If `tan2theta.tan4theta=1`, find `sin2theta+cos4theta=?`

Answer

To solve the problem where \( \tan(2\theta) \cdot \tan(4\theta) = 1 \) and we need to find \( \sin(2\theta) + \cos(4\theta) \), we can follow these steps:

### Step 1: Rewrite the equation
Given that \( \tan(2\theta) \cdot \tan(4\theta) = 1 \), we can express this as:
\[ \tan(2\theta) = \frac{1}{\tan(4\theta)} \]
This implies:
\[ \tan(2\theta) = \cot(4\theta) \]

### Step 2: Use the cotangent identity
Using the identity \( \cot(x) = \tan(90^\circ - x) \), we can rewrite the equation:
\[ \tan(2\theta) = \tan(90^\circ - 4\theta) \]

### Step 3: Set the angles equal
Since the tangent function is periodic, we can set the angles equal to each other:
\[ 2\theta = 90^\circ - 4\theta \]

### Step 4: Solve for \( \theta \)
Rearranging the equation gives:
\[ 2\theta + 4\theta = 90^\circ \]
\[ 6\theta = 90^\circ \]
Dividing both sides by 6:
\[ \theta = 15^\circ \]

### Step 5: Find \( 2\theta \) and \( 4\theta \)
Now we can find \( 2\theta \) and \( 4\theta \):
\[ 2\theta = 2 \times 15^\circ = 30^\circ \]
\[ 4\theta = 4 \times 15^\circ = 60^\circ \]

### Step 6: Calculate \( \sin(2\theta) \) and \( \cos(4\theta) \)
Now we can find \( \sin(2\theta) \) and \( \cos(4\theta) \):
\[ \sin(2\theta) = \sin(30^\circ) = \frac{1}{2} \]
\[ \cos(4\theta) = \cos(60^\circ) = \frac{1}{2} \]

### Step 7: Add the results
Finally, we can add these two results together:
\[ \sin(2\theta) + \cos(4\theta) = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
Thus, the value of \( \sin(2\theta) + \cos(4\theta) \) is:
\[ \boxed{1} \]