Find the value of `cos theta. cos 2theta. cos 2^(2)theta . cos 2^(3)theta"……." cos 2^(n-1) theta` is

Options

A. `(cos2^(n)theta)/(2^(n)sintheta)`
B. `(sin2^(n)theta)/(2^(n)costheta)`
C. `(cos2^(n)theta)/(2^(n)costheta)`
D. `(sin2^(n)theta)/(2^(n)sin theta)`

Answer

To find the value of the expression \( m = \cos \theta \cdot \cos 2\theta \cdot \cos 2^2\theta \cdots \cos 2^{n-1}\theta \), we can follow these steps:

### Step 1: Define the expression
Let
\[ m = \cos \theta \cdot \cos 2\theta \cdot \cos 2^2\theta \cdots \cos 2^{n-1}\theta \]

### Step 2: Use the sine double angle formula
We know from trigonometry that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
This implies:
\[ \sin \theta \cos \theta = \frac{\sin 2\theta}{2} \]

### Step 3: Multiply both sides by \(\sin \theta\)
Multiplying both sides of the equation by \(\sin \theta\):
\[ \sin \theta \cdot m = \sin \theta \cdot \cos \theta \cdot \cos 2\theta \cdots \cos 2^{n-1}\theta \]
Using the sine double angle formula, we can rewrite the right-hand side:
\[ \sin \theta \cdot m = \frac{\sin 2\theta}{2} \cdot \cos 2\theta \cdots \cos 2^{n-1}\theta \]

### Step 4: Apply the sine double angle formula repeatedly
Continuing this process, we can apply the sine double angle formula again:
\[ \sin 2\theta \cdot \cos 2\theta = \frac{\sin 4\theta}{2} \]
Thus, we can express \( \sin \theta \cdot m \) as:
\[ \sin \theta \cdot m = \frac{\sin 4\theta}{2^2} \cdots \cos 2^{n-1}\theta \]

### Step 5: Generalize the pattern
Continuing this process, we notice a pattern emerging. After \( n \) iterations, we have:
\[ \sin \theta \cdot m = \frac{\sin 2^n \theta}{2^n} \]

### Step 6: Solve for \( m \)
Now, we can solve for \( m \):
\[ m = \frac{\sin 2^n \theta}{2^n \sin \theta} \]

### Conclusion
Thus, the final value of the expression is:
\[ \boxed{m = \frac{\sin 2^n \theta}{2^n \sin \theta}} \]