Proving $1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$, for $vartheta in [0, frac{pi}{2}]$
$begingroup$
For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
48.7k870156
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
$endgroup$
add a comment |
$begingroup$
For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
48.7k870156
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
$endgroup$
add a comment |
$begingroup$
For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
48.7k870156
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
$endgroup$
For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
48.7k870156
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
edited Dec 14 '18 at 10:51
Blue
48.7k870156
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
edited Dec 14 '18 at 10:51
Blue
48.7k870156
edited Dec 14 '18 at 10:51
Blue
48.7k870156
edited Dec 14 '18 at 10:51
Blue
48.7k870156
48.7k870156
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
426
426
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039193%2fproving-1-frac2-vartheta-pi-sin-vartheta-leq-cos-vartheta-for%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
$begingroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
$begingroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
19.7k92352
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
1
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039193%2fproving-1-frac2-vartheta-pi-sin-vartheta-leq-cos-vartheta-for%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
