2.15 Use \( \cos (\alpha-\beta) \) to derive a formula for \( \sin (\alpha-\beta) \). (Hint: use co-function) \( \qquad \) \( \qquad \) (3) 21.6 Use \( \cos (\alpha-\beta) \) to derive a formula for \( \sin (\alpha+\beta) \). (Hint: use ce-function) \( \qquad \) \( \qquad \) (3) [22] Cond 12 Mathenutics SBA 2005 Prge 18 of 33

To derive a formula for \( \sin (\alpha-\beta) \) using \( \cos (\alpha-\beta) \), recall that \( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \). Using the co-function identity, we know that \( \sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right) \). By exploiting that relationship, we can derive the sine difference formula: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] Now for \( \sin(\alpha+\beta) \), we can manipulate the expression for \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \). Again using co-function identities, and knowing that \( \sin(\alpha+\beta) = \cos\left(\frac{\pi}{2} - (\alpha+\beta)\right) \), we can derive the sum of angles as: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]