Proving a cosine inequality












5












$begingroup$


Given $0 < Theta < frac{pi}{2}$, $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$



Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.



Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is



$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$



We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.



At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.



However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.






calculus trigonometry proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
    $endgroup$
    – Riemann
    1 hour ago










  • $begingroup$
    As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
    $endgroup$
    – AnatolyVorobey
    1 hour ago








  • 1




    $begingroup$
    Yes, it is right.
    $endgroup$
    – Riemann
    1 hour ago
















5












$begingroup$


Given $0 < Theta < frac{pi}{2}$, $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$



Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.



Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is



$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$



We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.



At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.



However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.






calculus trigonometry proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
    $endgroup$
    – Riemann
    1 hour ago










  • $begingroup$
    As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
    $endgroup$
    – AnatolyVorobey
    1 hour ago








  • 1




    $begingroup$
    Yes, it is right.
    $endgroup$
    – Riemann
    1 hour ago














5












5








5





$begingroup$


Given $0 < Theta < frac{pi}{2}$, $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$



Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.



Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is



$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$



We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.



At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.



However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.






calculus trigonometry proof-verification







share|cite|improve this question











$endgroup$




Given $0 < Theta < frac{pi}{2}$, $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$



Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.



Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is



$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$



We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.



At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.



However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.






calculus trigonometry proof-verification




calculus trigonometry proof-verification






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Kemono Chen

3,0001741




3,0001741










asked 3 hours ago









AnatolyVorobeyAnatolyVorobey

1,632513




1,632513












  • $begingroup$
    if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
    $endgroup$
    – Riemann
    1 hour ago










  • $begingroup$
    As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
    $endgroup$
    – AnatolyVorobey
    1 hour ago








  • 1




    $begingroup$
    Yes, it is right.
    $endgroup$
    – Riemann
    1 hour ago


















  • $begingroup$
    if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
    $endgroup$
    – Riemann
    1 hour ago










  • $begingroup$
    As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
    $endgroup$
    – AnatolyVorobey
    1 hour ago








  • 1




    $begingroup$
    Yes, it is right.
    $endgroup$
    – Riemann
    1 hour ago
















$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
1 hour ago




$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
1 hour ago












$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
1 hour ago






$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
1 hour ago






1




1




$begingroup$
Yes, it is right.
$endgroup$
– Riemann
1 hour ago




$begingroup$
Yes, it is right.
$endgroup$
– Riemann
1 hour ago










1 Answer
1






active

oldest

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3












$begingroup$

Use Mean Value Theorem you can prove the following lemma:



suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.



So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's pretty, thank you! Can you also comment if the custom argument in my proof looks correct?
    $endgroup$
    – AnatolyVorobey
    2 hours ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

Use Mean Value Theorem you can prove the following lemma:



suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.



So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's pretty, thank you! Can you also comment if the custom argument in my proof looks correct?
    $endgroup$
    – AnatolyVorobey
    2 hours ago
















3












$begingroup$

Use Mean Value Theorem you can prove the following lemma:



suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.



So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's pretty, thank you! Can you also comment if the custom argument in my proof looks correct?
    $endgroup$
    – AnatolyVorobey
    2 hours ago














3












3








3





$begingroup$

Use Mean Value Theorem you can prove the following lemma:



suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.



So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!






share|cite|improve this answer











$endgroup$



Use Mean Value Theorem you can prove the following lemma:



suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.



So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 3 hours ago









RiemannRiemann

3,4051322




3,4051322












  • $begingroup$
    That's pretty, thank you! Can you also comment if the custom argument in my proof looks correct?
    $endgroup$
    – AnatolyVorobey
    2 hours ago


















  • $begingroup$
    That's pretty, thank you! Can you also comment if the custom argument in my proof looks correct?
    $endgroup$
    – AnatolyVorobey
    2 hours ago
















$begingroup$
That's pretty, thank you! Can you also comment if the custom argument in my proof looks correct?
$endgroup$
– AnatolyVorobey
2 hours ago




$begingroup$
That's pretty, thank you! Can you also comment if the custom argument in my proof looks correct?
$endgroup$
– AnatolyVorobey
2 hours ago


















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