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Prove inequality $
x^2-2xcos(x)+1>0
$ [closed]

-2

1

$begingroup$

How can we prove that
$$
x^2-2xcos(x)+1>0
$$
for all real $x$

real-analysis trigonometry inequality

share|cite|improve this question

asked Apr 1 at 13:07

avan1235avan1235

3628

$endgroup$

closed as off-topic by Martin R, Umberto P., José Carlos Santos, Carl Mummert, John Omielan Apr 1 at 20:09

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." – Martin R, Umberto P., José Carlos Santos, Carl Mummert, John Omielan

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

add a comment | 

-2

1

$begingroup$

How can we prove that
$$
x^2-2xcos(x)+1>0
$$
for all real $x$

real-analysis trigonometry inequality

share|cite|improve this question

asked Apr 1 at 13:07

avan1235avan1235

3628

$endgroup$

closed as off-topic by Martin R, Umberto P., José Carlos Santos, Carl Mummert, John Omielan Apr 1 at 20:09

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." – Martin R, Umberto P., José Carlos Santos, Carl Mummert, John Omielan

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

add a comment | 

$begingroup$

How can we prove that
$$
x^2-2xcos(x)+1>0
$$
for all real $x$

real-analysis trigonometry inequality

share|cite|improve this question

asked Apr 1 at 13:07

avan1235avan1235

3628

$endgroup$

share|cite|improve this question

asked Apr 1 at 13:07

avan1235avan1235

3628

share|cite|improve this question

asked Apr 1 at 13:07

avan1235avan1235

3628

asked Apr 1 at 13:07

avan1235avan1235

3628

asked Apr 1 at 13:07

avan1235avan1235

3628

3 Answers
3

active

oldest

votes

6

$begingroup$

HINT

Complete the the square for $$x^2 -2xcos x$$

$$(x- cos x)^2 - cos^2 x + 1$$
$$(x- cos x)^2 + sin^2 x$$ which is always greater than $0$?

share|cite|improve this answer

edited Apr 1 at 13:19

Dbchatto67

3,207625

answered Apr 1 at 13:10

AvkaAvka

1107

$endgroup$

add a comment | 

5

$begingroup$

$x^2-2x cos(x) + 1=(x-cos(x))^2+1-cos^2(x)=(x-cos(x))^2+sin^2xgeq0$

Further, it is easy to see that when $x=cos x$, $sin xneq 0$. So, the equality never holds. So $x^2-2x cos(x) + 1>0$.

share|cite|improve this answer

answered Apr 1 at 13:11

Holding ArthurHolding Arthur

1,753417

$endgroup$

add a comment | 

0

$begingroup$

Well, there is also AM-GM : $x^2+1geqslant 2|x|geqslant 2|xcos x|geqslant 2xcos x$. As equality is not possible ($x^2=1$ and $|cos x|=1$ cannot simultaneously be true), the inequality is strict.

share|cite|improve this answer

answered Apr 1 at 14:05

MacavityMacavity

35.7k52554

$endgroup$

add a comment | 

6

$begingroup$

HINT

Complete the the square for $$x^2 -2xcos x$$

$$(x- cos x)^2 - cos^2 x + 1$$
$$(x- cos x)^2 + sin^2 x$$ which is always greater than $0$?

share|cite|improve this answer

edited Apr 1 at 13:19

Dbchatto67

3,207625

answered Apr 1 at 13:10

AvkaAvka

1107

$endgroup$

add a comment | 

6

$begingroup$

HINT

Complete the the square for $$x^2 -2xcos x$$

$$(x- cos x)^2 - cos^2 x + 1$$
$$(x- cos x)^2 + sin^2 x$$ which is always greater than $0$?

share|cite|improve this answer

edited Apr 1 at 13:19

Dbchatto67

3,207625

answered Apr 1 at 13:10

AvkaAvka

1107

$endgroup$

add a comment | 

$begingroup$

HINT

Complete the the square for $$x^2 -2xcos x$$

$$(x- cos x)^2 - cos^2 x + 1$$
$$(x- cos x)^2 + sin^2 x$$ which is always greater than $0$?

