Patterns in inequalities of triangle involving angles.












2












$begingroup$


I was reading this page and wondered as why, inequalities for $cos A$ (with argument $A$) become the same inequality for $sinfrac{A}{2}$ (with argument $frac{A}{2}$), similarly for $tan$ and $cot$.



Examples,




$$sinfrac{A}{2}sinfrac{B}{2}sinfrac{C}{2}lefrac{1}{8}$$$$cos Acos Bcos Clefrac{1}{8}$$




and




$$cos (A)+cos (B)+cos (C)lefrac{3}{2}$$$$displaystylesinfrac{A}{2}+sinfrac{B}{2}+sinfrac{C}{2}lefrac{3}{2}$$




Is there some greater Mathematics involved or just a pretty coincidence?






trigonometry inequality triangle geometric-inequalities







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago


















2












$begingroup$


I was reading this page and wondered as why, inequalities for $cos A$ (with argument $A$) become the same inequality for $sinfrac{A}{2}$ (with argument $frac{A}{2}$), similarly for $tan$ and $cot$.



Examples,




$$sinfrac{A}{2}sinfrac{B}{2}sinfrac{C}{2}lefrac{1}{8}$$$$cos Acos Bcos Clefrac{1}{8}$$




and




$$cos (A)+cos (B)+cos (C)lefrac{3}{2}$$$$displaystylesinfrac{A}{2}+sinfrac{B}{2}+sinfrac{C}{2}lefrac{3}{2}$$




Is there some greater Mathematics involved or just a pretty coincidence?






trigonometry inequality triangle geometric-inequalities







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago
















2












2








2





$begingroup$


I was reading this page and wondered as why, inequalities for $cos A$ (with argument $A$) become the same inequality for $sinfrac{A}{2}$ (with argument $frac{A}{2}$), similarly for $tan$ and $cot$.



Examples,




$$sinfrac{A}{2}sinfrac{B}{2}sinfrac{C}{2}lefrac{1}{8}$$$$cos Acos Bcos Clefrac{1}{8}$$




and




$$cos (A)+cos (B)+cos (C)lefrac{3}{2}$$$$displaystylesinfrac{A}{2}+sinfrac{B}{2}+sinfrac{C}{2}lefrac{3}{2}$$




Is there some greater Mathematics involved or just a pretty coincidence?






trigonometry inequality triangle geometric-inequalities







share|cite|improve this question











$endgroup$




I was reading this page and wondered as why, inequalities for $cos A$ (with argument $A$) become the same inequality for $sinfrac{A}{2}$ (with argument $frac{A}{2}$), similarly for $tan$ and $cot$.



Examples,




$$sinfrac{A}{2}sinfrac{B}{2}sinfrac{C}{2}lefrac{1}{8}$$$$cos Acos Bcos Clefrac{1}{8}$$




and




$$cos (A)+cos (B)+cos (C)lefrac{3}{2}$$$$displaystylesinfrac{A}{2}+sinfrac{B}{2}+sinfrac{C}{2}lefrac{3}{2}$$




Is there some greater Mathematics involved or just a pretty coincidence?






trigonometry inequality triangle geometric-inequalities




trigonometry inequality triangle geometric-inequalities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 27 mins ago









Michael Rozenberg

102k1791195




102k1791195










asked 2 hours ago









mnulbmnulb

1,408823




1,408823












  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago




















  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    1 hour ago


















$begingroup$
math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
$endgroup$
– lab bhattacharjee
1 hour ago






$begingroup$
math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
$endgroup$
– lab bhattacharjee
1 hour ago












1 Answer
1






active

oldest

votes


















6












$begingroup$

Great observation, I never noticed that before. It's not a coincidence, here is an explanation.



