How to prove $sin(x)arccos(x) + xcos(x) > 0$ [closed]
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How to solve this: $sin(x)arccos(x) + xcos(x) > 0$? I have drawn some graphs and investigated that $0<xleq1$. But how to prove it?
algebra-precalculus trigonometry
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
$endgroup$
closed as off-topic by Saad, user10354138, Gibbs, Davide Giraudo, amWhy Dec 20 '18 at 23:03
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." – Saad, user10354138, Gibbs, Davide Giraudo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to solve this: $sin(x)arccos(x) + xcos(x) > 0$? I have drawn some graphs and investigated that $0<xleq1$. But how to prove it?
algebra-precalculus trigonometry
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
$endgroup$
closed as off-topic by Saad, user10354138, Gibbs, Davide Giraudo, amWhy Dec 20 '18 at 23:03
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." – Saad, user10354138, Gibbs, Davide Giraudo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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Do you mean $text{acos}(x)=arccos(x)$?
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– Yadati Kiran
Dec 20 '18 at 7:05
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@YadatiKiran Yes!
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– Tortik2020
Dec 20 '18 at 7:06
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There must be an interval in which this inequality holds, put $x = -4pi$ and expression becomes negative. In what interval this inequality holds? I mean is the problem statement complete
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– K.K.McDonald
Dec 20 '18 at 7:13
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@K.K.McDonald. Do you want to compute $arccos(-4pi)$ ?
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– Claude Leibovici
Dec 20 '18 at 7:16
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Can any part of this expression be negative in $[0,1]$?
$endgroup$
– A.Γ.
Dec 20 '18 at 7:29
add a comment |
$begingroup$
How to solve this: $sin(x)arccos(x) + xcos(x) > 0$? I have drawn some graphs and investigated that $0<xleq1$. But how to prove it?
algebra-precalculus trigonometry
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
$endgroup$
How to solve this: $sin(x)arccos(x) + xcos(x) > 0$? I have drawn some graphs and investigated that $0<xleq1$. But how to prove it?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
edited Dec 20 '18 at 7:18
A.Γ.
22.8k32656
22.8k32656
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
asked Dec 20 '18 at 7:04
Tortik2020Tortik2020
42
42
closed as off-topic by Saad, user10354138, Gibbs, Davide Giraudo, amWhy Dec 20 '18 at 23:03
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." – Saad, user10354138, Gibbs, Davide Giraudo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, user10354138, Gibbs, Davide Giraudo, amWhy Dec 20 '18 at 23:03
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." – Saad, user10354138, Gibbs, Davide Giraudo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Do you mean $text{acos}(x)=arccos(x)$?
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– Yadati Kiran
Dec 20 '18 at 7:05
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@YadatiKiran Yes!
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– Tortik2020
Dec 20 '18 at 7:06
$begingroup$
There must be an interval in which this inequality holds, put $x = -4pi$ and expression becomes negative. In what interval this inequality holds? I mean is the problem statement complete
$endgroup$
– K.K.McDonald
Dec 20 '18 at 7:13
$begingroup$
@K.K.McDonald. Do you want to compute $arccos(-4pi)$ ?
$endgroup$
– Claude Leibovici
Dec 20 '18 at 7:16
$begingroup$
Can any part of this expression be negative in $[0,1]$?
$endgroup$
– A.Γ.
Dec 20 '18 at 7:29
add a comment |
$begingroup$
Do you mean $text{acos}(x)=arccos(x)$?
$endgroup$
– Yadati Kiran
Dec 20 '18 at 7:05
$begingroup$
@YadatiKiran Yes!
$endgroup$
– Tortik2020
Dec 20 '18 at 7:06
$begingroup$
There must be an interval in which this inequality holds, put $x = -4pi$ and expression becomes negative. In what interval this inequality holds? I mean is the problem statement complete
$endgroup$
– K.K.McDonald
Dec 20 '18 at 7:13
$begingroup$
@K.K.McDonald. Do you want to compute $arccos(-4pi)$ ?
