Question

For a triangle ABC prove that: $\cos A + \cos B + \cos C \leqslant \dfrac{3}{2}$ and for the case of equality the triangle will be equilateral.

Answer 1

Hint : First we save the inequality part of the problem by using standard trigonometric formulae and then by deducing the result of the first part we try to get to the required condition of an equilateral triangle i.e. each angle of it is of 60 degrees.

Complete step-by-step answer :
Firstly we write down the inequality to see how we proceed
$\cos A + \cos B + \cos C \leqslant \dfrac{3}{2}$
We assume this inequality to be
${\rm A} = \cos A + \cos B + \cos C \leqslant \dfrac{3}{2}$
Now we subtract $\dfrac{3}{2}$ from both sides of the inequality i.e.
${\rm A} = \cos A + \cos B + \cos C - \dfrac{3}{2} \leqslant 0\,{\text{ - - - - - - - - equation(1)}}$
Now using these formulae to solve it further

$$\operatorname{Cos} \alpha + \operatorname{Cos} \beta = 2\operatorname{Cos} \dfrac{{\alpha + \beta }}{2}\operatorname{Cos} \dfrac{{\alpha - \beta }}{2} \\\ {\text{and}} \\\ \operatorname{Cos} \phi = 1 - 2{\operatorname{Sin} ^2}\dfrac{\phi }{2} \\\$$

Applying the formulae
${\rm A} = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} + 1 - 2{\sin ^2}\dfrac{C}{2} - \dfrac{3}{2}\,{\text{ - - - - - - equation(2)}}$
We know that in $\vartriangle ABC,\angle A + \angle B + \angle C = 180^\circ$ so we have
$\angle A + \angle B = 180^\circ - \angle C \\\ \dfrac{{\angle A + \angle B}}{2} = 90^\circ - \dfrac{{\angle C}}{2} \\\$
Taking ‘Cosine’ function on both sides of the equation gives us

$$\operatorname{Cos} \left( {\dfrac{{\angle A + \angle B}}{2}} \right) = \operatorname{Cos} \left( {90^\circ - \dfrac{{\angle C}}{2}} \right) \\\ \Rightarrow \operatorname{Cos} \left( {\dfrac{{\angle A + \angle B}}{2}} \right) = \;\operatorname{Sin} \left( {\dfrac{{\angle C}}{2}} \right)\,{\text{ - - - - - - - - - }}\\{ \because \operatorname{Cos} (90^\circ - \theta ) = \operatorname{Sin} \theta \\} \\\$$

Putting the value in equation (2)
${\rm A} = 2\sin \dfrac{C}{2}\cos \dfrac{{A - B}}{2} + 1 - 2{\sin ^2}\dfrac{C}{2} - \dfrac{3}{2} \\\ \Rightarrow {\rm A} = - 2{\sin ^2}\dfrac{C}{2} + 2\sin \dfrac{C}{2}\cos \dfrac{{A - B}}{2} - \dfrac{1}{2} \;$
We assume the value of $\sin \dfrac{C}{2} = p$
$\Rightarrow {\rm A} = - 2{p^2} + 2p\cos \dfrac{{A - B}}{2} - \dfrac{1}{2}\,{\text{ - - - - - - - - - equation(3)}}$
We know that by equation (1): ${\rm A} < 0$ , This means the sign of the quadratic equation will be negative thus the delta which is given by $\sqrt {{b^2} - 4ac}$ will be negative i.e.
$\sqrt {{b^2} - 4ac} < 0$
Comparing equation (3) with the standard equation we have
$a = - 2 \\\ b = 2\cos \dfrac{{A - B}}{2} \\\ c = - \dfrac{1}{2} \;$
Calculating the delta $\Delta$

$$\sqrt {\left( {2\cos \dfrac{{A - B}}{2}} \right) - 4 \times \left( { - 2} \right) \times \left( { - \dfrac{1}{2}} \right)} \\\ \Rightarrow \sqrt {\left( {4{{\cos }^2}\dfrac{{A - B}}{2}} \right) - 4} \;$$

We know that $- 1 \leqslant \cos \theta \leqslant 1$
$\Rightarrow {\cos ^2}\dfrac{{A - B}}{2} \leqslant 1 \\\ \Rightarrow {\rm A} \leqslant 0 \;$
From equation (1) we can say now that
${\rm A} = \cos A + \cos B + \cos C \leqslant \dfrac{3}{2}$
Now coming to the second part of the problem which says that for the case of equality the triangle will be equilateral so for that the quadratic equation that we have in equation (3) i.e.
${\rm A} = - 2{p^2} + 2p\cos \dfrac{{A - B}}{2} - \dfrac{1}{2}$
The delta $\Delta$ here should be zero so we have
$\sqrt {\left( {4{{\cos }^2}\dfrac{{A - B}}{2}} \right) - 4} = 0 \\\ \Rightarrow 2\left( {\cos \dfrac{{A - B}}{2} - 1} \right) = 0 \\\ \Rightarrow \cos \left( {\dfrac{{A - B}}{2}} \right) = 1 \;$
We know that
$\cos 0^\circ = 1 \\\ \Rightarrow \cos \dfrac{{A - B}}{2} = \cos 0^\circ \\\ \Rightarrow A = B \;$
Putting these values in equation (3) and solving the quadratic equation
${\rm A} = - 2{p^2} + 2p\cos \dfrac{{A - A}}{2} - \dfrac{1}{2} = 0 \\\ \Rightarrow - 4{p^2} + 4p - 1 = 0 \\\ \Rightarrow 4{p^2} - 4p + 1 = 0 \;$
Solving the quadratic equation using grouping method gives us
$4{p^2} - 2p - 2p + 1 = 0 \\\ \Rightarrow 2p(2p - 1) - 1(2p - 1) = 0 \\\ \Rightarrow {(2p - 1)^2} = 0 \\\$
We know that $p = \sin \dfrac{C}{2}\,{\text{also}}\,p = \dfrac{1}{2}$ so after equating the two we have
$\sin \dfrac{C}{2} = \dfrac{1}{2} \\\ \Rightarrow C = 60^\circ \; \Rightarrow A + B = 120^\circ \;(\therefore A + B + C = 180^\circ ) \\\$
Also $A = B$ so we can say that
$A = B = C = 60^\circ$
This is the required condition of an equilateral triangle so $\vartriangle ABC\;$ is equilateral.
So, the correct answer is “ $\vartriangle ABC\;$ is equilateral”.

Note : Inequalities are solved by applying basic mathematical operations on both sides of it. We cannot just shift elements on one side in the given problem like we do in case of an ‘equals to’ sign. It works a bit different and we changed its sign by multiplying a negative number on both sides of it.