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Find the smallest value of $f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$ on the interval $(0,pi/2)$

Showing $leftlvert frac{sin(nx)}{nsin(x)} + frac{cos(nx)}{ncos(x)} rightrvert leleftlvertfrac{n+1}{n}rightrvert $Handy way to find the $x$ value where $sin x cos left( frac{pi}{2} sin x right)$ is maximum?Find the exact value of $frac{1-2sinleft(αright)cosleft(αright)}{1-2sin^2left(αright)}$.If $left(1+sin phiright)cdot left(1+cos phiright) = frac{5}{4};,$ Then $left(1-sin phiright)cdot left(1-cos phiright)$Max value of trignometric function $sin left(x+fracpi6right)+cos left(x+fracpi6right)$Is there a better way to find the minimum value of $displaystylefrac{1}{sin theta cos theta left ( sin theta + cos theta right )}$How to compare $left(sin left(xright)right)^{cos left(xright)}$ and $ left(cos left(xright)right)^{sin left(xright)}$Find the exact value of $cosleft(sin^{-1}left(-frac{3}{5}right)right)$.If $m$ denote the minimum value of $f(x)= left (frac{cos x}{sin^2x(cos x-sin x)}right)$ where $xinleft(0,frac{pi}{4}right)$Number of solutions to $cosleft(A cosleft(dfrac{s}{A} right)right) = sinleft(A cosleft(dfrac{s}{A} right)right)$

4

$begingroup$

There's a function defined as:
$$f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$$
In interval $$left(0,frac{pi}{2}right)$$
Find the smallest value (Save only its integer value)

I've managed to come to this
$${1over288}+{{2(sin(x)+cos(x))+576}over(sin(x)+cos(x))^2-1}$$
How can I find the smallest value now?

trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals

share|cite|improve this question

edited 1 hour ago

Blue

48.6k870154

asked 7 hours ago

a_man_with_no_namea_man_with_no_name

333

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

4

$begingroup$

There's a function defined as:
$$f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$$
In interval $$left(0,frac{pi}{2}right)$$
Find the smallest value (Save only its integer value)

I've managed to come to this
$${1over288}+{{2(sin(x)+cos(x))+576}over(sin(x)+cos(x))^2-1}$$
How can I find the smallest value now?

trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals

share|cite|improve this question

edited 1 hour ago

Blue

48.6k870154

asked 7 hours ago

a_man_with_no_namea_man_with_no_name

333

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

There's a function defined as:
$$f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$$
In interval $$left(0,frac{pi}{2}right)$$
Find the smallest value (Save only its integer value)

I've managed to come to this
$${1over288}+{{2(sin(x)+cos(x))+576}over(sin(x)+cos(x))^2-1}$$
How can I find the smallest value now?

trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals

share|cite|improve this question

edited 1 hour ago

Blue

48.6k870154

asked 7 hours ago

a_man_with_no_namea_man_with_no_name

333

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 1 hour ago

Blue

48.6k870154

asked 7 hours ago

a_man_with_no_namea_man_with_no_name

333

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 1 hour ago

Blue

48.6k870154

asked 7 hours ago

a_man_with_no_namea_man_with_no_name

333

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 1 hour ago

Blue

48.6k870154

edited 1 hour ago

Blue

48.6k870154

asked 7 hours ago

a_man_with_no_namea_man_with_no_name

333

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 7 hours ago

a_man_with_no_namea_man_with_no_name

333

3 Answers
3

active

oldest

votes

6

$begingroup$

Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:

$$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$

$$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$

We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

greedoidgreedoid

44.4k1156111

$endgroup$

add a comment | 

5

$begingroup$

$$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric

$$1+frac{288}{cos x}$$ is increasing.

Hence the minimum occurs ar $x=dfracpi4$.

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

129k675227

$endgroup$

add a comment | 

1

$begingroup$

Here's how to continue with your idea:
Let $sin{x}+cos{x}=t$.
Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
$$
f(x)
=frac{1}{288}+frac{2(t+288)}{t^2-1}
ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
=576 + frac{1}{288} + 2sqrt{2}$$
and we are done!

share|cite|improve this answer

edited 1 hour ago

Viktor Glombik

9911527

answered 6 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

add a comment | 

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6

$begingroup$

Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:

$$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$

$$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$

We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

greedoidgreedoid

44.4k1156111

$endgroup$

add a comment | 

6

$begingroup$

Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:

$$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$

$$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$

We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

greedoidgreedoid

44.4k1156111

$endgroup$

add a comment | 

$begingroup$

Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:

$$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$

$$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$

We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

greedoidgreedoid

44.4k1156111

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

greedoidgreedoid

44.4k1156111

answered 7 hours ago

greedoidgreedoid

44.4k1156111

answered 7 hours ago

greedoidgreedoid

44.4k1156111

5

$begingroup$

$$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric

$$1+frac{288}{cos x}$$ is increasing.

Hence the minimum occurs ar $x=dfracpi4$.

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

129k675227

$endgroup$

add a comment | 

5

$begingroup$

$$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric

$$1+frac{288}{cos x}$$ is increasing.

Hence the minimum occurs ar $x=dfracpi4$.

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

129k675227

$endgroup$

add a comment | 

$begingroup$

$$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric

$$1+frac{288}{cos x}$$ is increasing.

Hence the minimum occurs ar $x=dfracpi4$.

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

129k675227

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

129k675227

answered 7 hours ago

Yves DaoustYves Daoust

129k675227

answered 7 hours ago

Yves DaoustYves Daoust

129k675227

1

$begingroup$

Here's how to continue with your idea:
Let $sin{x}+cos{x}=t$.
Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
$$
f(x)
=frac{1}{288}+frac{2(t+288)}{t^2-1}
ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
=576 + frac{1}{288} + 2sqrt{2}$$
and we are done!

share|cite|improve this answer

edited 1 hour ago

Viktor Glombik

9911527

answered 6 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

add a comment | 

1

$begingroup$

Here's how to continue with your idea:
Let $sin{x}+cos{x}=t$.
Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
$$
f(x)
=frac{1}{288}+frac{2(t+288)}{t^2-1}
ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
=576 + frac{1}{288} + 2sqrt{2}$$
and we are done!

share|cite|improve this answer

edited 1 hour ago

Viktor Glombik

9911527

answered 6 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

add a comment | 

$begingroup$

Here's how to continue with your idea:
Let $sin{x}+cos{x}=t$.
Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
$$
f(x)
=frac{1}{288}+frac{2(t+288)}{t^2-1}
ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
=576 + frac{1}{288} + 2sqrt{2}$$
and we are done!

share|cite|improve this answer

edited 1 hour ago

Viktor Glombik

9911527

answered 6 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

share|cite|improve this answer

edited 1 hour ago

Viktor Glombik

9911527

answered 6 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

edited 1 hour ago

Viktor Glombik

9911527

edited 1 hour ago

Viktor Glombik

9911527

answered 6 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

answered 6 hours ago

Michael RozenbergMichael Rozenberg

105k1892198