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







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  • $begingroup$
    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 54 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
    55 mins ago











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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
    55 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
    55 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
    55 mins ago














  • 1




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










  • $begingroup$
    Very nice! I agree. +1
    $endgroup$
    – Michael Rozenberg
    55 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
55 mins ago




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


















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