$endgroup$
– Claude Leibovici
Dec 20 '18 at 7:16
$begingroup$
Can any part of this expression be negative in $[0,1]$?
$endgroup$
– A.Γ.
Dec 20 '18 at 7:29
$begingroup$
Do you mean $text{acos}(x)=arccos(x)$?
$endgroup$
– Yadati Kiran
Dec 20 '18 at 7:05
$begingroup$
Do you mean $text{acos}(x)=arccos(x)$?
$endgroup$
– Yadati Kiran
Dec 20 '18 at 7:05
$begingroup$
@YadatiKiran Yes!
$endgroup$
– Tortik2020
Dec 20 '18 at 7:06
$begingroup$
@YadatiKiran Yes!
$endgroup$
– Tortik2020
Dec 20 '18 at 7:06
$begingroup$
There must be an interval in which this inequality holds, put $x = -4pi$ and expression becomes negative. In what interval this inequality holds? I mean is the problem statement complete
$endgroup$
– K.K.McDonald
Dec 20 '18 at 7:13
$begingroup$
There must be an interval in which this inequality holds, put $x = -4pi$ and expression becomes negative. In what interval this inequality holds? I mean is the problem statement complete
$endgroup$
– K.K.McDonald
Dec 20 '18 at 7:13
$begingroup$
@K.K.McDonald. Do you want to compute $arccos(-4pi)$ ?
$endgroup$
– Claude Leibovici
Dec 20 '18 at 7:16
$begingroup$
@K.K.McDonald. Do you want to compute $arccos(-4pi)$ ?
$endgroup$
– Claude Leibovici
Dec 20 '18 at 7:16
$begingroup$
Can any part of this expression be negative in $[0,1]$?
$endgroup$
– A.Γ.
Dec 20 '18 at 7:29
$begingroup$
Can any part of this expression be negative in $[0,1]$?
$endgroup$
– A.Γ.
Dec 20 '18 at 7:29
add a comment |
2 Answers
2
active
oldest
votes
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The domain here is $[-1,1]$. Clearly, when $x = 0$, the left hand side evaluates to $0$, so it doesn't hold.
When $xin[-1,0)$, we have $sin(x) < 0$ and $arccos(x) > 0$, so the first term is negative; similarly, $x < 0$ and $cos(x) > 0$, so the second term is negative. Thus the left hand side is negative, and so it doesn't satisfy the inequality.
Finally, when $xin(0,1]$, we have $sin(x),cos(x),x > 0$ and $arccos(x) geq 0$, so the left hand side must be positive, thus satisfying the inequality.
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
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1
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@egreg I don't think this is quite true, as $arccos$ is not an even function.
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– AlexanderJ93
Dec 20 '18 at 9:50
add a comment |
$begingroup$
Not a sufficient proof.
Because of the $cos ^{-1}(x)$, the domain is limited to $-1 leq x leq 1$. Using a Taylor expansion built around $x=0$, we have
$$f(x)=sin (x) cos ^{-1}(x)+x cos (x)=left(1+frac{pi }{2}right) x-x^2-left(frac{1}{2}+frac{pi }{12}right) x^3+Oleft(x^5right)$$ and the rhs does not show, in the considered interval, any root except the trivial $x=0$. The rhs goes through a maximum at $$x_*=frac{sqrt{160+64 pi +8 pi ^2}-8}{12+2 pi }approx 0.709757 implies f(x_*) approx 1.04772$$ and $f(1)=cos(1) >0$.
For the fun of it, plot the function and the Taylor series on the same graph.
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
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1
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i like the "not a sufficient proof" part, but this builds up pretty well ;) +1
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– Ahmad Bazzi
Dec 20 '18 at 7:37
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The “domain” is limited, rather than the range.
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– egreg
Dec 20 '18 at 9:26
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@egreg. Thanks for pointing out my bad English ! Cheers :-(
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– Claude Leibovici
Dec 20 '18 at 9:27
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False friends? :-)
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– egreg
Dec 20 '18 at 9:30
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@egreg. At the time, in French, we did not use to make the distinction. The key problem is that this was almost 65 years ago and probably I did not update my vocabulary. Cheers and now :-)
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:33
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The domain here is $[-1,1]$. Clearly, when $x = 0$, the left hand side evaluates to $0$, so it doesn't hold.
