Discussion
Let \(f(x) = cosx\). Then the value of \(\frac{1}{2}[f(x + y) + f(y - x)] - f(x)f(y)\) is equal to
Answer: E
The value is 0.
Use the sum formula \(\cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}\) with \(C = x + y\), \(D = y - x\):
\[ \begin{aligned} \cos(x + y) + \cos(y - x) &= 2\cos y \cos x \\ \tfrac{1}{2}[\ldots] &= \cos x \cos y \end{aligned} \]
Subtracting \(f(x)f(y) = \cos x \cos y\) leaves 0 for every \(x\) and \(y\). Since the expression is identically zero, the constants 2, -2, 1 and -1 cannot be right; for example \(x = y = 0\) gives \(\tfrac{1}{2}(1 + 1) - 1 = 0\).
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