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Find the smallest value of the function

Finding the local extrema of this trigonometric, multivariate functionHow to find the following sums.Smallest positive integral value of $a$ such that ${sin}^2 x+acos x+{a}^2>1+cos x$ holds for all real $x$Find $theta$ for which $sin m theta = cos n theta$Find the value of $theta$Inversion of Trigonometric EquationsFind the Fourier series of, $f(x) = cos Big(frac{x}{3}Big)$ on the interval $- pi leq x leq pi. $Given trigonometric function, find largest and smallest valuesIf $sin x+sin^2x+sin^3x=1$, then find $cos^6x-4cos^4x+8cos^2x$Fourier sine series simplification

3

$begingroup$

There's a function defined as:
$$f(x) := bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg).$$
In interval $$Big(0,frac{pi}{2}Big)$$
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 3 hours ago

greedoid

44.4k1156110

asked 4 hours ago

a_man_with_no_namea_man_with_no_name

283

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

add a comment | 

3

$begingroup$

There's a function defined as:
$$f(x) := bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg).$$
In interval $$Big(0,frac{pi}{2}Big)$$
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 3 hours ago

greedoid

44.4k1156110

asked 4 hours ago

a_man_with_no_namea_man_with_no_name

283

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:
$$f(x) := bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg).$$
In interval $$Big(0,frac{pi}{2}Big)$$
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 3 hours ago

greedoid

44.4k1156110

asked 4 hours ago

a_man_with_no_namea_man_with_no_name

283

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 3 hours ago

greedoid

44.4k1156110

asked 4 hours ago

a_man_with_no_namea_man_with_no_name

283

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 3 hours ago

greedoid

44.4k1156110

asked 4 hours ago

a_man_with_no_namea_man_with_no_name

283

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 3 hours ago

greedoid

44.4k1156110

edited 3 hours ago

greedoid

44.4k1156110

asked 4 hours ago

a_man_with_no_namea_man_with_no_name

283

New contributor

a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago

a_man_with_no_namea_man_with_no_name

283

3 Answers
3

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3

$begingroup$

$$1+frac{288}{sin x}$$ 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 4 hours ago

Yves DaoustYves Daoust

129k675227

$endgroup$

add a comment | 

3

$begingroup$

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 4 hours ago

answered 4 hours ago

greedoidgreedoid

44.4k1156110

$endgroup$

add a comment | 

0

$begingroup$

You can end your idea.

Indeed, let $sin{x}+cos{x}=t$.

Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
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:
$$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!

share|cite|improve this answer

answered 3 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

add a comment | 

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3

$begingroup$

$$1+frac{288}{sin x}$$ 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 4 hours ago

Yves DaoustYves Daoust

129k675227

$endgroup$

add a comment | 

3

$begingroup$

$$1+frac{288}{sin x}$$ 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 4 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 4 hours ago

Yves DaoustYves Daoust

129k675227

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Yves DaoustYves Daoust

129k675227

answered 4 hours ago

Yves DaoustYves Daoust

129k675227

answered 4 hours ago

Yves DaoustYves Daoust

129k675227

3

$begingroup$

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 4 hours ago

answered 4 hours ago

greedoidgreedoid

44.4k1156110

$endgroup$

add a comment | 

3

$begingroup$

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 4 hours ago

answered 4 hours ago

greedoidgreedoid

44.4k1156110

$endgroup$

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 4 hours ago

answered 4 hours ago

greedoidgreedoid

44.4k1156110

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

greedoidgreedoid

44.4k1156110

answered 4 hours ago

greedoidgreedoid

44.4k1156110

answered 4 hours ago

greedoidgreedoid

44.4k1156110

0

$begingroup$

You can end your idea.

Indeed, let $sin{x}+cos{x}=t$.

Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
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:
$$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!

share|cite|improve this answer

answered 3 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

add a comment | 

0

$begingroup$

You can end your idea.

Indeed, let $sin{x}+cos{x}=t$.

Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
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:
$$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!

share|cite|improve this answer

answered 3 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

add a comment | 

$begingroup$

You can end your idea.

Indeed, let $sin{x}+cos{x}=t$.

Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
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:
$$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!

share|cite|improve this answer

answered 3 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

answered 3 hours ago

Michael RozenbergMichael Rozenberg

105k1892198

answered 3 hours ago

Michael RozenbergMichael Rozenberg

105k1892198