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)

Concept: Use trigonometric identities: \[ \cos(A+B) + \cos(A-B) = 2\cos A \cos B \]

Step 1: Substitute function.
\[ \frac{1}{2}[\cos(x+y) + \cos(y-x)] - \cos x \cos y \]

Step 2: Use identity.
\[ \cos(x+y) + \cos(y-x) = 2\cos x \cos y \]

Step 3: Simplify.
\[ \frac{1}{2}(2\cos x \cos y) - \cos x \cos y \] \[ = \cos x \cos y - \cos x \cos y = 0 \]