Consider the following for next three) items:
It given that `cos(theta-alpha ) = a, cos(theta-beta) = b`.
What is cos `(alpha-beta)` equal to ?

Options

A. `ab+sqrt(1-a^2)sqrt(1-b^2)`
B. `ab-(1-a^2)(1-b^2)`
C. `asqrt(1-b^2)-bsqrt(1-a^2)`
D. `a sqrt (1-b^2)+bsqrt(1-a^2)`

Answer

To find \( \cos(\alpha - \beta) \) given that \( \cos(\theta - \alpha) = a \) and \( \cos(\theta - \beta) = b \), we can follow these steps:

### Step 1: Express \( \sin(\theta - \alpha) \) and \( \sin(\theta - \beta) \)

Using the Pythagorean identity, we know that:
\[ \sin^2(x) + \cos^2(x) = 1 \]
Thus, we can express \( \sin(\theta - \alpha) \) and \( \sin(\theta - \beta) \) as:
\[ \sin(\theta - \alpha) = \sqrt{1 - a^2} \]
\[ \sin(\theta - \beta) = \sqrt{1 - b^2} \]

### Step 2: Use the cosine subtraction formula

The cosine of the difference of two angles can be expressed using the formula:
\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]
In our case, we want to find \( \cos(\alpha - \beta) \), which can be rewritten as:
\[ \cos(\alpha - \beta) = \cos((\theta - \alpha) - (\theta - \beta)) \]
This leads us to:
\[ \cos(\alpha - \beta) = \cos(\theta - \alpha) \cos(\theta - \beta) + \sin(\theta - \alpha) \sin(\theta - \beta) \]

### Step 3: Substitute the known values

Now substituting the values we have:
\[ \cos(\alpha - \beta) = a \cdot b + \sqrt{1 - a^2} \cdot \sqrt{1 - b^2} \]

### Step 4: Simplify the expression

Thus, we can express \( \cos(\alpha - \beta) \) as:
\[ \cos(\alpha - \beta) = ab + \sqrt{(1 - a^2)(1 - b^2)} \]

### Final Answer

The final expression for \( \cos(\alpha - \beta) \) is:
\[ \cos(\alpha - \beta) = ab + \sqrt{(1 - a^2)(1 - b^2)} \]

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