If `1^0=alpha` radians, then find the approximate value of `cos60^0 1^(prime)dot`

Answer

To find the approximate value of \( \cos(60^\circ 1') \) given that \( 1^\circ = \alpha \) radians, we can follow these steps:

### Step 1: Understand the problem
We need to find the value of \( \cos(60^\circ 1') \). Here, \( 1' \) (1 minute) is a small angle, and we can express it in degrees. Since \( 1^\circ = \alpha \) radians, we have:
\[ 1' = \frac{1}{60}^\circ = \frac{\alpha}{60} \text{ radians} \]

### Step 2: Set up the function
Let \( f(x) = \cos(x) \). We want to find \( f(60^\circ + 1') \), which can be expressed as:
\[ f(60^\circ + 1') = f(60^\circ + \frac{\alpha}{60}) \]

### Step 3: Use the derivative for approximation
Using the derivative, we can approximate \( f(60^\circ + \frac{\alpha}{60}) \) using:
\[ f(a + h) \approx f(a) + f'(a) \cdot h \]
where \( a = 60^\circ \) and \( h = \frac{\alpha}{60} \).

### Step 4: Calculate \( f(60^\circ) \) and \( f'(60^\circ) \)
We know:
\[ f(60^\circ) = \cos(60^\circ) = \frac{1}{2} \]
Now, we need to find the derivative \( f'(x) \):
\[ f'(x) = -\sin(x) \]
Thus,
\[ f'(60^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2} \]

### Step 5: Substitute into the approximation formula
Now we can substitute into our approximation:
\[ f(60^\circ + \frac{\alpha}{60}) \approx f(60^\circ) + f'(60^\circ) \cdot \frac{\alpha}{60} \]
Substituting the values we found:
\[ f(60^\circ + \frac{\alpha}{60}) \approx \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\alpha}{60} \]

### Step 6: Final expression
Thus, the approximate value of \( \cos(60^\circ 1') \) is:
\[ \cos(60^\circ 1') \approx \frac{1}{2} - \frac{\sqrt{3}}{120} \cdot \alpha \]