Prove that `cos theta cos 2 theta cos 4 theta…. Cos 2 ^(n-1) theta = ( sin ( 2 ^(n) theta))/( 2 ^(n) (sin theta)) .`

Answer

To prove that

\[ \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta = \frac{\sin(2^n \theta)}{2^n \sin \theta}, \]

we will start from the left-hand side (LHS) and manipulate it step by step to reach the right-hand side (RHS).

### Step 1: Write the LHS
We start with the left-hand side:

\[ \text{LHS} = \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta. \]

### Step 2: Multiply by \( \frac{2 \sin \theta}{2 \sin \theta} \)
To manipulate the expression, we multiply and divide by \( 2 \sin \theta \):

\[ \text{LHS} = \frac{2 \sin \theta \cdot \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta}{2 \sin \theta}. \]

### Step 3: Use the identity \( 2 \sin \theta \cos \theta = \sin(2\theta) \)
Using the trigonometric identity \( 2 \sin \theta \cos \theta = \sin(2\theta) \), we can rewrite the numerator:

\[ \text{LHS} = \frac{\sin(2\theta) \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta}{2 \sin \theta}. \]

### Step 4: Repeat the process
Now, we can apply the same technique to the remaining cosines. We multiply and divide by \( 2 \sin(2\theta) \):

\[ \text{LHS} = \frac{2 \sin(2\theta) \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta}{4 \sin \theta}. \]

Using the identity again:

\[ 2 \sin(2\theta) \cos(2\theta) = \sin(4\theta), \]

we have:

\[ \text{LHS} = \frac{\sin(4\theta) \cos 4\theta \ldots \cos 2^{n-1}\theta}{4 \sin \theta}. \]

### Step 5: Continue the process
Continuing this process, we will keep multiplying by \( 2 \sin(2^k \theta) \) for \( k = 1, 2, \ldots, n-1 \):

After \( n \) steps, we will have:

\[ \text{LHS} = \frac{\sin(2^n \theta)}{2^n \sin \theta}. \]

### Conclusion
Thus, we have shown that:

\[ \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{n-1}\theta = \frac{\sin(2^n \theta)}{2^n \sin \theta}, \]

which proves the statement.