To find the area of the region enclosed by the curves $y = \sin(x)$ and $y = \cos(x)$ on the interval $[0, \frac{\pi}{2}]$, follow these steps:

Step 1: Determine the Points of Intersection

Set $\sin(x) = \cos(x)$ to find the points where the curves intersect.

$\sin(x) = \cos(x)$

Divide both sides by $\cos(x)$ (assuming $\cos(x) \neq 0$):

$\tan(x) = 1$

The solution for $x$ within the interval $[0, \frac{\pi}{2}]$ is:

$x = \frac{\pi}{4}$

Step 2: Set Up the Integral

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $a$ to $b$ is given by:

$A = \int_{a}^{b} \left| f(x) - g(x) \right| \, dx$

For the interval $[0, \frac{\pi}{4}]$:

$A_1 = \int_{0}^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx$

For the interval $[\frac{\pi}{4}, \frac{\pi}{2}]$:

$A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin(x) - \cos(x)) \, dx$

Step 3: Compute the Integrals

$A_1 = \int_{0}^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx = \left[\sin(x) + \cos(x)\right]_{0}^{\frac{\pi}{4}}$ $= \left(\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)\right) - \left(\sin(0) + \cos(0)\right)$ $= \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1$

$A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin(x) - \cos(x)) \, dx = \left[-\cos(x) - \sin(x)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$ [ = \left[-\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\right] - \left[-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right] ] [ = \left[-0 - 1\right] - \left[-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right] = -1 + \sqrt{2} ]

Step 4: Add the Two Areas Together

$A = A_1 + A_2 = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2$

Thus, the area of the region enclosed by the curves $y = \sin(x)$ and $y = \cos(x)$ on the interval $[0, \frac{\pi}{2}]$ is:

$\boxed{2\sqrt{2} - 2}$

Would you like further details or have any questions?

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Here are 5 related questions to explore:

1. How would the area change if the interval was $[0, \pi]$?
2. Can you find the area between $y = \sin(x)$ and $y = \cos(x)$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$?
3. What happens to the area if we consider the interval $[0, \pi/4]$ only?
4. How do we calculate the area between $y = \sin(2x)$ and $y = \cos(2x)$ on $[0, \frac{\pi}{2}]$?
5. What if the curves were $y = \sin(x)$ and $y = \sin(2x)$ on $[0, \pi]$?

Tip: When comparing areas under curves, always identify the points of intersection to correctly set up your integral limits.