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 40 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


















7












$begingroup$

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
    2 hours ago










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    41 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









7












$begingroup$

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
    2 hours ago










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
















7












$begingroup$

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
    2 hours ago










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














7












7








7





$begingroup$

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 2 hours ago









DavidDavid

68.5k664126




68.5k664126








  • 1




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    2 hours ago










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














  • 1




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    2 hours ago










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








1




1




$begingroup$
Nice explanation.
$endgroup$
– marty cohen
2 hours ago




$begingroup$
Nice explanation.
$endgroup$
– marty cohen
2 hours ago












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




$begingroup$
Very nice! I agree. +1
$endgroup$
– Michael Rozenberg
41 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

Liste der Baudenkmale in Friedland (Mecklenburg)

Image
In der Liste der Baudenkmale in Friedland sind alle denkmalgeschützten Bauten der Stadt Friedland (Mecklenburg-Vorpommern) und ihrer Ortsteile aufgelistet. Grundlage ist die Veröffentlichung der Denkmalliste des Landkreises Mecklenburgische Seenplatte (Auszug) vom 28. September 2016. Inhaltsverzeichnis 1 Baudenkmale nach Ortsteilen 1.1 Friedland 1.2 Bresewitz 1.3 Brohm 1.4 Cosa 1.5 Dishley 1.6 Dishley-Ausbau 1.7 Eichhorst 1.8 Glienke 1.9 Heinrichshöh 1.10 Heinrichswalde 1.11 Jatzke 1.12 Jatzke-Meierei 1.13 Liepen 1.14 Ramelow 1.15 Schwanbeck 2 Vorgesehener Denkmalbereich 3 Quelle 4 Weblinks Baudenkmale nach Ortsteilen | Friedland | ID Lage Bezeichnung Beschreibung Bild 281 An der Kleinbahn 2 Windmühle mit Umgang 282 Anklamer Straße alter Friedhof mit Portalwand 282 Anklamer Straße drei Erbbegräbnissen und Gräbern u. a. H.A. Riemann 284
Read more

Single-Malt-Whisky

Image
Fünf verschiedene schottische Single-Malt-Whiskys Als Single-Malt-Whisky oder kürzer Single Malt werden Whiskys bezeichnet, die zwei besondere Bedingungen erfüllen: Sie stammen zum ersten aus einer einzigen Brennerei, sind also kein Verschnitt aus mehreren Whiskysorten (daher Single , deutsch: einzeln). Zum zweiten wird als Getreide ausschließlich gemälzte Gerste verwendet (deswegen Malt , deutsch: Malz). Der Zusatz Single Malt wird als Prädikat verstanden, die so ausgezeichneten Whiskys sind in aller Regel besonders hochwertig. Hinzu kommt ihre besonders lange Lagerung, die Reifungszeit währt oft ein Jahrzehnt und länger. Inhaltsverzeichnis 1 Allgemeines 2 Herkunftsländer 2.1 Schottland 2.2 Irland 2.3 Japan 2.4 Kontinentaleuropa 3 Produktionsschritte 3.1 Mälzen 3.2 Brauen 3.3 Brennen 3.4 Reifung und Finish 3.5 Abfüllung 4 Altersangaben 5 Lagerung von Whiskyflaschen 6 Geschmacksklassifikation 7 Sonstig
Read more

Czorneboh

Image
Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
Read more