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To find the equation of the tangent line to the curve \( y = \cos(2x) \sin(4x) \) at the point \( \left( \frac{\pi}{8}, \frac{\sqrt{2}}{2} \right) \), we can follow these steps:

### Step 1: Find the derivative of \( y \)
The equation of the tangent line requires the slope at the point, which is given by the derivative of \( y \) with respect to \( x \). Using the product rule, since \( y(x) = \cos(2x) \sin(4x) \), we have:

\[
\frac{dy}{dx} = \frac{d}{dx} \left( \cos(2x) \sin(4x) \right)
\]

Applying the product rule:
\[
\frac{dy}{dx} = \frac{d}{dx} \left( \cos(2x) \right) \cdot \sin(4x) + \cos(2x) \cdot \frac{d}{dx} \left( \sin(4x) \right)
\]

Now, using the chain rule to find the derivatives of \( \cos(2x) \) and \( \sin(4x) \):

\[
\frac{d}{dx} \left( \cos(2x) \right) = -2 \sin(2x)
\]
\[
\frac{d}{dx} \left( \sin(4x) \right) = 4 \cos(4x)
\]

Thus, the derivative becomes:

\[
\frac{dy}{dx} = -2 \sin(2x) \sin(4x) + 4 \cos(2x) \cos(4x)
\]

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{8} \)
Substitute \( x = \frac{\pi}{8} \) into the derivative:

First, calculate the necessary trigonometric values:
\[
\sin\left( 2 \cdot \frac{\pi}{8} \right) = \sin\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}
\]
\[
\sin\left( 4 \cdot \frac{\pi}{8} \right) = \sin\left( \frac{\pi}{2} \right) = 1
\]
\[
\cos\left( 2 \cdot \frac{\pi}{8} \right) = \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}
\]
\[
\cos\left( 4 \cdot \frac{\pi}{8} \right) = \cos\left( \frac{\pi}{2} \right) = 0
\]

Substituting these into the derivative expression:

\[
\frac{dy}{dx} \Bigg|_{x = \frac{\pi}{8}} = -2 \left( \frac{\sqrt{2}}{2} \right) \cdot 1 + 4 \left( \frac{\sqrt{2}}{2} \right) \cdot 0
\]
\[
\frac{dy}{dx} \Bigg|_{x = \frac{\pi}{8}} = -\sqrt{2}
\]

### Step 3: Equation of the tangent line
The equation of the tangent line is given by the point-slope form:

\[
y - y_1 = m(x - x_1)
\]

where \( m = -\sqrt{2} \) (the slope), \( x_1 = \frac{\pi}{8} \), and \( y_1 = \frac{\sqrt{2}}{2} \). Substituting these values:

\[
y - \frac{\sqrt{2}}{2} = -\sqrt{2} \left( x - \frac{\pi}{8} \right)
\]

Simplifying:

\[
y = -\sqrt{2}x + \frac{\pi \sqrt{2}}{8} + \frac{\sqrt{2}}{2}
\]

Thus, the equation of the tangent line is:

\[
y = -\sqrt{2}x + \frac{\pi \sqrt{2}}{8} + \frac{\sqrt{2}}{2}
\]

Would you like more details or have any questions? 

Here are some related questions:
1. How do you apply the product rule to different types of functions?
2. What is the significance of the chain rule in derivative calculations?
3. Can you explain why evaluating trigonometric functions at specific angles is useful in this context?
4. How would you find the normal line to the curve at the same point?
5. What changes if the given point was not on the curve?

**Tip:** When working with trigonometric derivatives, always simplify your trigonometric expressions as much as possible before differentiating!

Find an equation of the tangent line to the curve y = cos(2x)sin(4x) at the point (π/8, √2/2).

Find the equation of the tangent line to the curve y = cos(2x)sin(4x) at the point (π/8, √2/2). This solution involves the product rule, chain rule, and applying trigonometric derivatives. Suitable for calculus students in college or advanced high school.

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