Here is what I did, tell me whether I did correct or not: \begin{align*} y &= \cos A + \cos B + \cos C\\ y &= \cos A + 2\cos\left(\frac{B+C}2\right)\cos\left(\frac {B-C}2\right)\\ y &= \cos A + 2\sin\left(\frac A2\right)\cos\left(\frac {BC}2\right) && \text{since $A+B+C = \pi$} \end{align*}

Now for maximum value of $y$ if we put $\cos\left(\frac {B-C}2\right) = 1$ then \begin{align*} y &\le \cos A + 2\sin\left(\frac A2\right)\\ y &\le 1-2\sin^2\left(\frac A2\right) + 2\sin\left(\frac A2\right) \end{align*}

By completing the square we get

$$y \le \frac 32 - 2\left(\sin\frac A2 - \frac 12\right)^2$$

$y_{\max} = \frac 32$ at $\sin\frac A2 = \frac 12$ and $y_{\min} > 1$ at $\sin \frac A2>0$ because it is a ratio of two sides of a triangle.

Is this solution correct? If there is a better solution then please post it here. Help!


We can prove $$\cos A+\cos B+\cos C=1+4\sin\frac A2\sin\frac B2\sin\frac C2$$

Now, $0<\dfrac A2<90^\circ\implies\sin\dfrac A2>0$

For the other part,

$$y=1-2\sin^2\frac A2+2\sin\frac A2\cos\frac{B-C}2$$

$$\iff2\sin^2\frac A2-\sin\frac A2\cdot2\cos\frac{B-C}2+y-1=0$$ which is a Quadratic equation in $\sin\dfrac A2$ which is real

So, the discriminant must be $\ge0$

i.e., $\left(2\cos\dfrac{B-C}2\right)^2-4\cdot2(y-1)\ge0$

$\iff 4y\le4+2\cos^2\dfrac{B-C}2=4+1+\cos(B-C)\le4+1+1$

The equality occurs iff $\cos(B-C)=1\iff B=C$ as $0<B,C<180^\circ$

where $\sin\dfrac A2=\dfrac12\implies\dfrac A2=30^\circ$ as $0<\dfrac A2<90^\circ$

  • $\begingroup$ @Shubham, See also : math.stackexchange.com/questions/639890/… $\endgroup$ – lab bhattacharjee Sep 28 '14 at 16:06
  • $\begingroup$ This step isn't clear $4y≤4+2cos^2\frac{B−C}2=5+cos(B−C)≤6$ Can you please rewrite this for better understanding? :) $\endgroup$ – Shubham Sep 28 '14 at 16:13
  • $\begingroup$ @Shubham, $\cos2x=2\cos^2x-1\iff2\cos^2x=1+\cos2x$ right? And for real $y,\cos y\le1$ $\endgroup$ – lab bhattacharjee Sep 28 '14 at 16:15
  • $\begingroup$ Thank you we proved $y\le3/2$ but how did we prove that $y \gt 1 $? $\endgroup$ – Shubham Sep 28 '14 at 16:19
  • $\begingroup$ @Shubham, My God! Please read the part before "For the other part" $\endgroup$ – lab bhattacharjee Sep 28 '14 at 16:20

better is to show that $\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$ where $r$ is the inradius and $R$ is circumradius of the given triangle. Your theorem follows from this equation

  • $\begingroup$ Graubner can you please elaborate? $\endgroup$ – Shubham Sep 28 '14 at 14:01
  • $\begingroup$ use the theorem of cosines and $A=rs$ and $A=\frac{abc}{4R}$ $\endgroup$ – Dr. Sonnhard Graubner Sep 28 '14 at 14:03
  • $\begingroup$ Im not really following up. I just know the basics of law of cosines. Do you have any other method? :( $\endgroup$ – Shubham Sep 28 '14 at 14:16
  • $\begingroup$ show that $\cos(A)+\cos(B)+\cos(C)=1+4\sin(A/2)\sin(B/2)\sin(C/2)$ $\endgroup$ – Dr. Sonnhard Graubner Sep 28 '14 at 14:23
  • $\begingroup$ Yes I can do that (did that) but couldn't prove the question. Can you please post more? $\endgroup$ – Shubham Sep 28 '14 at 14:54

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