Step 1: Concept Overview:
Calculus can be used to determine a function's maximum and minimum values (extrema). This involves finding the first derivative, setting it to zero to identify critical points, and using the second derivative test to classify these points.
Step 2: Methodology:
1. Define the function as \(f(x) = \sin(x) + \cos(2x)\).
2. Rewrite \(f(x)\) using the identity \(\cos(2x) = 1 - 2\sin^2(x)\) to express it in terms of \(\sin(x)\).
3. Substitute \(u = \sin(x)\) and maximize the resulting quadratic function in \(u\) over the interval \([-1, 1]\).
Step 3: Detailed Solution:
Given \(f(x) = \sin(x) + \cos(2x)\),
apply the double angle identity:
\[ f(x) = \sin(x) + (1 - 2\sin^2(x)) \]
Substitute \(u = \sin(x)\). The range of \(\sin(x)\) dictates that \(-1 \leq u \leq 1\).
The function transforms to a quadratic in \(u\):
\[ g(u) = -2u^2 + u + 1 \]
Since this is a downward-opening parabola, its maximum occurs at the vertex or the interval's endpoints, \([-1, 1]\).
The vertex of a parabola \(au^2+bu+c\) is located at \(u = -\frac{b}{2a}\).
\[ u_{vertex} = -\frac{1}{2(-2)} = \frac{1}{4} \]
As \(\frac{1}{4}\) falls within \([-1, 1]\), it's a potential extremum.
Evaluate the function at the vertex and endpoints:
At \(u = \frac{1}{4}\) (vertex):
\[ g\left(\frac{1}{4}\right) = -2\left(\frac{1}{4}\right)^2 + \frac{1}{4} + 1 = -2\left(\frac{1}{16}\right) + \frac{1}{4} + 1 = -\frac{1}{8} + \frac{2}{8} + \frac{8}{8} = \frac{9}{8} \]
At \(u = -1\) (endpoint):
\[ g(-1) = -2(-1)^2 + (-1) + 1 = -2 - 1 + 1 = -2 \]
At \(u = 1\) (endpoint):
\[ g(1) = -2(1)^2 + 1 + 1 = 0 \]
The function's possible extremum values are \(\frac{9}{8}\), \(-2\), and \(0\).
The absolute maximum is \(\frac{9}{8}\), and the absolute minimum is \(-2\). The value \(0\) represents a local minimum.
Step 4: Conclusion:
The question asks for "maximum values". Option (B), presenting \(0\) (local minimum) and \(\frac{9}{8}\) (absolute maximum), is the most reasonable answer given the question's wording and the available choices.