Consider the function fn(x)=cos(2x)(cos4x)...cos(2^nx), n equal to integers. By using Mathematical Induction, prove that fn(x)=sin2^(n+1)x/2^nsin2x where x is not mpi/2

We are tasked with proving by mathematical induction that for a given integer \( n \), the function

\[
f_n(x) = \cos(2x) \cdot \cos(4x) \cdot \cos(8x) \cdots \cos(2^n x)
\]

is equal to:

\[
f_n(x) = \frac{\sin(2^{n+1} x)}{2^n \sin(2x)}
\]

for \( x \neq \frac{m\pi}{2} \), where \( m \) is an integer.

### Step 1: Base Case (n = 1)

Let's first verify the base case where \( n = 1 \).

For \( n = 1 \), the given product function \( f_1(x) \) is:

\[
f_1(x) = \cos(2x)
\]

We need to show that:

\[
f_1(x) = \frac{\sin(2^{1+1} x)}{2^1 \sin(2x)} = \frac{\sin(4x)}{2 \sin(2x)}
\]

Using the identity for the sine of a multiple angle:

\[
\frac{\sin(4x)}{2 \sin(2x)} = \cos(2x)
\]

We observe that this is indeed true because the identity holds for this specific case.

Thus, the base case is true.

### Step 2: Inductive Hypothesis

Now, we assume that the formula holds for some \( n = k \). That is, we assume:

\[
f_k(x) = \cos(2x) \cdot \cos(4x) \cdot \cos(8x) \cdots \cos(2^k x)
\]
is equal to:

\[
f_k(x) = \frac{\sin(2^{k+1} x)}{2^k \sin(2x)}
\]

### Step 3: Inductive Step

Now, we must prove that the formula holds for \( n = k + 1 \).

For \( n = k + 1 \), the function \( f_{k+1}(x) \) is:

\[
f_{k+1}(x) = \cos(2x) \cdot \cos(4x) \cdot \cos(8x) \cdots \cos(2^k x) \cdot \cos(2^{k+1} x)
\]

By the inductive hypothesis, we know:

\[
f_k(x) = \frac{\sin(2^{k+1} x)}{2^k \sin(2x)}
\]

Therefore, \( f_{k+1}(x) \) becomes:

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

Now, using the product-to-sum trigonometric identity:

\[
\sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)]
\]

We can apply this identity to the term \( \sin(2^{k+1}x) \cos(2^{k+1}x) \):

\[
\sin(2^{k+1}x) \cos(2^{k+1}x) = \frac{1}{2}[\sin(2^{k+2}x) + \sin(0)] = \frac{1}{2} \sin(2^{k+2} x)
\]

Thus, we now have:

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

Hence, we have shown that:

\[
f_{k+1}(x) = \frac{\sin(2^{k+2} x)}{2^{k+1} \sin(2x)}
\]

### Step 4: Conclusion

By the principle of mathematical induction, we conclude that the formula

\[
f_n(x) = \frac{\sin(2^{n+1} x)}{2^n \sin(2x)}
\]

holds for all \( n \geq 1 \), provided \( x \neq \frac{m\pi}{2} \) for any integer \( m \).

### Do you need further clarification on this proof or any specific part of it? Here are some related questions:

1. How can trigonometric identities like product-to-sum help simplify complex expressions in induction proofs?
2. What is the general form of the product-to-sum identity in trigonometry, and how can it be derived?
3. Could the formula \( f_n(x) = \frac{\sin(2^{n+1} x)}{2^n \sin(2x)} \) be generalized further, and what would that look like?
4. What are some common pitfalls to watch for when proving trigonometric identities using induction?
5. How can we apply this formula to solve real-world problems involving periodic functions?

**Tip**: When dealing with mathematical induction proofs, always double-check your base case and make sure the inductive step follows logically from your hypothesis.

Learn how to prove by mathematical induction that the function fn(x) = cos(2x) * cos(4x) * ... * cos(2^n x) simplifies to fn(x) = sin(2^(n+1) x) / (2^n * sin(2x)), for x not equal to mπ/2. This detailed solution involves trigonometric identities and induction principles, suitable for students in Grades 11-12.

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