The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area $$A$$. Then $$A^4$$ is equal to ______

The graphs of $$\sin x$$ and $$\cos x$$ intersect where $$\sin x = \cos x$$, i.e., at $$x = \frac{\pi}{4} + n\pi$$ for integer $$n$$.

Between two consecutive intersection points, say from $$x = \frac{\pi}{4}$$ to $$x = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$$, one function is above the other. In the interval $$\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$$, $$\sin x > \cos x$$ (we can verify at $$x = \frac{\pi}{2}$$: $$\sin\frac{\pi}{2} = 1 > 0 = \cos\frac{\pi}{2}$$).

The enclosed area is $$A = \int_{\pi/4}^{5\pi/4}(\sin x - \cos x)\,dx$$.

Evaluating: $$A = \left[-\cos x - \sin x\right]_{\pi/4}^{5\pi/4} = \left(-\cos\frac{5\pi}{4} - \sin\frac{5\pi}{4}\right) - \left(-\cos\frac{\pi}{4} - \sin\frac{\pi}{4}\right)$$.

Computing: $$\cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$$, $$\sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$$, $$\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$, $$\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$.

So $$A = \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$$.

Therefore, $$A^4 = (2\sqrt{2})^4 = 2^4 \cdot (\sqrt{2})^4 = 16 \cdot 4 = 64$$.