Given that $\sin\alpha = \dfrac{3}{5}$ and $\cos\beta = \dfrac{12}{13}$, show that one possible value of $\cos(\alpha + \beta)$ is $\dfrac{33}{65}$, and find all the other possible values.

1. Find all possible values for $\cos\alpha$ and $\sin\beta$ using the Pythagorean identity.

Given $\sin\alpha = \dfrac{3}{5}$:
$\cos^2\alpha = 1 - \sin^2\alpha = 1 - \left(\dfrac{3}{5}\right)^2 = \dfrac{16}{25}$
$\cos\alpha = \pm\dfrac{4}{5}$

Given $\cos\beta = \dfrac{12}{13}$:
$\sin^2\beta = 1 - \cos^2\beta = 1 - \left(\dfrac{12}{13}\right)^2 = \dfrac{25}{169}$
$\sin\beta = \pm\dfrac{5}{13}$

2. State the compound angle formula for $\cos(\alpha + \beta)$.

$\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta$

3. Show that $\dfrac{33}{65}$ is one possible value by selecting the appropriate signs.

Taking $\cos\alpha = \dfrac{4}{5}$ and $\sin\beta = \dfrac{5}{13}$:
$\cos(\alpha + \beta) = \left(\dfrac{4}{5}\right)\left(\dfrac{12}{13}\right) - \left(\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right)$
$= \dfrac{48}{65} - \dfrac{15}{65} = \dfrac{33}{65} \quad \text{(Shown)}$

4. Calculate all other possible values for $\cos(\alpha + \beta)$ by exhausting the remaining sign combinations.

Case 2: $\cos\alpha = \dfrac{4}{5}$ and $\sin\beta = -\dfrac{5}{13}$
$\cos(\alpha + \beta) = \left(\dfrac{4}{5}\right)\left(\dfrac{12}{13}\right) - \left(\dfrac{3}{5}\right)\left(-\dfrac{5}{13}\right) = \dfrac{48}{65} + \dfrac{15}{65} = \dfrac{63}{65}$

Case 3: $\cos\alpha = -\dfrac{4}{5}$ and $\sin\beta = \dfrac{5}{13}$
$\cos(\alpha + \beta) = \left(-\dfrac{4}{5}\right)\left(\dfrac{12}{13}\right) - \left(\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right) = -\dfrac{48}{65} - \dfrac{15}{65} = -\dfrac{63}{65}$

Case 4: $\cos\alpha = -\dfrac{4}{5}$ and $\sin\beta = -\dfrac{5}{13}$
$\cos(\alpha + \beta) = \left(-\dfrac{4}{5}\right)\left(\dfrac{12}{13}\right) - \left(\dfrac{3}{5}\right)\left(-\dfrac{5}{13}\right) = -\dfrac{48}{65} + \dfrac{15}{65} = -\dfrac{33}{65}$