se \( \cos (\alpha-\beta) \) to derive a formula for \( \sin (\alpha-\beta) \) (Hint: use co-function

To derive a formula for \( \sin(\alpha - \beta) \) using the co-function identity, we can start with the cosine difference identity: \[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] Using the co-function identity, we know that: \[ \sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right) \] Thus, we can express \( \sin(\alpha - \beta) \) in terms of cosine: \[ \sin(\alpha - \beta) = \cos\left(\frac{\pi}{2} - (\alpha - \beta)\right) \] This simplifies to: \[ \sin(\alpha - \beta) = \cos\left(\frac{\pi}{2} - \alpha + \beta\right) \] Now, we can use the cosine of a sum identity: \[ \cos(x + y) = \cos x \cos y - \sin x \sin y \] Applying this to our expression: \[ \sin(\alpha - \beta) = \cos\left(\frac{\pi}{2} - \alpha\right) \cos(\beta) - \sin\left(\frac{\pi}{2} - \alpha\right) \sin(\beta) \] Using the co-function identities again: \[ \cos\left(\frac{\pi}{2} - \alpha\right) = \sin(\alpha) \] \[ \sin\left(\frac{\pi}{2} - \alpha\right) = \cos(\alpha) \] Substituting these into our equation gives: \[ \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta) \] Thus, the derived formula for \( \sin(\alpha - \beta) \) is: \[ \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta) \]