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Show the inequality $lvert sin(z)rvert > frac2pi$ for $z$ on the circle of radius $(n+1/2) pi$.

0

0

$begingroup$

I want to show the inequality $lvertsin(z)rvert > frac2pi$ for $z$ on the circle of radius $(n+1/2) pi$.
What I have gotten so far is using the fact that

$lvertsin(x+iy)rvert^2 = sin^2(x)+ sinh^2(y)$

to get

$sin^2((n+frac12)pi cos(theta)) + sinh^2((n+frac12)pi sin(theta)) > frac4pi^2$ for $theta in [0,2pi]$

I verified on a graphing calculator that such statement is true, but have not been able to present any proper argument.

Please help me, thanks in advance.

complex-analysis trigonometry upper-lower-bounds

share|cite|improve this question

edited Mar 31 at 21:59

egreg

186k1486208

asked Mar 31 at 19:23

William AmbroseWilliam Ambrose

213

$endgroup$

add a comment | 

0

0

$begingroup$

I want to show the inequality $lvertsin(z)rvert > frac2pi$ for $z$ on the circle of radius $(n+1/2) pi$.
What I have gotten so far is using the fact that

$lvertsin(x+iy)rvert^2 = sin^2(x)+ sinh^2(y)$

to get

$sin^2((n+frac12)pi cos(theta)) + sinh^2((n+frac12)pi sin(theta)) > frac4pi^2$ for $theta in [0,2pi]$

I verified on a graphing calculator that such statement is true, but have not been able to present any proper argument.

Please help me, thanks in advance.

complex-analysis trigonometry upper-lower-bounds

share|cite|improve this question

edited Mar 31 at 21:59

egreg

186k1486208

asked Mar 31 at 19:23

William AmbroseWilliam Ambrose

213

$endgroup$

add a comment | 

$begingroup$

I want to show the inequality $lvertsin(z)rvert > frac2pi$ for $z$ on the circle of radius $(n+1/2) pi$.
What I have gotten so far is using the fact that

$lvertsin(x+iy)rvert^2 = sin^2(x)+ sinh^2(y)$

to get

$sin^2((n+frac12)pi cos(theta)) + sinh^2((n+frac12)pi sin(theta)) > frac4pi^2$ for $theta in [0,2pi]$

I verified on a graphing calculator that such statement is true, but have not been able to present any proper argument.

Please help me, thanks in advance.

complex-analysis trigonometry upper-lower-bounds

share|cite|improve this question

edited Mar 31 at 21:59

egreg

186k1486208

asked Mar 31 at 19:23

William AmbroseWilliam Ambrose

213

$endgroup$

share|cite|improve this question

edited Mar 31 at 21:59

egreg

186k1486208

asked Mar 31 at 19:23

William AmbroseWilliam Ambrose

213

share|cite|improve this question

edited Mar 31 at 21:59

egreg

186k1486208

asked Mar 31 at 19:23

William AmbroseWilliam Ambrose

213

edited Mar 31 at 21:59

egreg

186k1486208

edited Mar 31 at 21:59

egreg

186k1486208

asked Mar 31 at 19:23

William AmbroseWilliam Ambrose

213

asked Mar 31 at 19:23

William AmbroseWilliam Ambrose

213

1 Answer
1

active

oldest

votes

1

$begingroup$

WLOG we can consider only $x, y ge 0$, so $y = sqrtleft(n+frac12right)^2 pi^2 - x^2$

Observe (by staring at the graph) that
$f(x) := sin^2x + sinh^2y = sin^2x + sinh^2sqrtleft(n+frac12right)^2 pi^2 - x^2$ is decreasing from $x = 0$ to $x = left(n+frac12right)pi $. To prove it, I will use the following two inequalities (you can come up with a proof):

(1) $sinx le x$

(2) $sinhx ge x$

Now for $x ge 0$, we have

$beginalign
f'(x) &= 2sinxcosx + 2sinhycoshy cdot frac-xy \
&= sin2x - 2x cdot fracsinh2y2y \
&le 2xleft(1 - fracsinh2y2yright) text ...... by (1) \
&le 2xleft(1 - 1right) text ...... by (2) \
&le 0.
endalign$

Therefore $f(x)$ must attain minimum at $x = left(n+frac12right)pi$, and we actually proved a stronger bound:

$|sinz| ge sqrtfleft(left(n+frac12right)piright) = 1$

share|cite|improve this answer

edited Apr 1 at 11:03

answered Mar 31 at 20:50

diagonal2diagonal2

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Derivative of y w.r.t x is =x/y. So $f’(x) = sin(2x) - 2x* sinh(2y)/2y leq 2x(1-1) = 0$
  $endgroup$
  – William Ambrose
  Mar 31 at 23:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops edited thanks a lot
  $endgroup$
  – diagonal2
  Apr 1 at 11:03
  