Rewrite the sine terms as cosines, for example,
$$cos(90^circ-tfrac A2)+cos(90^circ-tfrac B2)+cos(90^circ-tfrac C2)letfrac32 .$$
Now
$$eqalign{
A,B,C& hbox{are the angles of a triangle}cr
&Leftrightarrowquad A+B+C=180^circcr
&Leftrightarrowquad tfrac A2+tfrac B2+tfrac C2=90^circcr
&Leftrightarrowquad (90^circ-tfrac A2)+(90^circ-tfrac B2)+(90^circ-tfrac C2)=180^circcr
&Leftrightarrowquad (90^circ-tfrac A2),(90^circ-tfrac B2),(90^circ-tfrac C2) hbox{are the angles of a triangle}cr}$$
So, in this context,




  • anything true for all triangles that you can say about $A,B,C$ will also be true about $(90^circ-frac A2),(90^circ-frac B2),(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $cos(90^circ-frac A2),cos(90^circ-frac B2),cos(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $sinfrac A2,sinfrac B2,sinfrac C2$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    1 hour ago










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    29 mins ago











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3104700%2fpatterns-in-inequalities-of-triangle-involving-angles%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









6












$begingroup$

Great observation, I never noticed that before. It's not a coincidence, here is an explanation.



Rewrite the sine terms as cosines, for example,
$$cos(90^circ-tfrac A2)+cos(90^circ-tfrac B2)+cos(90^circ-tfrac C2)letfrac32 .$$
Now
$$eqalign{
A,B,C& hbox{are the angles of a triangle}cr
&Leftrightarrowquad A+B+C=180^circcr
&Leftrightarrowquad tfrac A2+tfrac B2+tfrac C2=90^circcr
&Leftrightarrowquad (90^circ-tfrac A2)+(90^circ-tfrac B2)+(90^circ-tfrac C2)=180^circcr
&Leftrightarrowquad (90^circ-tfrac A2),(90^circ-tfrac B2),(90^circ-tfrac C2) hbox{are the angles of a triangle}cr}$$
So, in this context,




  • anything true for all triangles that you can say about $A,B,C$ will also be true about $(90^circ-frac A2),(90^circ-frac B2),(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $cos(90^circ-frac A2),cos(90^circ-frac B2),cos(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $sinfrac A2,sinfrac B2,sinfrac C2$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    1 hour ago










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    29 mins ago
















6












$begingroup$

Great observation, I never noticed that before. It's not a coincidence, here is an explanation.



Rewrite the sine terms as cosines, for example,
$$cos(90^circ-tfrac A2)+cos(90^circ-tfrac B2)+cos(90^circ-tfrac C2)letfrac32 .$$
Now
$$eqalign{
A,B,C& hbox{are the angles of a triangle}cr
&Leftrightarrowquad A+B+C=180^circcr
&Leftrightarrowquad tfrac A2+tfrac B2+tfrac C2=90^circcr
&Leftrightarrowquad (90^circ-tfrac A2)+(90^circ-tfrac B2)+(90^circ-tfrac C2)=180^circcr
&Leftrightarrowquad (90^circ-tfrac A2),(90^circ-tfrac B2),(90^circ-tfrac C2) hbox{are the angles of a triangle}cr}$$
So, in this context,




  • anything true for all triangles that you can say about $A,B,C$ will also be true about $(90^circ-frac A2),(90^circ-frac B2),(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $cos(90^circ-frac A2),cos(90^circ-frac B2),cos(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $sinfrac A2,sinfrac B2,sinfrac C2$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    1 hour ago










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    29 mins ago














6












6








6





$begingroup$

Great observation, I never noticed that before. It's not a coincidence, here is an explanation.



