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\begin{tabular}{l}2.1 .6 Use \( \cos (\alpha-\beta) \) to derive a formula for \( \sin (\alpha+\beta) \). (Hint: use co-function) in degreec \( (\alpha) \) \\ \hline\end{tabular}

Ask by Rojas Olson. in South Africa
Feb 03,2025

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Answer

\[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

Solution

Certainly! Let's derive the formula for \( \sin (\alpha + \beta) \) using the cosine of a difference angle \( \cos (\alpha - \beta) \) and co-function identities. ### **Objective:** Derive the formula for \( \sin (\alpha + \beta) \) using \( \cos (\alpha - \beta) \) and co-function identities. ### **Definitions and Identities:** 1. **Cosine of Difference Identity:** \[ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] 2. **Co-function Identity:** \[ \sin \theta = \cos \left(90^\circ - \theta \right) \] \[ \cos \theta = \sin \left(90^\circ - \theta \right) \] ### **Step-by-Step Derivation:** 1. **Express \( \sin (\alpha + \beta) \) Using the Co-function Identity:** The co-function identity allows us to rewrite sine in terms of cosine: \[ \sin (\alpha + \beta) = \cos \left(90^\circ - (\alpha + \beta)\right) \] 2. **Simplify the Argument inside the Cosine:** Distribute the negative sign inside the parentheses: \[ 90^\circ - (\alpha + \beta) = (90^\circ - \alpha) - \beta \] So, \[ \sin (\alpha + \beta) = \cos \left( (90^\circ - \alpha) - \beta \right) \] 3. **Apply the Cosine of a Difference Identity:** Using the identity \( \cos (A - B) = \cos A \cos B + \sin A \sin B \), let: \[ A = 90^\circ - \alpha \quad \text{and} \quad B = \beta \] Then, \[ \cos \left( (90^\circ - \alpha) - \beta \right) = \cos (90^\circ - \alpha) \cos \beta + \sin (90^\circ - \alpha) \sin \beta \] 4. **Simplify Using Co-function Identities Again:** Recall the co-function identities: \[ \cos (90^\circ - \alpha) = \sin \alpha \] \[ \sin (90^\circ - \alpha) = \cos \alpha \] Substitute these into the equation: \[ \cos (90^\circ - \alpha) \cos \beta + \sin (90^\circ - \alpha) \sin \beta = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] 5. **Combine the Results:** Therefore, \[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] ### **Final Formula:** \[ \boxed{ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta } \] ### **Summary:** By expressing \( \sin (\alpha + \beta) \) as a cosine function using the co-function identity and then applying the cosine of a difference identity, we successfully derived the well-known sine addition formula: \[ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

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To derive the formula for \( \sin(\alpha + \beta) \) using \( \cos(\alpha - \beta) \), we start with the co-function identity for sine and cosine, which states that \( \sin(x) = \cos(90^\circ - x) \). We can express \( \sin(\alpha + \beta) \) as follows: 1. We know that \( \cos(\alpha - \beta) = \sin(90^\circ - (\alpha - \beta)) = \sin(90^\circ - \alpha + \beta) \). 2. Rearranging, this gives us \( \cos(\alpha - \beta) = \sin(90^\circ - \alpha + \beta) \). 3. Now let's take the complementary angles: \( 90^\circ - \alpha + \beta = 90^\circ - (\beta - \alpha) \). 4. Therefore, we conclude that \( \sin(\alpha + \beta) = \cos(90^\circ - (\alpha + \beta)) \). However, to directly derive \( \sin(\alpha + \beta) \) without confusion, we can also use the sine addition formula: \[ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \] This gives us the direct relation we’re looking for through geometric interpretations of these trigonometric functions.

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