Question

$2^2 \sin(\frac{x}{2^2}) \cos(\frac{x}{2}) \cos(\frac{x}{2^2}) =$

• A. $\sin 2x$
• B. $\sin x$
• C. $\cos 2x$
• D. $\cos^2 x$
• E. $\sin \frac{x}{2}$

Hint:

When you see a chain of cosines with halving angles, look for the sine double angle identity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks to simplify a trigonometric expression. The expression is in the form of the right-hand side of the sine double-angle identity.
Step 2: Key Formula or Approach:
The double-angle identity for sine is:
\[ \sin(2A) = 2 \sin(A) \cos(A) \] We can apply this formula by setting \( A = \frac{x}{2} \).
Step 3: Detailed Explanation:
The given expression is \( 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \).
Let's compare this to the double-angle formula \( 2 \sin(A) \cos(A) = \sin(2A) \).
If we let \( A = \frac{x}{2} \), the expression perfectly matches the left side of the identity.
So, we can replace it with the right side, \( \sin(2A) \):
\[ 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) = \sin\left(2 \cdot \frac{x}{2}\right) \] Simplifying the angle:
\[ \sin(x) \] Step 4: Final Answer:
The expression simplifies to \( \sin x \).