If `a = sin^(-1) (sin 5)` and `b = cos^(-1) (cos 5)` then value of `a^2 + b^2` is

Options

A. `(2pi - 5)^2`
B. `(3pi - 7)^2`
C. `2(2pi - 5)^2`
D. `2(3pi - 7)^2`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given by:

1. \( a = \sin^{-1}(\sin 5) \)
2. \( b = \cos^{-1}(\cos 5) \)

### Step 1: Find \( a \)

The function \( \sin^{-1}(x) \) (or arcsin) returns a value in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\). Since \( 5 \) radians is outside this range, we need to find an equivalent angle within this range.

We can express \( 5 \) radians in terms of \( 2\pi \):

\[ 5 = 2\pi - (5 - 2\pi) = 2\pi - 5 \]

Since \( \sin(5) = \sin(2\pi - 5) \), we have:

\[ a = \sin^{-1}(\sin 5) = \sin^{-1}(\sin(2\pi - 5)) = 2\pi - 5 \]

However, since \( 2\pi - 5 \) is not in the range of \( \sin^{-1} \), we can also express \( 5 \) as:

\[ 5 = \pi + (5 - \pi) = \pi + (5 - \pi) \]

Thus, we can write:

\[ \sin(5) = -\sin(5 - \pi) \]

This gives:

\[ a = - (5 - \pi) = \pi - 5 \]

### Step 2: Find \( b \)

The function \( \cos^{-1}(x) \) (or arccos) returns a value in the range \([0, \pi]\). Since \( 5 \) radians is outside this range, we can express \( 5 \) in terms of \( 2\pi \):

\[ b = \cos^{-1}(\cos 5) = 2\pi - 5 \]

### Step 3: Calculate \( a^2 + b^2 \)

Now we have:

\[ a = \pi - 5 \]
\[ b = 2\pi - 5 \]

Now we can calculate \( a^2 + b^2 \):

\[ a^2 = (\pi - 5)^2 = \pi^2 - 10\pi + 25 \]
\[ b^2 = (2\pi - 5)^2 = 4\pi^2 - 20\pi + 25 \]

Adding these two results:

\[ a^2 + b^2 = (\pi^2 - 10\pi + 25) + (4\pi^2 - 20\pi + 25) \]
\[ = \pi^2 + 4\pi^2 - 10\pi - 20\pi + 25 + 25 \]
\[ = 5\pi^2 - 30\pi + 50 \]

### Final Result

Thus, the value of \( a^2 + b^2 \) is:

\[ \boxed{5\pi^2 - 30\pi + 50} \]