Question

Evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. \(f(x)=\cos 2 x\) (a) \(f(0)\) (b) \(f(-\pi / 4)\) (c) \(f(\pi / 3)\)

Short answer

The evaluations of the function at the given values are: (a) \(f(0) = 1\), (b) \(f(-\pi / 4) = 0\), and (c) \(f(\pi / 3) = -1/2\).

Step-by-step solution

Evaluate (a) f(0)

For \(f(0)\), substitutute \(x = 0\) in the function \(f(x)=\cos 2x\), which gives \(f(0) = \cos 2(0) = \cos (0)\). The function value of cos at 0 is 1, so \(f(0) = 1\).

Evaluate (b) f(-\(\pi / 4\))

For \(f(-\pi / 4)\), substituting \(x = -\pi / 4\) in the function \(f(x)=\cos 2x\) gives \(f(-\pi / 4) = \cos 2(-\pi/4) = \cos (-\pi/2)\). Now, \(cos (-\theta)\) is equal to \(cos (\theta)\), thus \(cos (-\pi / 2)\) = 0. So, \(f(-\pi / 4) = 0\).

Evaluate (c) f(\(\pi / 3\))

For \(f(\pi / 3)\), substituting \(x = \pi / 3\) in the function \(f(x)=\cos 2x\) gives \(f(\pi / 3) = \cos 2(\pi/3) = \cos (2\pi/3)\). This is a commonly known value, which equals to -1/2. So, \(f(\pi / 3) = -1/2\).