Question

If A + B + C = $\pi ,$ the maximum value of cos A + cos B + cos C can have is --

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

Answer 1

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

If A + B + C =$\pi ,$

$\Rightarrow C=\pi -(A+B)$......(i)

Consider given function f(A, B)=cos A + cos B + cos C

$\Rightarrow f(A,B)=\cos A+\cos B+\cos (\pi -(A+B\left)\right)$

$f(A,B)=\cos A+\cos B-\cos (A+B)$

Partially differentiating with respect to $A$ and $B$, we get

$f_A(A,B)=\sin (A+B)-\sin A$=0......(ii) and $f_B(A,B)=\sin (A+B)-\sin B=0$.......(iii)

Solving equation(ii) and (iii) to get the critical points.

$\Rightarrow \sin A=\sin B$

$\therefore A=B$

Substitute A=B in equation(ii), we get

$\sin (A+A)-\sin A=0$

$\Rightarrow \sin 2A=\sin A$

$\Rightarrow 2\sin A\cos A=\sin A$

$\Rightarrow \sin A(2\cos A-1)=0$

$\Rightarrow (2\cos A-1)=0$ and $\sin A=0$ represents the boundary of the given function that can't be consider.

$\Rightarrow \cos A=\frac{1}{2}$

$\Rightarrow cosA=\cos \frac{\pi }{3}$

$\therefore A=\frac{\pi }{3}$

So, $A=B=\frac{\pi }{3}$ and $C=\pi -(A+A)\Rightarrow C=\pi -2A$

$\Rightarrow C=\pi -2\left(\frac{\pi }{3}\right)$

$=\frac{3\pi -2\pi }{3}$

$C=\frac{\pi }{3}$

Therefore, the maximum value occurs at $A=B=C=\frac{\pi }{3}$ and its value $f(\frac{\pi }{3},\frac{\pi }{3})=\cos \frac{\pi }{3}+\cos \frac{\pi }{3}+\cos \frac{\pi }{3}$

$=3\cos \frac{\pi }{3}$

$=3\times \frac{1}{2}$

$=\frac{3}{2}$