7.4 Q-8

Question Statement

Prove the trigonometric identity:

$$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$

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Background and Explanation

This problem involves proving a well-known trigonometric identity using vectors. The identity itself is a standard result in trigonometry, known as the sine subtraction formula.

To prove this identity, we will use the concept of unit vectors in the $xy$-plane, making angles $\alpha$ and $\beta$ with the x-axis, and applying the cross product of two vectors. The goal is to relate the sine of the angle difference between two vectors to the components of the vectors in the $x$ and $y$ directions.

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Solution

Step 1: Define the unit vectors

Let $\hat{a}$ and $\hat{b}$ be the unit vectors in the $xy$-plane, making angles $\alpha$ and $\beta$ with the $x$-axis, respectively, where $\alpha > \beta$. These unit vectors can be expressed as:

$$\hat{a} = \cos \alpha , \underline{i} + \sin \alpha , \underline{j}$$ $$\hat{b} = \cos \beta , \underline{i} + \sin \beta , \underline{j}$$

where $\underline{i}$ and $\underline{j}$ are the unit vectors along the $x$- and $y$-axes, respectively.

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Step 2: Compute the cross product $\hat{a} \times \hat{b}$

The next step is to compute the cross product of $\hat{a}$ and $\hat{b}$, which gives us the magnitude of the sine of the angle between them.

The cross product is given by the determinant of the matrix:

$$\hat{a} \times \hat{b} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \cos \beta & \sin \beta & 0 \cos \alpha & \sin \alpha & 0 \end{array} \right|$$

Expanding this determinant:

$$\hat{a} \times \hat{b} = \hat{i}(0) - \hat{j}(0) + \hat{k}(\cos \beta \sin \alpha - \sin \beta \cos \alpha)$$ $$\Rightarrow \hat{a} \times \hat{b} = (\cos \beta \sin \alpha - \sin \beta \cos \alpha) \hat{k}$$

This expression represents the magnitude of the cross product along the $\hat{k}$-direction (perpendicular to the plane formed by $\hat{i}$ and $\hat{j}$).

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Step 3: Use the magnitude of the cross product

The magnitude of the cross product can also be written as:

$$\hat{a} \times \hat{b} = |\hat{a}| |\hat{b}| \sin (\alpha - \beta) \hat{k}$$

Since $\hat{a}$ and $\hat{b}$ are both unit vectors, their magnitudes are 1:

$$|\hat{a}| = |\hat{b}| = 1$$

Thus:

$$\hat{a} \times \hat{b} = \sin (\alpha - \beta) \hat{k}$$

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Step 4: Compare the two expressions for $\hat{a} \times \hat{b}$

Now, comparing the two expressions for $\hat{a} \times \hat{b}$, we have:

$$\sin (\alpha - \beta) \hat{k} = (\cos \beta \sin \alpha - \sin \beta \cos \alpha) \hat{k}$$

Since the $\hat{k}$-components are equal, we conclude:

$$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$

Thus, the identity is proved.

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Key Formulas or Methods Used

1. Cross Product of Two Vectors:

$$\hat{a} \times \hat{b} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} a_1 & a_2 & a_3 b_1 & b_2 & b_3 \end{array} \right|$$

2. Magnitude of Cross Product:

$$|\hat{a} \times \hat{b}| = |\hat{a}| |\hat{b}| \sin (\theta)$$

where $\theta$ is the angle between the two vectors.

3. Trigonometric Identity:

$$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$

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Summary of Steps

1. Define the unit vectors $\hat{a}$ and $\hat{b}$ in terms of the angles $\alpha$ and $\beta$.
2. Compute the cross product $\hat{a} \times \hat{b}$ and simplify the result.
3. Use the formula for the magnitude of the cross product to express it in terms of $\sin (\alpha - \beta)$.
4. Compare the two expressions for the cross product and conclude the identity.