When $xin[-1,0)$, we have $sin(x) < 0$ and $arccos(x) > 0$, so the first term is negative; similarly, $x < 0$ and $cos(x) > 0$, so the second term is negative. Thus the left hand side is negative, and so it doesn't satisfy the inequality.
Finally, when $xin(0,1]$, we have $sin(x),cos(x),x > 0$ and $arccos(x) geq 0$, so the left hand side must be positive, thus satisfying the inequality.
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
$endgroup$
1
$begingroup$
@egreg I don't think this is quite true, as $arccos$ is not an even function.
$endgroup$
– AlexanderJ93
Dec 20 '18 at 9:50
add a comment |
$begingroup$
The domain here is $[-1,1]$. Clearly, when $x = 0$, the left hand side evaluates to $0$, so it doesn't hold.
When $xin[-1,0)$, we have $sin(x) < 0$ and $arccos(x) > 0$, so the first term is negative; similarly, $x < 0$ and $cos(x) > 0$, so the second term is negative. Thus the left hand side is negative, and so it doesn't satisfy the inequality.
Finally, when $xin(0,1]$, we have $sin(x),cos(x),x > 0$ and $arccos(x) geq 0$, so the left hand side must be positive, thus satisfying the inequality.
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
$endgroup$
1
$begingroup$
@egreg I don't think this is quite true, as $arccos$ is not an even function.
$endgroup$
– AlexanderJ93
Dec 20 '18 at 9:50
add a comment |
$begingroup$
The domain here is $[-1,1]$. Clearly, when $x = 0$, the left hand side evaluates to $0$, so it doesn't hold.
When $xin[-1,0)$, we have $sin(x) < 0$ and $arccos(x) > 0$, so the first term is negative; similarly, $x < 0$ and $cos(x) > 0$, so the second term is negative. Thus the left hand side is negative, and so it doesn't satisfy the inequality.
Finally, when $xin(0,1]$, we have $sin(x),cos(x),x > 0$ and $arccos(x) geq 0$, so the left hand side must be positive, thus satisfying the inequality.
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
$endgroup$
The domain here is $[-1,1]$. Clearly, when $x = 0$, the left hand side evaluates to $0$, so it doesn't hold.
When $xin[-1,0)$, we have $sin(x) < 0$ and $arccos(x) > 0$, so the first term is negative; similarly, $x < 0$ and $cos(x) > 0$, so the second term is negative. Thus the left hand side is negative, and so it doesn't satisfy the inequality.
Finally, when $xin(0,1]$, we have $sin(x),cos(x),x > 0$ and $arccos(x) geq 0$, so the left hand side must be positive, thus satisfying the inequality.
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
answered Dec 20 '18 at 9:09
AlexanderJ93AlexanderJ93
6,173823
6,173823
1
$begingroup$
@egreg I don't think this is quite true, as $arccos$ is not an even function.
$endgroup$
– AlexanderJ93
Dec 20 '18 at 9:50
add a comment |
1
$begingroup$
@egreg I don't think this is quite true, as $arccos$ is not an even function.
$endgroup$
– AlexanderJ93
Dec 20 '18 at 9:50
1
1
$begingroup$
@egreg I don't think this is quite true, as $arccos$ is not an even function.
$endgroup$
– AlexanderJ93
Dec 20 '18 at 9:50
$begingroup$
@egreg I don't think this is quite true, as $arccos$ is not an even function.
$endgroup$
– AlexanderJ93
Dec 20 '18 at 9:50
add a comment |
$begingroup$
Not a sufficient proof.
