Find `sin 2x + cos 4x` if `tan 2x. Tan 4x = 1`

Answer

To solve the problem, we need to find the value of \( \sin 2x + \cos 4x \) given that \( \tan 2x \cdot \tan 4x = 1 \).

### Step 1: Understand the condition \( \tan 2x \cdot \tan 4x = 1 \)

From the property of tangent, we know that if \( \tan A \cdot \tan B = 1 \), then \( A + B = 90^\circ \) or \( A + B = \frac{\pi}{2} \) radians.

In our case, let \( A = 2x \) and \( B = 4x \). Therefore, we have:
\[ 2x + 4x = 90^\circ \]
This simplifies to:
\[ 6x = 90^\circ \]

### Step 2: Solve for \( x \)

Dividing both sides by 6 gives:
\[ x = \frac{90^\circ}{6} = 15^\circ \]

### Step 3: Calculate \( \sin 2x \)

Now we need to find \( \sin 2x \):
\[ 2x = 2 \cdot 15^\circ = 30^\circ \]
Using the sine value:
\[ \sin 30^\circ = \frac{1}{2} \]

### Step 4: Calculate \( \cos 4x \)

Next, we calculate \( \cos 4x \):
\[ 4x = 4 \cdot 15^\circ = 60^\circ \]
Using the cosine value:
\[ \cos 60^\circ = \frac{1}{2} \]

### Step 5: Add \( \sin 2x \) and \( \cos 4x \)

Now we can add the two results:
\[ \sin 2x + \cos 4x = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer

Thus, the final answer is:
\[ \sin 2x + \cos 4x = 1 \]

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