The value of `sin^(-1) (sin 12) + cos^(-1) (cos 12)` is equal to

Options

A. zero
B. `24 - 2pi`
C. `4pi -24`
D. none of these

Explanation

To solve the problem \( \sin^{-1}(\sin 12) + \cos^{-1}(\cos 12) \), we can follow these steps:

### Step 1: Understanding the Range of Inverse Functions
The function \( \sin^{-1}(x) \) (or arcsin) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) and \( \cos^{-1}(x) \) (or arccos) has a range of \( [0, \pi] \).

### Step 2: Simplifying \( \sin^{-1}(\sin 12) \)
Since \( 12 \) radians is greater than \( \frac{\pi}{2} \) (approximately 1.57 radians), we need to find an equivalent angle within the range of \( \sin^{-1} \). We can express \( 12 \) in terms of \( 2\pi \):

\[ 12 \text{ radians} = 12 - 4\pi \text{ radians} \quad (\text{since } 4\pi \approx 12.566 \text{ radians}) \]

Now, we can find \( \sin^{-1}(\sin 12) \):

\[ \sin^{-1}(\sin 12) = \sin^{-1}(\sin(12 - 4\pi)) = 12 - 4\pi \]

### Step 3: Simplifying \( \cos^{-1}(\cos 12) \)
Similarly, since \( 12 \) radians is also greater than \( \pi \), we can express \( \cos^{-1}(\cos 12) \) as:

\[ \cos^{-1}(\cos 12) = \cos^{-1}(\cos(12 - 2\pi)) = 12 - 2\pi \]

### Step 4: Adding the Two Results
Now we can add the two results together:

\[ \sin^{-1}(\sin 12) + \cos^{-1}(\cos 12) = (12 - 4\pi) + (12 - 2\pi) \]

Combining these gives:

\[ = 12 - 4\pi + 12 - 2\pi = 24 - 6\pi \]

### Step 5: Final Result
Thus, the value of \( \sin^{-1}(\sin 12) + \cos^{-1}(\cos 12) \) is:

\[ \boxed{24 - 6\pi} \]