Trigonometry! #33

Note that $\cos 3A + \cos 3B + \cos 3C = 1-4\sin\frac{3A}{2}\sin\frac{3B}{2}\sin\frac{3C}{2}$

$\therefore 1 = 1-4\sin\frac{3A}{2}\sin\frac{3B}{2}\sin\frac{3C}{2}$

$\sin\frac{3A}{2}\sin\frac{3B}{2}\sin\frac{3C}{2}=0$

Therefore one of $\sin\frac{3A}{2}$ , $\sin\frac{3B}{2}$ and $\sin\frac{3C}{2}$ is zero.

Now, we know that $\sin0^{\circ}=\sin180^{\circ}=0$ . But as $A$ , $B$ and $C$ are angles of a triangle, they cannot be zero.

Therefore one of $\frac{3A}{2}$ , $\frac{3B}{2}$ , $\frac{3C}{2}$ is $180^{\circ}$ .

And hence one of $A$ , $B$ and $C$ is $120^{\circ}$ .

If any of the other two angles were greater than $120^{\circ}$ , the sum of the three angles would exceed $180^{\circ}$ , which is not possible. Thus the greatest angle is $\boxed{120^{\circ}}$ .

Even though the question asks for a degree measure, I'm going to do the bulk of the calculations in radians seeing as there's going to be a lot of trigonometry.

Since this is a triangle, we have $A+B+C=\pi$ and so $3A+3B+3C=3\pi$ . Setting $x=3A, y=3B, z=3C$ , we have:

$1=\cos x+\cos y+\cos z$ $=\cos x+\cos y+\cos (3\pi-x-y)$ $=\cos x+\cos y-\cos(x+y)$ $=\cos x+\cos y-\cos x\cos y+\sin x\sin y=1$

Therefore:

$\sin x\sin y = (1-\cos x)(1-\cos y)$

$\sin x\sin y (1+\cos x)(1+\cos y)= (1-\cos^2 x)(1-\cos^2 y)=\sin^2 x\sin^2 y$

Suppose $\sin x\sin y\neq 0$ ; in that case we get

$(1+\cos x)(1+\cos y)=\sin x\sin y=(1-\cos x)(1-\cos y)$

which leads to $\cos x+\cos y=0$ . But then, by symmetry arguments, this equation would also hold for the pairs $(y,z)$ and $(z,x)$ , meaning that $\cos x=\cos y=\cos z=0$ . This would contradict our initial identity; therefore our assumption was wrong and we can conclude that $\sin x\sin y=0$ .

Assume without loss of generality that $\sin x=0$ . Since $0<x<3\pi$ , $x$ is equal to either $\pi$ or $2\pi$ . If we suppose that $x=\pi$ , the original equation turns into:

$1=\cos \pi+\cos y-\cos(\pi+y)=-1+2\cos y$

Therefore $\cos y=1$ and $y=2\pi$ . This would mean that $z=0$ , giving a somewhat degenerate triangle with biggest angle $B=\frac13 y=\frac23\pi$ .

If, on the other hand, $x=2\pi$ , we have $y+z=\pi$ which would mean that $A$ is the biggest angle, equal to $\frac13 x=\frac23\pi$ .

Either way, our final answer is $\frac23\pi$ radians or $\boxed{120^\circ}$ .

Interestingly, any triangle where one of the angles is equal to $120^\circ$ satisfies the original identity. The proof of this is left as an exercise to the reader.