Question

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

Ask by Munoz Hampton. in South Africa
Feb 03,2025

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Tutor-Verified Answer

Answer

The formula for \( \sin(\alpha + \beta) \) is: \[ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \]

Solution

To derive a formula for \( \sin(\alpha + \beta) \) using the cosine of the difference of angles, we can follow these steps: 1. **Recall the co-function identity**: The sine and cosine functions are co-functions, which means that: \[ \sin(\theta) = \cos(90^\circ - \theta) \] 2. **Express \( \sin(\alpha + \beta) \)**: We can express \( \sin(\alpha + \beta) \) in terms of cosine using the co-function identity: \[ \sin(\alpha + \beta) = \cos(90^\circ - (\alpha + \beta)) \] 3. **Simplify the expression**: We can rewrite the argument of the cosine function: \[ 90^\circ - (\alpha + \beta) = 90^\circ - \alpha - \beta \] Thus, we have: \[ \sin(\alpha + \beta) = \cos(90^\circ - \alpha - \beta) \] 4. **Use the cosine of a difference**: Now, we can use the cosine of the difference of angles: \[ \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \] However, we need to express \( \sin(\alpha + \beta) \) in terms of \( \cos(\alpha - \beta) \). 5. **Relate \( \sin(\alpha + \beta) \) to \( \cos(\alpha - \beta) \)**: We can use the identity: \[ \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \] to derive the sine addition formula. 6. **Final formula**: The sine addition formula can be derived as follows: \[ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \] Thus, the derived formula for \( \sin(\alpha + \beta) \) is: \[ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \]

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To derive a formula for \( \sin (\alpha + \beta) \), we start by using the co-function identity that states \( \sin(\theta) = \cos(90^\circ - \theta) \). Applying this to \( \sin(\alpha + \beta) \): \[ \sin(\alpha + \beta) = \cos(90^\circ - (\alpha + \beta)) = \cos(90^\circ - \alpha - \beta) \] Next, we simplify \( \cos(90^\circ - \alpha - \beta) \) using the cosine subtraction formula: \[ \cos(90^\circ - (\alpha + \beta)) = \cos(90^\circ - \alpha - \beta) = \sin(\alpha + \beta) \quad \text{(using the co-function identity)} \] Now, applying the cosine subtraction identity: \[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] Therefore, we can conclude that: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] Thus, the derived formula is: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

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