Question

If in a triangle $\vartriangle ABC$ , $\cos 3A+\cos 3B+\cos 3C=1$, then one angle must be exactly equal to:
A. ${{90}^{o}}$
B. ${{45}^{o}}$
C. ${{120}^{0}}$
D. None of these.

Answer 1

Hint: We will use the angle sum property of the triangle according to which the total sum of the angles in a triangle is ${{180}^{o}}$.
We will then multiply the equation by $3$, and then write the equation in terms of one of the angles. Using a given equation, we will use a substitution method and trigonometric formulas to derive the value of one of the angles.

Formula Used: The half angle formula of sine and cosine are:
$\begin{align}
  & \cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}-1 \\
 & \sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}
\end{align}$
The other formula of cosine which will be used is:
$\cos (a+b)=\cos a\cos b-\sin a\sin b$

Complete step by step solution: We are given a triangle $\vartriangle ABC$ for which $\cos 3A+\cos 3B+\cos 3C=1$ and we have to find the value of one of the angles.
First we will use the angle sum property for the triangle $\vartriangle ABC$,
So,
$A+B+C={{180}^{0}}$………(i)
We will multiply this equation by $3$ on both sides,
So the equation will be,
$3A+3B+3C={{540}^{0}}$
Now we will write this equation in terms of any of the angle,
$3C={{540}^{0}}-(3A+3B)$
We will take cosine on the both side,
$\cos \,3C=\cos \,\,({{540}^{0}}-(3A+3B))$
$\cos \,3C=\cos \,\,(3\pi -(3A+3B))$
$\cos \,3C=-\cos \,\,(3A+3B)$……(ii)
Now we will substitute equation (ii) in $\cos 3A+\cos 3B+\cos 3C=1$.
\[\begin{align}
  & \cos 3A+\cos 3B+(-\cos (3A+3B))=1 \\
 & \cos 3A+\cos 3B-\cos (3A+3B)=1 \\
 & \cos 3A+\cos 3B-(\cos 3A\cos 3B-\sin 3A\sin 3B)=1 \\
 & \cos 3A+\cos 3B-\cos 3A\cos 3B+\sin 3A\sin 3B=1
\end{align}\]
Taking $\cos 3A$ common,
\[\begin{align}
  & \cos 3A-\cos 3A\cos 3B+\cos 3B+\sin 3A\sin 3B=1 \\
 & \cos 3A(1-\cos 3B)+\cos 3B+\sin 3A\sin 3B=1 \\
 & \cos 3A(1-\cos 3B)+\cos 3B+\sin 3A\sin 3B-1=0
\end{align}\]
Now we will take $\cos 3B$ common,
\[\begin{align}
  & \cos 3A(1-\cos 3B)-1+\cos 3B+\sin 3A\sin 3B=0 \\
 & \cos 3A(1-\cos 3B)-1(1-\cos 3B)+\sin 3A\sin 3B=0 \\
 & (1-\cos 3B)(\cos 3A-1)+\sin 3A\sin 3B=0 \\
 & -(1-\cos 3B)(1-\cos 3A)+\sin 3A\sin 3B=0
\end{align}\]
Now,
\[\sin 3A\sin 3B=(1-\cos 3B)(1-\cos 3A)\]
\[\dfrac{\sin 3A}{(1-\cos 3A)}=\dfrac{(1-\cos 3B)}{\sin 3B}\]
We will now use half angles formula on both the sides,
\[\dfrac{2\sin \dfrac{3A}{2}\cos \dfrac{3A}{2}}{1+2{{\cos }^{2}}\dfrac{3A}{2}-1}=\dfrac{1+2{{\cos }^{2}}\dfrac{3B}{2}-1}{2\sin \dfrac{3B}{2}\cos \dfrac{3B}{2}}\]
\[\dfrac{2\sin \dfrac{3A}{2}\cos \dfrac{3A}{2}}{2{{\cos }^{2}}\dfrac{3A}{2}}=\dfrac{2{{\cos }^{2}}\dfrac{3B}{2}}{2\sin \dfrac{3B}{2}\cos \dfrac{3B}{2}}\]
\[\dfrac{\sin \dfrac{3A}{2}}{\cos \dfrac{3A}{2}}=\dfrac{\cos \dfrac{3B}{2}}{\sin \dfrac{3B}{2}}\]
\[\tan \dfrac{3A}{2}=\cot \dfrac{3B}{2}\]
We can write $\cot $ in terms of $\tan $,
\[\begin{align}
  & \tan \dfrac{3A}{2}=\tan (\dfrac{\pi }{2}-\dfrac{3B}{2}) \\
 & \dfrac{3A}{2}=\dfrac{\pi }{2}-\dfrac{3B}{2}
\end{align}\]
$\begin{align}
  & 3A=\pi -3B \\
 & 3A+3B=\pi \\
 & 3(A+B)=\pi \\
 & A+B=\dfrac{\pi }{3}.................(iii)
\end{align}$
Substituting the equation (iii) in equation (i).
$\begin{align}
  & \dfrac{\pi }{3}+C={{180}^{0}} \\
 & {{60}^{0}}+C={{180}^{0}} \\
 & C={{120}^{0}}
\end{align}$

The value of one of the angles of a triangle $\vartriangle ABC$ for which $\cos 3A+\cos 3B+\cos 3C=1$, then one of the angles is $C={{120}^{0}}$. Hence the correct option is (C).

Note: The formula of the cosine $\cos (3\pi -\theta )$is $-\cos \theta $ that is $\cos (3\pi -\theta )=-\cos \theta $.