The number of ordered pairs $(\alpha, \beta)$ , where $\alpha, \beta \in (-\pi, \pi)$ satisfying $\cos(\alpha - \beta)=1$...

Question

The number of ordered pairs $(\alpha, \beta)$ , where $\alpha, \beta \in (-\pi, \pi)$ satisfying $\cos(\alpha - \beta)=1 $ and $\cos(\alpha + \beta) = \frac{1}{e}$ is...

What I have tried: $$\cos(\alpha - \beta)=1 \implies \alpha - \beta = 0$$ $$\cos(\alpha + \beta) = \frac{1}{e} \implies \cos^{-1} \frac{1}{e}$$ From here onwards I do not know what else to do. Please Help.

Answer 1

Given that $$\cos(\alpha-\beta)=1$$ $$\cos(\alpha-\beta)=\cos0$$ $$\alpha-\beta=0$$ $$\alpha=\beta$$

Now, we have $\cos(\alpha+\beta)=\dfrac1e$

Note that we have $\alpha=\beta$, so $$cos(\alpha+\alpha)=\dfrac1e$$ $$\cos2\alpha=\dfrac1e$$

Since, the angle is between $[\pi,\pi]$ $$-\pi\le\alpha\le\pi$$ $$-2\pi\le2\alpha\le2\pi$$

So, we have angles from $-2\pi$ to $0$ and from $0$ to $2\pi$.

So, the value of $\cos$ is positive and is in the quadrants $I,IV$.

So, there are $4$ pairs of $(\alpha,\beta)$ in total.

Answer 2

We have $\alpha=\beta$ so $\cos2\alpha=\frac1e$. Since $0<\frac1e<1$, there are four possible values of $2\alpha=\cos^{-1}\frac1e$ (since $2\alpha\in(-2\pi,2\pi)$), and the same number of solutions $(\alpha,\beta)$.

Answer 3

Hint The condition $\cos \theta = 1$ holds iff $\theta \in 2 \pi \Bbb Z$. So, since the open interval $(-\pi, \pi)$ has length $2 \pi$, we have $\alpha = \beta$. The remaining condition thus simplifies to $\cos 2 \alpha = \frac{1}{e}$.

Now, use the fact that $\cos \tau$ achieves each value in $(-1, 1)$ exactly twice per period in $\tau$.