Because of the $cos ^{-1}(x)$, the domain is limited to $-1 leq x leq 1$. Using a Taylor expansion built around $x=0$, we have
$$f(x)=sin (x) cos ^{-1}(x)+x cos (x)=left(1+frac{pi }{2}right) x-x^2-left(frac{1}{2}+frac{pi }{12}right) x^3+Oleft(x^5right)$$ and the rhs does not show, in the considered interval, any root except the trivial $x=0$. The rhs goes through a maximum at $$x_*=frac{sqrt{160+64 pi +8 pi ^2}-8}{12+2 pi }approx 0.709757 implies f(x_*) approx 1.04772$$ and $f(1)=cos(1) >0$.
For the fun of it, plot the function and the Taylor series on the same graph.
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
1
$begingroup$
i like the "not a sufficient proof" part, but this builds up pretty well ;) +1
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– Ahmad Bazzi
Dec 20 '18 at 7:37
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The “domain” is limited, rather than the range.
$endgroup$
– egreg
Dec 20 '18 at 9:26
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@egreg. Thanks for pointing out my bad English ! Cheers :-(
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:27
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False friends? :-)
$endgroup$
– egreg
Dec 20 '18 at 9:30
$begingroup$
@egreg. At the time, in French, we did not use to make the distinction. The key problem is that this was almost 65 years ago and probably I did not update my vocabulary. Cheers and now :-)
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:33
add a comment |
$begingroup$
Not a sufficient proof.
Because of the $cos ^{-1}(x)$, the domain is limited to $-1 leq x leq 1$. Using a Taylor expansion built around $x=0$, we have
$$f(x)=sin (x) cos ^{-1}(x)+x cos (x)=left(1+frac{pi }{2}right) x-x^2-left(frac{1}{2}+frac{pi }{12}right) x^3+Oleft(x^5right)$$ and the rhs does not show, in the considered interval, any root except the trivial $x=0$. The rhs goes through a maximum at $$x_*=frac{sqrt{160+64 pi +8 pi ^2}-8}{12+2 pi }approx 0.709757 implies f(x_*) approx 1.04772$$ and $f(1)=cos(1) >0$.
For the fun of it, plot the function and the Taylor series on the same graph.
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
1
$begingroup$
i like the "not a sufficient proof" part, but this builds up pretty well ;) +1
$endgroup$
– Ahmad Bazzi
Dec 20 '18 at 7:37
$begingroup$
The “domain” is limited, rather than the range.
$endgroup$
– egreg
Dec 20 '18 at 9:26
$begingroup$
@egreg. Thanks for pointing out my bad English ! Cheers :-(
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:27
$begingroup$
False friends? :-)
$endgroup$
– egreg
Dec 20 '18 at 9:30
$begingroup$
@egreg. At the time, in French, we did not use to make the distinction. The key problem is that this was almost 65 years ago and probably I did not update my vocabulary. Cheers and now :-)
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:33
add a comment |
$begingroup$
Not a sufficient proof.
Because of the $cos ^{-1}(x)$, the domain is limited to $-1 leq x leq 1$. Using a Taylor expansion built around $x=0$, we have
$$f(x)=sin (x) cos ^{-1}(x)+x cos (x)=left(1+frac{pi }{2}right) x-x^2-left(frac{1}{2}+frac{pi }{12}right) x^3+Oleft(x^5right)$$ and the rhs does not show, in the considered interval, any root except the trivial $x=0$. The rhs goes through a maximum at $$x_*=frac{sqrt{160+64 pi +8 pi ^2}-8}{12+2 pi }approx 0.709757 implies f(x_*) approx 1.04772$$ and $f(1)=cos(1) >0$.
For the fun of it, plot the function and the Taylor series on the same graph.
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
Not a sufficient proof.
Because of the $cos ^{-1}(x)$, the domain is limited to $-1 leq x leq 1$. Using a Taylor expansion built around $x=0$, we have
$$f(x)=sin (x) cos ^{-1}(x)+x cos (x)=left(1+frac{pi }{2}right) x-x^2-left(frac{1}{2}+frac{pi }{12}right) x^3+Oleft(x^5right)$$ and the rhs does not show, in the considered interval, any root except the trivial $x=0$. The rhs goes through a maximum at $$x_*=frac{sqrt{160+64 pi +8 pi ^2}-8}{12+2 pi }approx 0.709757 implies f(x_*) approx 1.04772$$ and $f(1)=cos(1) >0$.
