Question

Let \( f(x) = \cos x \). Then the value of \( \frac{1}{2}[f(x+y) + f(y-x)] - f(x)f(y) \) is equal to

• A. \(2 \)
• B. \(-2 \)
• C. \(1 \)
• D. \(-1 \)
• E. \(0 \)

Hint:

Always remember: \(\cos(A+B)+\cos(A-B)=2\cos A \cos B\) — very high-frequency identity.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This problem requires substituting the given function f(x) = cosx into the expression and then simplifying it using trigonometric identities.
Step 2: Key Formula or Approach:
The key trigonometric identity needed here is the product-to-sum formula:
\[ \cos(A)\cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)] \] We will also use the property that cosine is an even function, i.e., cos(-\(\theta\)) = cos(\(\theta\)).
Step 3: Detailed Explanation:
First, substitute f(x) = cosx into the given expression:
\[ \frac{1}{2}[f(x + y) + f(y - x)] - f(x)f(y) = \frac{1}{2}[\cos(x + y) + \cos(y - x)] - \cos(x)\cos(y) \] Since cos is an even function, cos(y - x) = cos(-(x - y)) = cos(x - y). So the expression becomes:
\[ \frac{1}{2}[\cos(x + y) + \cos(x - y)] - \cos(x)\cos(y) \] Now, apply the product-to-sum identity. We can see that the first part of the expression, \(\frac{1}{2}[\cos(x + y) + \cos(x - y)]\), is exactly the expansion of \(\cos(x)\cos(y)\).
So, we can replace the first part with \(\cos(x)\cos(y)\):
\[ \cos(x)\cos(y) - \cos(x)\cos(y) \] This simplifies to 0.
Step 4: Final Answer:
The value of the expression is 0.