Question

$2^{11}\sin(\frac{x}{2^{10}})\cos(\frac{x}{2})\cos(\frac{x}{2^{2}})\cos(\frac{x}{2^{3}})\dots\cos(\frac{x}{2^{10}})$ is equal to

• A. $\sin x$
• B. $2\sin x$
• C. $-2\sin x$
• D. $4\sin x$
• E. $8\sin x$

Hint:

Math Tip: This is a famous telescoping product. The standard formula is $\cos A \cos 2A \dots \cos(2^{n-1}A) = \frac{\sin(2^nA)}{2^n \sin A}$. You can bypass all steps by identifying $n=10$, multiplying the initial sine term, and directly canceling out the powers of 2.

Answer 1

The Correct Option is B

Concept:
Trigonometry - Multiple Angle Formulas (Continuous Halving).
Step 1: Identify the core trigonometric identity to use.
We will repeatedly use the double angle formula for sine: $$ \sin(2\theta) = 2\sin\theta\cos\theta \implies \sin\theta\cos\theta = \frac{1}{2}\sin(2\theta) $$
Step 2: Rearrange the given expression to pair terms.
Bring the smallest angles together first to set up the domino effect: $$ 2^{11} \left[ \sin\left(\frac{x}{2^{10}}\right) \cos\left(\frac{x}{2^{10}}\right) \right] \cos\left(\frac{x}{2^9}\right) \dots \cos\left(\frac{x}{2}\right) $$
Step 3: Apply the identity to the first pair.
Combine the terms inside the square bracket where $\theta = \frac{x}{2^{10}}$: $$ \sin\left(\frac{x}{2^{10}}\right) \cos\left(\frac{x}{2^{10}}\right) = \frac{1}{2}\sin\left(2 \cdot \frac{x}{2^{10}}\right) = \frac{1}{2}\sin\left(\frac{x}{2^9}\right) $$ Substitute this back into the sequence: $$ 2^{11} \cdot \frac{1}{2} \left[ \sin\left(\frac{x}{2^9}\right) \cos\left(\frac{x}{2^9}\right) \right] \dots \cos\left(\frac{x}{2}\right) $$
Step 4: Recognize the cascading pattern.
Notice that every time we combine a sine and cosine of the same angle, it produces a new sine term with double the angle and an extra factor of $\frac{1}{2}$. This process will repeat for every cosine term in the sequence.
Step 5: Count the number of iterations.
The terms go from $2^{10}$ down to $2^1$ (which is $2$). There are exactly $10$ cosine terms. Therefore, the $\frac{1}{2}$ factor will be applied $10$ times, resulting in $\left(\frac{1}{2}\right)^{10} = \frac{1}{2^{10}}$.
Step 6: Evaluate the final expression.
After 10 iterations, the angle multiplies by $2^{10}$, taking it from $\frac{x}{2^{10}}$ back to $x$. $$ = 2^{11} \cdot \left(\frac{1}{2^{10}}\right) \cdot \sin\left(2^{10} \cdot \frac{x}{2^{10}}\right) $$ $$ = \frac{2^{11}}{2^{10}} \cdot \sin(x) $$ $$ = 2\sin x $$