B2. Let $\sin \alpha + \sin \beta = 1$ and $\cos \alpha + \cos \beta = -\sqrt{3}$.

(a) Calculate the value of the expression $\cos (\alpha - \beta)$.

(b) Find all pairs of real numbers $\alpha$ and $\beta$ that satisfy the given equations.

## 61st Mathematical Competition for High School Students in Slovenia

Selection Competition, March 16, 2017

## Problems for 4th Year

## Time for solving: 45 minutes.
B2. (a) The addition theorem for cosine gives us $\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$. We square the given equations to obtain

$$
\begin{array}{r}
\sin ^{2} \alpha+2 \sin \alpha \sin \beta+\sin ^{2} \beta=1 \\
\cos ^{2} \alpha+2 \cos \alpha \cos \beta+\cos ^{2} \beta=3
\end{array}
$$

We now add the obtained equations and use the identity $\sin ^{2} x+\cos ^{2} x=1$, to get

$$
1+2(\sin \alpha \sin \beta+\cos \alpha \cos \beta)+1=4
$$

from which it follows that $\sin \alpha \sin \beta+\cos \alpha \cos \beta=1$ or $\cos (\alpha-\beta)=1$.

(b) Since $\cos (\alpha-\beta)=1$, it follows that $\alpha-\beta=2 k \pi$ or $\alpha=\beta+2 k \pi, k \in \mathbb{Z}$. Therefore, $\cos \alpha=\cos (\beta+2 k \pi)=\cos \beta$ and similarly $\sin \alpha=\sin \beta$. When we substitute this into both given equations, we can express $\sin \beta=\frac{1}{2}$ and $\cos \beta=-\frac{\sqrt{3}}{2}$. From this, we conclude that $\beta=\frac{5 \pi}{6}+2 n \pi, n \in \mathbb{Z}$. Thus, $\alpha=\frac{5 \pi}{6}+2 n \pi+2 k \pi=\frac{5 \pi}{6}+2(n+k) \pi$. If we let $k+n=m$, we can write $\alpha=\frac{5 \pi}{6}+2 m \pi, m \in \mathbb{Z}$. The pairs that satisfy the given equations are therefore $\alpha=\frac{5 \pi}{6}+2 m \pi, \beta=\frac{5 \pi}{6}+2 n \pi$, where $m$ and $n$ are any integers.

(a) Writing or using the addition theorem. 3 points Squaring the equations. 2 points
Writing or using the identity $\sin ^{2} x+\cos ^{2} x=1$ ..... 1 point
Writing the equation $1+2(\sin \alpha \sin \beta+\cos \alpha \cos \beta)+1=4$ ..... 2 points
Transforming the equation into $\cos \alpha \cos \beta+\sin \alpha \sin \beta=1$ ..... 2 points
(b) Writing the solution $\alpha-\beta=2 k \pi$. ..... 1 point
Determining that $\cos \alpha=\cos \beta$ and $\sin \alpha=\sin \beta$. ..... 2 points
Determining that $\sin \beta=\frac{1}{2}$ and $\cos \beta=-\frac{\sqrt{3}}{2}$.... ..... 2 points
Calculating $\beta=\frac{5 \pi}{6}+2 n \pi, n \in \mathbb{Z}$ ..... 1 point
Calculating $\alpha=\frac{5 \pi}{6}+2 n \pi+2 k \pi=\frac{5 \pi}{6}+2(n+k) \pi$... ..... 2 points
Writing the solutions $\alpha=\frac{5 \pi}{6}+2 m \pi, \beta=\frac{5 \pi}{6}+2 n \pi$, where $m, n \in \mathbb{Z}$ ..... 2 points
(If the contestant omits $m, n \in \mathbb{Z}$, 1 point is deducted)

## 61st Mathematical Competition for High School Students in Slovenia

Selection Competition, March 16, 2017

Each mathematically correct and complete solution is worth $\mathbf{2 0}$ points, even if the solving procedure is different from the official solution.

## Solutions for 4th Year