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

Examples,

$$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\le\frac{1}{8}$$$$\cos A\cos B\cos C\le\frac{1}{8}$$

and

$$\cos (A)+\cos (B)+\cos (C)\le\frac{3}{2}$$$$\displaystyle\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\le\frac{3}{2}$$

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

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

Rewrite the sine terms as cosines, for example, $$\cos(90^\circ-\tfrac A2)+\cos(90^\circ-\tfrac B2)+\cos(90^\circ-\tfrac C2)\le\tfrac32\ .$$ Now $$\eqalign{ A,B,C&\ \hbox{are the angles of a triangle}\cr &\Leftrightarrow\quad A+B+C=180^\circ\cr &\Leftrightarrow\quad \tfrac A2+\tfrac B2+\tfrac C2=90^\circ\cr &\Leftrightarrow\quad (90^\circ-\tfrac A2)+(90^\circ-\tfrac B2)+(90^\circ-\tfrac C2)=180^\circ\cr &\Leftrightarrow\quad (90^\circ-\tfrac A2),(90^\circ-\tfrac B2),(90^\circ-\tfrac C2)\ \hbox{are the angles of a triangle}\cr}$$ 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 $\sin\frac A2,\sin\frac B2,\sin\frac C2$.
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  • 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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