share|cite|improve this answer

edited Apr 1 at 13:19

Dbchatto67

3,207625

answered Apr 1 at 13:10

AvkaAvka

1107

$endgroup$

share|cite|improve this answer

edited Apr 1 at 13:19

Dbchatto67

3,207625

answered Apr 1 at 13:10

AvkaAvka

1107

edited Apr 1 at 13:19

Dbchatto67

3,207625

edited Apr 1 at 13:19

Dbchatto67

3,207625

answered Apr 1 at 13:10

AvkaAvka

1107

answered Apr 1 at 13:10

AvkaAvka

1107

5

$begingroup$

$x^2-2x cos(x) + 1=(x-cos(x))^2+1-cos^2(x)=(x-cos(x))^2+sin^2xgeq0$

Further, it is easy to see that when $x=cos x$, $sin xneq 0$. So, the equality never holds. So $x^2-2x cos(x) + 1>0$.

share|cite|improve this answer

answered Apr 1 at 13:11

Holding ArthurHolding Arthur

1,753417

$endgroup$

add a comment | 

5

$begingroup$

$x^2-2x cos(x) + 1=(x-cos(x))^2+1-cos^2(x)=(x-cos(x))^2+sin^2xgeq0$

Further, it is easy to see that when $x=cos x$, $sin xneq 0$. So, the equality never holds. So $x^2-2x cos(x) + 1>0$.

share|cite|improve this answer

answered Apr 1 at 13:11

Holding ArthurHolding Arthur

1,753417

$endgroup$

add a comment | 

$begingroup$

$x^2-2x cos(x) + 1=(x-cos(x))^2+1-cos^2(x)=(x-cos(x))^2+sin^2xgeq0$

Further, it is easy to see that when $x=cos x$, $sin xneq 0$. So, the equality never holds. So $x^2-2x cos(x) + 1>0$.

share|cite|improve this answer

answered Apr 1 at 13:11

Holding ArthurHolding Arthur

1,753417

$endgroup$

share|cite|improve this answer

answered Apr 1 at 13:11

Holding ArthurHolding Arthur

1,753417

answered Apr 1 at 13:11

Holding ArthurHolding Arthur

1,753417

answered Apr 1 at 13:11

Holding ArthurHolding Arthur

1,753417

0

$begingroup$

Well, there is also AM-GM : $x^2+1geqslant 2|x|geqslant 2|xcos x|geqslant 2xcos x$. As equality is not possible ($x^2=1$ and $|cos x|=1$ cannot simultaneously be true), the inequality is strict.

share|cite|improve this answer

answered Apr 1 at 14:05

MacavityMacavity

35.7k52554

$endgroup$

add a comment | 

0

$begingroup$

Well, there is also AM-GM : $x^2+1geqslant 2|x|geqslant 2|xcos x|geqslant 2xcos x$. As equality is not possible ($x^2=1$ and $|cos x|=1$ cannot simultaneously be true), the inequality is strict.

share|cite|improve this answer

answered Apr 1 at 14:05

MacavityMacavity

35.7k52554

$endgroup$

add a comment | 

$begingroup$

Well, there is also AM-GM : $x^2+1geqslant 2|x|geqslant 2|xcos x|geqslant 2xcos x$. As equality is not possible ($x^2=1$ and $|cos x|=1$ cannot simultaneously be true), the inequality is strict.

share|cite|improve this answer

answered Apr 1 at 14:05

MacavityMacavity

35.7k52554

$endgroup$

share|cite|improve this answer

answered Apr 1 at 14:05

MacavityMacavity

35.7k52554

answered Apr 1 at 14:05

MacavityMacavity

35.7k52554

answered Apr 1 at 14:05

MacavityMacavity

35.7k52554