Patterns in inequalities of triangle involving angles.












9












$begingroup$


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 triangles geometric-inequalities







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    Feb 8 at 5:27


















9












$begingroup$


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 triangles geometric-inequalities







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    Feb 8 at 5:27
















9












9








9


1



$begingroup$


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 triangles 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 triangles geometric-inequalities




trigonometry inequality triangles geometric-inequalities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 8 at 6:13









Michael Rozenberg

109k1896201




109k1896201










asked Feb 8 at 4:28









mnulbmnulb

1,482823




1,482823












  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    Feb 8 at 5:27




















  • $begingroup$
    math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
    $endgroup$
    – lab bhattacharjee
    Feb 8 at 5:27


















$begingroup$
math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
$endgroup$
– lab bhattacharjee
Feb 8 at 5:27






$begingroup$
math.stackexchange.com/questions/952893/…. math.stackexchange.com/questions/1374163/…
$endgroup$
– lab bhattacharjee
Feb 8 at 5:27












1 Answer
1






active

oldest

votes


















15












$begingroup$

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









$endgroup$









  • 2




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    Feb 8 at 4:53










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    Feb 8 at 6:12










  • $begingroup$
    This post reduces my effort of remembering these inequalities to half, Thanks!
    $endgroup$
    – mnulb
    Feb 9 at 4:11












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









15












$begingroup$

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









$endgroup$









  • 2




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    Feb 8 at 4:53










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    Feb 8 at 6:12










  • $begingroup$
    This post reduces my effort of remembering these inequalities to half, Thanks!
    $endgroup$
    – mnulb
    Feb 9 at 4:11
















15












$begingroup$

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









$endgroup$









  • 2




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    Feb 8 at 4:53










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    Feb 8 at 6:12










  • $begingroup$
    This post reduces my effort of remembering these inequalities to half, Thanks!
    $endgroup$
    – mnulb
    Feb 9 at 4:11














15












15








15





$begingroup$

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 Feb 8 at 4:48









DavidDavid

69.6k668131




69.6k668131








  • 2




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    Feb 8 at 4:53










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    Feb 8 at 6:12










  • $begingroup$
    This post reduces my effort of remembering these inequalities to half, Thanks!
    $endgroup$
    – mnulb
    Feb 9 at 4:11














  • 2




    $begingroup$
    Nice explanation.
    $endgroup$
    – marty cohen
    Feb 8 at 4:53










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    Feb 8 at 6:12










  • $begingroup$
    This post reduces my effort of remembering these inequalities to half, Thanks!
    $endgroup$
    – mnulb
    Feb 9 at 4:11








2




2




$begingroup$
Nice explanation.
$endgroup$
– marty cohen
Feb 8 at 4:53




$begingroup$
Nice explanation.
$endgroup$
– marty cohen
Feb 8 at 4:53












$begingroup$
Very nice! I agree. +1
$endgroup$
– Michael Rozenberg
Feb 8 at 6:12




$begingroup$
Very nice! I agree. +1
$endgroup$
– Michael Rozenberg
Feb 8 at 6:12












$begingroup$
This post reduces my effort of remembering these inequalities to half, Thanks!
$endgroup$
– mnulb
Feb 9 at 4:11




$begingroup$
This post reduces my effort of remembering these inequalities to half, Thanks!
$endgroup$
– mnulb
Feb 9 at 4:11


















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