2.1.5 Use \( \cos (\alpha-\beta) \) to derive a formula for \( \sin (\alpha-\beta) \). (Hint: use co-function) in degrees

To derive a formula for \( \sin(\alpha - \beta) \) using the cosine of the difference of angles, we can start with the known identity for \( \cos(\alpha - \beta) \): \[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] Now, we can use the co-function identity, which states that: \[ \sin(\theta) = \cos(90^\circ - \theta) \] Using this identity, we can express \( \sin(\alpha - \beta) \) in terms of cosine: \[ \sin(\alpha - \beta) = \cos(90^\circ - (\alpha - \beta)) = \cos(90^\circ - \alpha + \beta) \] Now, we can apply the cosine of a sum identity: \[ \cos(90^\circ - \alpha + \beta) = \cos(90^\circ - \alpha) \cos(\beta) - \sin(90^\circ - \alpha) \sin(\beta) \] Using the co-function identities again, we have: \[ \cos(90^\circ - \alpha) = \sin(\alpha) \quad \text{and} \quad \sin(90^\circ - \alpha) = \cos(\alpha) \] Substituting these into our expression gives: \[ \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta) \] Thus, the derived formula for \( \sin(\alpha - \beta) \) is: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]