prove that `sin2x+2sin4x+sin6x=4cos^(2)x.sinx`

Answer

To prove that \( \sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin x \), we will start with the left-hand side and manipulate it to reach the right-hand side.

### Step 1: Start with the left-hand side
We have:
\[ \sin 2x + 2\sin 4x + \sin 6x \]

### Step 2: Use the sine addition formula
We can express \( \sin 6x \) in terms of \( \sin 2x \) and \( \sin 4x \):
\[ \sin 6x = \sin(2x + 4x) = \sin 2x \cos 4x + \cos 2x \sin 4x \]

### Step 3: Substitute \( \sin 6x \) back into the equation
Substituting this into the left-hand side gives:
\[ \sin 2x + 2\sin 4x + (\sin 2x \cos 4x + \cos 2x \sin 4x) \]

### Step 4: Combine like terms
Now we can combine the terms involving \( \sin 2x \):
\[ \sin 2x (1 + \cos 4x) + 2\sin 4x + \cos 2x \sin 4x \]

### Step 5: Factor out common terms
Notice that we can factor \( \sin 4x \) from the last two terms:
\[ \sin 2x (1 + \cos 4x) + \sin 4x (2 + \cos 2x) \]

### Step 6: Use double angle identities
Using the double angle identity, we know that:
\[ 1 + \cos 4x = 2\cos^2 2x \]
and
\[ 2 + \cos 2x = 2(1 + \frac{1}{2}\cos 2x) = 2\cos^2 x \]

### Step 7: Substitute back into the equation
Now substituting these identities back, we have:
\[ \sin 2x (2\cos^2 2x) + \sin 4x (2\cos^2 x) \]

### Step 8: Factor out \( 2 \)
Factoring out 2 gives:
\[ 2(\sin 2x \cos^2 2x + \sin 4x \cos^2 x) \]

### Step 9: Recognize \( \sin 2x \) and \( \sin 4x \)
Recall that:
\[ \sin 2x = 2\sin x \cos x \]
and
\[ \sin 4x = 2\sin 2x \cos 2x = 4\sin x \cos x \cos 2x \]

### Step 10: Substitute and simplify
Substituting these back into our equation:
\[ 2(2\sin x \cos x \cos^2 2x + 4\sin x \cos x \cos^2 x) \]

### Step 11: Factor out \( 4\sin x \)
Factoring out \( 4\sin x \) gives:
\[ 4\sin x (\cos^2 2x + \cos^2 x) \]

### Step 12: Use the identity
Using the identity \( \cos^2 2x + \cos^2 x = 1 \) leads us to:
\[ 4\sin x \]

### Conclusion
Thus, we have shown:
\[ \sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin x \]