attained at $theta= pi/4$.
Guessing $theta$
The expression you care about is
- periodic with period $pi/2$; and
- symmetric when you replace $theta$ by $pi/2 - theta$.
So you just need to look at the range $theta in (0,pi/2)$; the endpoints are ruled out since your function is infinite there. The symmetry implies that $theta = pi/4$ is a critical point.
$pi/4$ is the only critical point
Observe that $sin^2theta + cos^2theta = 1$. Observe further that $sec, csc, tan,cot$ are all
- non-negative on $(0,pi/2)$
- convex on $(0,pi/2)$
these imply that $tan^2$ etc. are all convex functions on $(0,pi/2)$. Hence their sum is convex. A convex function can only have at most one critical point, and the critical point, if exists, is the global minimum.
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
add a comment |Â
up vote
1
down vote
One can also answer using the "guess" that the value is attained at $theta= pi/4$.
Guessing $theta$
The expression you care about is
- periodic with period $pi/2$; and
- symmetric when you replace $theta$ by $pi/2 - theta$.
So you just need to look at the range $theta in (0,pi/2)$; the endpoints are ruled out since your function is infinite there. The symmetry implies that $theta = pi/4$ is a critical point.
$pi/4$ is the only critical point
Observe that $sin^2theta + cos^2theta = 1$. Observe further that $sec, csc, tan,cot$ are all
- non-negative on $(0,pi/2)$
- convex on $(0,pi/2)$
these imply that $tan^2$ etc. are all convex functions on $(0,pi/2)$. Hence their sum is convex. A convex function can only have at most one critical point, and the critical point, if exists, is the global minimum.
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One can also answer using the "guess" that the value is attained at $theta= pi/4$.
Guessing $theta$
The expression you care about is
- periodic with period $pi/2$; and
- symmetric when you replace $theta$ by $pi/2 - theta$.
So you just need to look at the range $theta in (0,pi/2)$; the endpoints are ruled out since your function is infinite there. The symmetry implies that $theta = pi/4$ is a critical point.
$pi/4$ is the only critical point
Observe that $sin^2theta + cos^2theta = 1$. Observe further that $sec, csc, tan,cot$ are all
- non-negative on $(0,pi/2)$
- convex on $(0,pi/2)$
these imply that $tan^2$ etc. are all convex functions on $(0,pi/2)$. Hence their sum is convex. A convex function can only have at most one critical point, and the critical point, if exists, is the global minimum.
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
One can also answer using the "guess" that the value is attained at $theta= pi/4$.
Guessing $theta$
The expression you care about is
- periodic with period $pi/2$; and
- symmetric when you replace $theta$ by $pi/2 - theta$.
So you just need to look at the range $theta in (0,pi/2)$; the endpoints are ruled out since your function is infinite there. The symmetry implies that $theta = pi/4$ is a critical point.
$pi/4$ is the only critical point
Observe that $sin^2theta + cos^2theta = 1$. Observe further that $sec, csc, tan,cot$ are all
- non-negative on $(0,pi/2)$
- convex on $(0,pi/2)$
these imply that $tan^2$ etc. are all convex functions on $(0,pi/2)$. Hence their sum is convex. A convex function can only have at most one critical point, and the critical point, if exists, is the global minimum.
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
answered Jul 19 at 14:10
Willie Wong
54.2k8103206
54.2k8103206
add a comment |Â
add a comment |Â
1
Might help to simplify the expression (using $sin^2theta +cos^2 theta =1$ and so on).
– lulu
Jul 19 at 13:33
To address the question of $theta$ being 45 degrees: the symmetry of replacing $theta$ by $90-theta$ in your expression implies that 45 degrees is a critical point. But if you follow down this direction you still need to find out whether the critical point is in fact the global minimum. So while 45 degrees does give you the correct answer, you will not know that for sure unless you do more work.
– Willie Wong
Jul 19 at 13:59