For all `n in N, cos theta cos 2theta cos 4theta ....cos2^(n-1)theta` equals to

Options

A. `(sin2^(n)theta)/(2^(n)sintheta)`
B. `(sin2^(n)theta)/(sintheta)`
C. `(cos2^(n)theta)/(2^(n)cos2theta)`
D. `(cos2^(n)theta)/(2^(n)sin2theta)`

Answer

To prove that for all \( n \in \mathbb{N} \):

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

we will use mathematical induction.

### Step 1: Base Case
Let's start with \( n = 1 \):

\[ \cos \theta = \frac{\sin 2^1 \theta}{2^1 \sin \theta} = \frac{\sin 2\theta}{2 \sin \theta} \]

Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):

\[ \frac{\sin 2\theta}{2 \sin \theta} = \frac{2 \sin \theta \cos \theta}{2 \sin \theta} = \cos \theta \]

Thus, the base case holds true.

### Step 2: Inductive Step
Assume the statement is true for \( n = k \):

\[ \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{k-1}\theta = \frac{\sin 2^k \theta}{2^k \sin \theta} \]

Now we need to prove it for \( n = k + 1 \):

\[ \cos \theta \cos 2\theta \cos 4\theta \ldots \cos 2^{k-1}\theta \cos 2^k\theta \]

Using the inductive hypothesis, we can write:

\[ = \frac{\sin 2^k \theta}{2^k \sin \theta} \cos 2^k \theta \]

Now, we can express \( \cos 2^k \theta \) using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):

\[ = \frac{\sin 2^k \theta}{2^k \sin \theta} \cdot \cos 2^k \theta = \frac{\sin 2^k \theta \cdot \cos 2^k \theta}{2^k \sin \theta} \]

Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):

\[ = \frac{\sin(2 \cdot 2^k \theta)}{2 \cdot 2^k \sin \theta} = \frac{\sin 2^{k+1} \theta}{2^{k+1} \sin \theta} \]

Thus, the statement holds for \( n = k + 1 \).

### Conclusion
By the principle of mathematical induction, the statement is true for all \( n \in \mathbb{N} \):

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