  
  
  
  

add a comment | 

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1

$begingroup$

WLOG we can consider only $x, y ge 0$, so $y = sqrtleft(n+frac12right)^2 pi^2 - x^2$

Observe (by staring at the graph) that
$f(x) := sin^2x + sinh^2y = sin^2x + sinh^2sqrtleft(n+frac12right)^2 pi^2 - x^2$ is decreasing from $x = 0$ to $x = left(n+frac12right)pi $. To prove it, I will use the following two inequalities (you can come up with a proof):

(1) $sinx le x$

(2) $sinhx ge x$

Now for $x ge 0$, we have

$beginalign
f'(x) &= 2sinxcosx + 2sinhycoshy cdot frac-xy \
&= sin2x - 2x cdot fracsinh2y2y \
&le 2xleft(1 - fracsinh2y2yright) text ...... by (1) \
&le 2xleft(1 - 1right) text ...... by (2) \
&le 0.
endalign$

Therefore $f(x)$ must attain minimum at $x = left(n+frac12right)pi$, and we actually proved a stronger bound:

$|sinz| ge sqrtfleft(left(n+frac12right)piright) = 1$

share|cite|improve this answer

edited Apr 1 at 11:03

answered Mar 31 at 20:50

diagonal2diagonal2

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Derivative of y w.r.t x is =x/y. So $f’(x) = sin(2x) - 2x* sinh(2y)/2y leq 2x(1-1) = 0$
  $endgroup$
  – William Ambrose
  Mar 31 at 23:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops edited thanks a lot
  $endgroup$
  – diagonal2
  Apr 1 at 11:03
  

  
  
  
  

add a comment | 

1

$begingroup$

WLOG we can consider only $x, y ge 0$, so $y = sqrtleft(n+frac12right)^2 pi^2 - x^2$

Observe (by staring at the graph) that
$f(x) := sin^2x + sinh^2y = sin^2x + sinh^2sqrtleft(n+frac12right)^2 pi^2 - x^2$ is decreasing from $x = 0$ to $x = left(n+frac12right)pi $. To prove it, I will use the following two inequalities (you can come up with a proof):

(1) $sinx le x$

(2) $sinhx ge x$

Now for $x ge 0$, we have

$beginalign
f'(x) &= 2sinxcosx + 2sinhycoshy cdot frac-xy \
&= sin2x - 2x cdot fracsinh2y2y \
&le 2xleft(1 - fracsinh2y2yright) text ...... by (1) \
&le 2xleft(1 - 1right) text ...... by (2) \
&le 0.
endalign$

Therefore $f(x)$ must attain minimum at $x = left(n+frac12right)pi$, and we actually proved a stronger bound:

$|sinz| ge sqrtfleft(left(n+frac12right)piright) = 1$

share|cite|improve this answer

edited Apr 1 at 11:03

answered Mar 31 at 20:50

diagonal2diagonal2

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Derivative of y w.r.t x is =x/y. So $f’(x) = sin(2x) - 2x* sinh(2y)/2y leq 2x(1-1) = 0$
  $endgroup$
  – William Ambrose
  Mar 31 at 23:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops edited thanks a lot
  $endgroup$
  – diagonal2
  Apr 1 at 11:03
  

  
  
  
  

add a comment | 

$begingroup$

WLOG we can consider only $x, y ge 0$, so $y = sqrtleft(n+frac12right)^2 pi^2 - x^2$

Observe (by staring at the graph) that
$f(x) := sin^2x + sinh^2y = sin^2x + sinh^2sqrtleft(n+frac12right)^2 pi^2 - x^2$ is decreasing from $x = 0$ to $x = left(n+frac12right)pi $. To prove it, I will use the following two inequalities (you can come up with a proof):

(1) $sinx le x$

(2) $sinhx ge x$

Now for $x ge 0$, we have

$beginalign
f'(x) &= 2sinxcosx + 2sinhycoshy cdot frac-xy \
&= sin2x - 2x cdot fracsinh2y2y \
&le 2xleft(1 - fracsinh2y2yright) text ...... by (1) \
&le 2xleft(1 - 1right) text ...... by (2) \
&le 0.
endalign$

Therefore $f(x)$ must attain minimum at $x = left(n+frac12right)pi$, and we actually proved a stronger bound:

$|sinz| ge sqrtfleft(left(n+frac12right)piright) = 1$

share|cite|improve this answer

edited Apr 1 at 11:03

answered Mar 31 at 20:50

diagonal2diagonal2

114

$endgroup$

share|cite|improve this answer

edited Apr 1 at 11:03

answered Mar 31 at 20:50

diagonal2diagonal2

114

answered Mar 31 at 20:50

diagonal2diagonal2

114

answered Mar 31 at 20:50

diagonal2diagonal2

114