If $A, B$ and $C$ are real angles such that $$\cos (B-C)+\cos (C-A)+\cos (A-B)=-3/2,$$find $$\cos (A)+\cos (B)+\cos (C)$$
Problem
Source: Canada RepĂȘchage 2021/7 CMOQR
Tags: Canada, repechage, algebra, trigonometry
booth
30.03.2021 18:27
Is $ABC$ a triangle?
baldeagle123
30.03.2021 19:37
@Above I don't think that knowledge is necessary to solve the problem, unless I am missing smth.
Let $\cos A + \cos B + \cos C = x$ and $\sin A + \sin B + \sin C = y$.
The cosine difference identity and some factoring on the left side of the given equation yields
\begin{align*} \cos A \cos B+ \sin A \sin B + \cos B \cos C+ \sin B \sin C + \cos A \cos C+ \sin A \sin C &= 2(\frac{\cos A \cos B+ \sin A \sin B}{2}) + 2(\frac{\cos A \cos C+ \sin A \sin C}{2}) + 2(\frac{\cos C \cos B+ \sin C \sin B}{2}),\\
&= \frac{\cos C(x - \cos C) + \cos B(x - \cos B) + \cos A(x - \cos A) + \sin C(y - \sin C) + \sin B (y - \sin B) + \sin A(y - \sin A)}{2},\\
&= \frac{x^2 + y^2 - 3}{2}.
\end{align*}
Equating this value to $-\frac{3}{2}$, we find that $x^2 + y^2 = 0$. Since $A, B, C$ are all real angles, their corresponding cosine and sine values are also necessarily real. Thus, for the sum of the two squares to be $0$, both $x^2$ and $y^2$ also must be $0$.
Thus $x^2 = x =\boxed{ \cos A + \cos B + \cos C= 0}$ and we are done.
sqing
31.03.2021 15:48
Let $A, B$ and $C$ are real angles such that $\cos (A-B)+\cos (B-C)+\cos (C-A)= -\frac{3}{2}.$ Find the value of $\cos A+\cos B+\cos C.$ Very good. CMOQR (Qualifying Repechage) Prove that for all $x_1, x_2,\ldots , x_n \in \mathbb R$ the following inequality holds: \[\sum_{n\geq i>j\geq 1} cos(x_i-x_j)\geq -\frac{n}{2}.\]