Rewrite the sine terms as cosines, for example,
$$cos(90^circ-tfrac A2)+cos(90^circ-tfrac B2)+cos(90^circ-tfrac C2)letfrac32 .$$
Now
$$eqalign{
A,B,C& hbox{are the angles of a triangle}cr
&Leftrightarrowquad A+B+C=180^circcr
&Leftrightarrowquad tfrac A2+tfrac B2+tfrac C2=90^circcr
&Leftrightarrowquad (90^circ-tfrac A2)+(90^circ-tfrac B2)+(90^circ-tfrac C2)=180^circcr
&Leftrightarrowquad (90^circ-tfrac A2),(90^circ-tfrac B2),(90^circ-tfrac C2) hbox{are the angles of a triangle}cr}$$
So, in this context,




  • anything true for all triangles that you can say about $A,B,C$ will also be true about $(90^circ-frac A2),(90^circ-frac B2),(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $cos(90^circ-frac A2),cos(90^circ-frac B2),cos(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $sinfrac A2,sinfrac B2,sinfrac C2$.






share|cite|improve this answer









$endgroup$



Great observation, I never noticed that before. It's not a coincidence, here is an explanation.



Rewrite the sine terms as cosines, for example,
$$cos(90^circ-tfrac A2)+cos(90^circ-tfrac B2)+cos(90^circ-tfrac C2)letfrac32 .$$
Now
$$eqalign{
A,B,C& hbox{are the angles of a triangle}cr
&Leftrightarrowquad A+B+C=180^circcr
&Leftrightarrowquad tfrac A2+tfrac B2+tfrac C2=90^circcr
&Leftrightarrowquad (90^circ-tfrac A2)+(90^circ-tfrac B2)+(90^circ-tfrac C2)=180^circcr
&Leftrightarrowquad (90^circ-tfrac A2),(90^circ-tfrac B2),(90^circ-tfrac C2) hbox{are the angles of a triangle}cr}$$
So, in this context,




  • anything true for all triangles that you can say about $A,B,C$ will also be true about $(90^circ-frac A2),(90^circ-frac B2),(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $cos(90^circ-frac A2),cos(90^circ-frac B2),cos(90^circ-frac C2)$;

  • hence, anything true for all triangles that you can say about $cos A,cos B,cos C$ will also be true about $sinfrac A2,sinfrac B2,sinfrac C2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









DavidDavid

68.5k664126




68.5k664126








  • 1




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    1 hour ago










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    29 mins ago














  • 1




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    1 hour ago










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    29 mins ago








1




1




$begingroup$
Nice explanation.
$endgroup$
– marty cohen
1 hour ago




$begingroup$
Nice explanation.
$endgroup$
– marty cohen
1 hour ago












$begingroup$
Very nice! I agree. +1
$endgroup$
– Michael Rozenberg
29 mins ago




$begingroup$
Very nice! I agree. +1
$endgroup$
– Michael Rozenberg
29 mins ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3104700%2fpatterns-in-inequalities-of-triangle-involving-angles%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Ponta tanko

Image
Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
Read more

Tantalo (mitologio)

Image
Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
Read more

Franz Schubert

Image
Franz Schubert Franz Schubert, pentrita en 1875 fare de Wilhelm August Rieder Bazaj informoj Naskiĝo 31-an de januaro 1797 ( 1797-01-31 ) en Himmelpfortgrund Morto 19-an de novembro 1828 ( 1828-11-19 ) (31-jara) en Vieno Ĝenroj Klasika muziko Profesio komponisto Aktivaj jaroj 1808-1828 v   •   d   •   r Franz SCHUBERT ([ˈfʁants ˈʃuːbɛɐ̯t], esperante Franz' Ŝuberto ; naskiĝis la 31-an de januaro 1797 je Himmelpfortgrund en Vieno, kiu nun estas parto de la 9-a viena distrikto Alsergrund, kaj mortis la 19-an de novembro 1828 en Vieno) estis aŭstra komponisto de klasika kaj fruromantisma muziko. Li komponis ĉirkaŭ 600 lidojn (germanaj kantoj) kaj estas nomata la "reĝo de lido". Enhavo 1 Vivo 1.1 Infaneco 1.2 Junaĝo 1.3 Zelis 1.4 Jaroj de maturiĝo kaj matureco 1.5 La lastaj jaroj 2 Postefiko kaj akceptado 3 Honoradoj 4 Verkoj 4.1 Eklezia muziko 4.2 Dram...
Read more