Find the approximate values of :
`cos(60^(@)30')," given "1^(@)=0.0175^( c )" and "sin60^(@)=0.8660`.

Answer

To find the approximate value of \( \cos(60^\circ 30') \), we can follow these steps:

### Step 1: Understand the angle
First, we need to convert \( 30' \) (30 minutes) into degrees. Since \( 1^\circ = 60' \), we have:
\[ 30' = \frac{30}{60}^\circ = 0.5^\circ \]
Thus, \( 60^\circ 30' = 60^\circ + 0.5^\circ = 60.5^\circ \).

### Step 2: Define the function
Let \( f(x) = \cos(x) \). We want to find \( f(60.5^\circ) \).

### Step 3: Use the first principle of derivatives
Using the first principle of derivatives, we can approximate \( f(60.5^\circ) \) using:
\[ f(60.5^\circ) \approx f(60^\circ) + h \cdot f'(60^\circ) \]
where \( h = 0.5^\circ \).

### Step 4: Calculate \( f(60^\circ) \)
From trigonometric values, we know:
\[ f(60^\circ) = \cos(60^\circ) = \frac{1}{2} = 0.5 \]

### Step 5: Calculate the derivative \( f'(x) \)
The derivative of \( f(x) = \cos(x) \) is:
\[ f'(x) = -\sin(x) \]
Thus, we need to evaluate \( f'(60^\circ) \):
\[ f'(60^\circ) = -\sin(60^\circ) \]
Given that \( \sin(60^\circ) = 0.8660 \), we have:
\[ f'(60^\circ) = -0.8660 \]

### Step 6: Substitute values into the approximation formula
Now we can substitute \( f(60^\circ) \), \( h \), and \( f'(60^\circ) \) into the approximation:
\[ f(60.5^\circ) \approx 0.5 + 0.5 \cdot (-0.8660) \]
Calculating this gives:
\[ f(60.5^\circ) \approx 0.5 - 0.4330 = 0.0670 \]

### Step 7: Final Calculation
To find the approximate value of \( \cos(60.5^\circ) \):
\[ \cos(60.5^\circ) \approx 0.5 - 0.4330 = 0.0670 \]

### Conclusion
Thus, the approximate value of \( \cos(60^\circ 30') \) is:
\[ \cos(60^\circ 30') \approx 0.0670 \]