Animated Solution for Mathematics - Trigonometry: $\cos (\alpha -\beta )=1$ and $\cos (\alpha +\beta )=1/e$ where $\alpha ,\beta \in [-\pi ,\pi ]$. Pairs of $\alpha ,\beta$ which satisfy both the equations is/are

We are tasked with solving a system of trigonometric equations for variables $\alpha$ and $\beta$ within the interval $[-\pi ,\pi ]$:

These constraints define the boundary of our mathematical domain.

First, we examine the equation $\cos (\alpha -\beta )=1$. The cosine function attains the value of $1$ only at even multiples of $\pi$.

Therefore, we have:

Given that $\alpha ,\beta \in [-\pi ,\pi ]$, the difference $\alpha -\beta$ is strictly restricted to the interval $[-2\pi ,2\pi ]$. This limits the possible values for the integer $n$ to $\{-1,0,1\}$.

We must test the boundary cases where $n=1$ and $n=-1$.

If $\alpha -\beta =2\pi$, then $\alpha =\pi$ and $\beta =-\pi$. Conversely, if $\alpha -\beta =-2\pi$, then $\alpha =-\pi$ and $\beta =\pi$.

In both scenarios, the sum $\alpha +\beta$ equals $0$. Substituting this into the second equation yields:

This contradicts our second given equation, $\cos (\alpha +\beta )=\frac{1}{e}$. Consequently, these boundary cases are rejected, leaving us with the elegant conclusion that $\alpha -\beta =0$, or simply $\alpha =\beta$.

With the simplification $\beta =\alpha$, the second equation becomes:

We seek solutions for $\alpha \in [-\pi ,\pi ]$, which implies that the argument $2\alpha$ lies in the interval $[-2\pi ,2\pi ]$.

By observing the graph of $y=\cos (2\alpha )$ over two full cycles, we note that since $0<\frac{1}{e}<1$, the horizontal line $y=\frac{1}{e}$ intersects the cosine wave at exactly four distinct points.

Each intersection corresponds to a valid value for $\alpha$. Since $\beta =\alpha$ for every solution, there are exactly four unique pairs $(\alpha ,\beta )$ that satisfy the system.