Question

If \( a = \sin^{-1} (\sin(5)) \) and \( b = \cos^{-1} (\cos(5)) \), then \( a^2 + b^2 \) is equal to

• A. \( 4\pi^2 + 25 \)
• B. \( 8\pi^2 - 40\pi + 50 \)
• C. \( 4\pi^2 - 20\pi + 50 \)
• D. \( 25 \)

Answer 1

The Correct Option is B

To solve the problem, we need to determine the values of \(a\) and \(b\), and subsequently find \(a^2 + b^2\).

Given:

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

First, consider \(a = \sin^{-1} (\sin(5))\):

The sine inverse function, \(\sin^{-1}(x)\), returns values within the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\). However, \(\sin(5)\) may not be directly within this interval. We should note that the sine function is periodic with period \(2\pi\), which means: \[ \sin(5) = \sin(5 - 2\pi) = \sin(5 - 2\pi k) \quad \text{for some integer } k. \]

Since \(5\) is approximately \(1.5908\pi\), the angle \(5\) is within the range \([\pi, 2\pi]\). Thus: \[ 5 - 2\pi = 5 - 6.2832 \approx -1.2832 \quad \text{which is within the interval } [-\frac{\pi}{2}, \frac{\pi}{2}]. \]

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

Now consider \(b = \cos^{-1} (\cos(5))\):

The cosine inverse function, \(\cos^{-1}(x)\), returns values within \([0, \pi]\): \[ b = \cos^{-1}(\cos(5)) = 2\pi - 5, \] since \(5\) is in the interval \([\pi, 2\pi]\) and \(\cos(x)\) has a symmetry in cosine when moved by \(2\pi\).

Now, calculate \(a^2 + b^2\):

Substitute the computed values of \(a\) and \(b\): \[ a = 5 - 2\pi,\quad b = 2\pi - 5. \]

We have: \[ a^2 = (5 - 2\pi)^2 = 25 - 20\pi + 4\pi^2, \] \[ b^2 = (2\pi - 5)^2 = 4\pi^2 - 20\pi + 25. \]

Thus: \[ a^2 + b^2 = (25 - 20\pi + 4\pi^2) + (4\pi^2 - 20\pi + 25) \] \[ = 8\pi^2 - 40\pi + 50. \]

Hence, the correct answer is \(8\pi^2 - 40\pi + 50\).

Answer 2

Calculate \( a = \sin^{-1}(\sin(5)) \). To find \( a \), note that \( \sin^{-1}(\sin(x)) \) gives a result in the range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Since 5 is outside this range, we need to adjust it. We have:

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

Thus,

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

Calculate \( b = \cos^{-1}(\cos(5)) \). To find \( b \), note that \( \cos^{-1}(\cos(x)) \) gives a result in the range \([0, \pi]\).

Since 5 is within this range, we can write:

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

Calculate \( a^2 + b^2 \). Now, substitute \( a = 5 - 2\pi \) and \( b = 2\pi - 5 \):

\[ a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2. \]

Expanding both terms:

\[ = (5 - 2\pi)^2 + (2\pi - 5)^2 = (25 - 20\pi + 4\pi^2) + (4\pi^2 - 20\pi + 25). \]

Combine like terms:

\[ = 8\pi^2 - 40\pi + 50. \]

Thus, the answer is:

\[ 8\pi^2 - 40\pi + 50 \]