Question

In $△ABC,$ if $A,B$ and $C$ represent the angles of a triangle, then the maximum value of $\cos A+\cos B+\cos C$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

The correct option is B $\frac{3}{2}$

Let $\cos A+\cos B+\cos C=x$
$\Rightarrow 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+1-2{\sin }^2\frac{C}{2}=x\Rightarrow 2\sin \frac{C}{2}\cos \frac{A-B}{2}+1-2{\sin }^2\frac{C}{2}=x\Rightarrow 2{\sin }^2\frac{C}{2}-2\sin \frac{C}{2}\cos \frac{A-B}{2}+x-1=0$

As $\sin \frac{C}{2}\in R$
$D\ge 0$
$\Rightarrow 4{\cos }^2\left(\frac{A-B}{2}\right)-4\times 2(x-1)\ge 0\Rightarrow 2(x-1)-{\cos }^2\left(\frac{A-B}{2}\right)\le 0\Rightarrow 2(x-1)\le {\cos }^2\left(\frac{A-B}{2}\right)\Rightarrow 2(x-1)\le 1\Rightarrow x\le \frac{3}{2}$
Therefore, the maximum occur when $A=B=C={60}^{\circ }$

Hence, the maximum value of the given expression is $\frac{3}{2}$