For the fun of it, plot the function and the Taylor series on the same graph.
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
edited Dec 20 '18 at 9:28
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
answered Dec 20 '18 at 7:34
Claude LeiboviciClaude Leibovici
123k1157135
123k1157135
1
$begingroup$
i like the "not a sufficient proof" part, but this builds up pretty well ;) +1
$endgroup$
– Ahmad Bazzi
Dec 20 '18 at 7:37
$begingroup$
The “domain” is limited, rather than the range.
$endgroup$
– egreg
Dec 20 '18 at 9:26
$begingroup$
@egreg. Thanks for pointing out my bad English ! Cheers :-(
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:27
$begingroup$
False friends? :-)
$endgroup$
– egreg
Dec 20 '18 at 9:30
$begingroup$
@egreg. At the time, in French, we did not use to make the distinction. The key problem is that this was almost 65 years ago and probably I did not update my vocabulary. Cheers and now :-)
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:33
add a comment |
1
$begingroup$
i like the "not a sufficient proof" part, but this builds up pretty well ;) +1
$endgroup$
– Ahmad Bazzi
Dec 20 '18 at 7:37
$begingroup$
The “domain” is limited, rather than the range.
$endgroup$
– egreg
Dec 20 '18 at 9:26
$begingroup$
@egreg. Thanks for pointing out my bad English ! Cheers :-(
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:27
$begingroup$
False friends? :-)
$endgroup$
– egreg
Dec 20 '18 at 9:30
$begingroup$
@egreg. At the time, in French, we did not use to make the distinction. The key problem is that this was almost 65 years ago and probably I did not update my vocabulary. Cheers and now :-)
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:33
1
1
$begingroup$
i like the "not a sufficient proof" part, but this builds up pretty well ;) +1
$endgroup$
– Ahmad Bazzi
Dec 20 '18 at 7:37
$begingroup$
i like the "not a sufficient proof" part, but this builds up pretty well ;) +1
$endgroup$
– Ahmad Bazzi
Dec 20 '18 at 7:37
$begingroup$
The “domain” is limited, rather than the range.
$endgroup$
– egreg
Dec 20 '18 at 9:26
$begingroup$
The “domain” is limited, rather than the range.
$endgroup$
– egreg
Dec 20 '18 at 9:26
$begingroup$
@egreg. Thanks for pointing out my bad English ! Cheers :-(
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:27
$begingroup$
@egreg. Thanks for pointing out my bad English ! Cheers :-(
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:27
$begingroup$
False friends? :-)
$endgroup$
– egreg
Dec 20 '18 at 9:30
$begingroup$
False friends? :-)
$endgroup$
– egreg
Dec 20 '18 at 9:30
$begingroup$
@egreg. At the time, in French, we did not use to make the distinction. The key problem is that this was almost 65 years ago and probably I did not update my vocabulary. Cheers and now :-)
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:33
$begingroup$
@egreg. At the time, in French, we did not use to make the distinction. The key problem is that this was almost 65 years ago and probably I did not update my vocabulary. Cheers and now :-)
$endgroup$
– Claude Leibovici
Dec 20 '18 at 9:33
add a comment |
$begingroup$
Do you mean $text{acos}(x)=arccos(x)$?
$endgroup$
– Yadati Kiran
Dec 20 '18 at 7:05
$begingroup$
@YadatiKiran Yes!
$endgroup$
– Tortik2020
Dec 20 '18 at 7:06
$begingroup$
There must be an interval in which this inequality holds, put $x = -4pi$ and expression becomes negative. In what interval this inequality holds? I mean is the problem statement complete
$endgroup$
– K.K.McDonald
Dec 20 '18 at 7:13
$begingroup$
@K.K.McDonald. Do you want to compute $arccos(-4pi)$ ?
$endgroup$
– Claude Leibovici
Dec 20 '18 at 7:16
$begingroup$
Can any part of this expression be negative in $[0,1]$?
$endgroup$
– A.Γ.
Dec 20 '18 at 7:29