Express 2cos4xsin2x as an algebraic sum of sines and cosines

Answer

To express \( 2 \cos 4x \sin 2x \) as an algebraic sum of sines and cosines, we can use the trigonometric identity for the product of sine and cosine. Here's a step-by-step solution:

### Step 1: Identify the Identity
We will use the identity:
\[ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \]
In our case, we can set \( A = 2x \) and \( B = 4x \).

### Step 2: Substitute into the Identity
Substituting \( A \) and \( B \) into the identity gives us:
\[ 2 \cos 4x \sin 2x = \sin(2x + 4x) + \sin(2x - 4x) \]

### Step 3: Simplify the Angles
Now, simplify the angles:
- \( 2x + 4x = 6x \)
- \( 2x - 4x = -2x \)

So we have:
\[ 2 \cos 4x \sin 2x = \sin(6x) + \sin(-2x) \]

### Step 4: Use the Property of Sine
We know that \( \sin(-\theta) = -\sin(\theta) \). Therefore:
\[ \sin(-2x) = -\sin(2x) \]

### Step 5: Write the Final Expression
Now we can write the final expression:
\[ 2 \cos 4x \sin 2x = \sin(6x) - \sin(2x) \]

This is the algebraic sum of sines.

### Final Answer
Thus, the expression \( 2 \cos 4x \sin 2x \) as an algebraic sum of sines is:
\[ \sin(6x) - \sin(2x) \]

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