Key Concepts and Formulas

• Area Between Curves: The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$ is given by $\int_a^b |f(x) - g(x)| dx$. If $f(x) \ge g(x)$ on $[a,b]$, then the area is $\int_a^b (f(x) - g(x)) dx$.
• Trigonometric Identities: The general solution to $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$, where $n$ is an integer.
• Fundamental Theorem of Calculus: $\int_a^b F'(x) dx = F(b) - F(a)$, where $F'(x)$ is the derivative of $F(x)$.

Step-by-Step Solution

Step 1: Find the Intersection Points of $\sin x$ and $\cos x$

We need to find where the curves $y = \sin x$ and $y = \cos x$ intersect. This occurs when $\sin x = \cos x$. To solve for $x$, we divide both sides by $\cos x$ (assuming $\cos x \neq 0$ at the points of intersection): $\frac{\sin x}{\cos x} = 1$ $\tan x = 1$ The general solution for $\tan x = 1$ is given by: $x = n\pi + \frac{\pi}{4}$ where $n$ is an integer. Consecutive intersection points are found by using consecutive integer values of $n$. For $n = 0$, $x = \frac{\pi}{4}$. For $n = 1$, $x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$. We'll use the interval $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$.

Step 2: Determine Which Function is Greater on the Interval $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$

We need to determine whether $\sin x \ge \cos x$ or $\cos x \ge \sin x$ on the interval $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$. Let's test a value in the interval, such as $x = \frac{\pi}{2}$. $\sin\left(\frac{\pi}{2}\right) = 1$ $\cos\left(\frac{\pi}{2}\right) = 0$ Since $1 > 0$, $\sin x > \cos x$ on the interval $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$. Thus, we will integrate $\sin x - \cos x$.

Step 3: Set Up the Definite Integral

The area $A$ between the curves is given by the definite integral: $A = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) dx$

Step 4: Evaluate the Definite Integral

We find the antiderivative of $\sin x - \cos x$: $\int (\sin x - \cos x) dx = -\cos x - \sin x + C$ Now we evaluate the definite integral: $A = \left[-\cos x - \sin x\right]_{\pi/4}^{5\pi/4}$ $A = \left(-\cos\left(\frac{5\pi}{4}\right) - \sin\left(\frac{5\pi}{4}\right)\right) - \left(-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right)$ We know that $\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\cos\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}$. Substituting these values: $A = \left(-\left(-\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right)\right) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)$ $A = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - \left(-\frac{2}{\sqrt{2}}\right)$ $A = \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$

Step 5: Calculate $A^4$

We have $A = 2\sqrt{2}$. We need to find $A^4$: $A^4 = (2\sqrt{2})^4 = (2 \cdot 2^{1/2})^4 = 2^4 \cdot (2^{1/2})^4 = 2^4 \cdot 2^2 = 2^6 = 64$

Common Mistakes & Tips

• Be careful with the signs of trigonometric functions in different quadrants.
• Always check which function is greater than the other in the given interval to avoid a negative area.
• Rationalize the denominator to simplify the expression.

Summary

The area $A$ between the curves $y = \sin x$ and $y = \cos x$ between two consecutive intersection points is $2\sqrt{2}$. We were asked to find $A^4$, which is equal to 64.

The final answer is \boxed{64}.