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

We need to find $$a^2 + b^2$$ where $$a = \sin^{-1}(\sin 5)$$ and $$b = \cos^{-1}(\cos 5)$$.

The inverse trigonometric functions return values in their principal ranges:

$$\sin^{-1}$$: range $$[-\pi/2, \pi/2]$$

$$\cos^{-1}$$: range $$[0, \pi]$$

Find $$a = \sin^{-1}(\sin 5)$$. Note that $$5$$ radians is in the interval $$(3\pi/2, 2\pi) \approx (4.712, 6.283)$$.

For $$x \in (3\pi/2, 2\pi)$$: $$\sin^{-1}(\sin x) = x - 2\pi$$.

$$a = 5 - 2\pi$$

Find $$b = \cos^{-1}(\cos 5)$$. Since $$5 \in (3\pi/2, 2\pi) \approx (4.712, 6.283)$$:

For $$x \in (\pi, 2\pi)$$: $$\cos^{-1}(\cos x) = 2\pi - x$$.

$$b = 2\pi - 5$$

Compute $$a^2 + b^2$$:

$$a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2 = 2(5 - 2\pi)^2$$

$$= 2(25 - 20\pi + 4\pi^2) = 50 - 40\pi + 8\pi^2$$

$$= 8\pi^2 - 40\pi + 50$$

The correct answer is Option (2): $$8\pi^2 - 40\pi + 50$$.