Finding the Maximum Value of the Trigonometric Function

We are asked to find the maximum values of the function defined by $ f(x) = \sin(x) + \cos(2x) $. Let's analyze this function to determine its highest possible value.

Rewriting the Function Using Trigonometric Identities

To make the function easier to work with, we can express $ \cos(2x) $ in terms of $ \sin(x) $. We use the double angle identity: $ \cos(2x) = 1 - 2\sin^2(x) $ Substituting this into our function, we get: $ f(x) = \sin(x) + (1 - 2\sin^2(x)) $ $ f(x) = -2\sin^2(x) + \sin(x) + 1 $

Analyzing the Function as a Quadratic

Let's make a substitution to simplify the expression. Let $ y = \sin(x) $. Since the range of the sine function is from -1 to 1, we know that $ -1 \le y \le 1 $. Our function now looks like a quadratic function in terms of $ y $: $ g(y) = -2y^2 + y + 1 $, where $ -1 \le y \le 1 $. We need to find the maximum value of this quadratic function within the interval $ [-1, 1] $.

Finding the Vertex of the Parabola

The graph of $ g(y) = -2y^2 + y + 1 $ is a parabola opening downwards (because the coefficient of $ y^2 $ is negative, $ a = -2 $). The maximum value of a downward-opening parabola occurs at its vertex. The y-coordinate of the vertex is given by the formula $ y = -\frac{b}{2a} $, where $ a = -2 $ and $ b = 1 $. $ y_{vertex} = -\frac{1}{2(-2)} = -\frac{1}{-4} = \frac{1}{4} $

Checking the Vertex's Validity

The value $ y = \frac{1}{4} $ lies within our allowed range for $ y $ (which is $ [-1, 1] $). Therefore, the maximum value of the quadratic function $ g(y) $ occurs at $ y = \frac{1}{4} $.

Calculating the Maximum Value

Now, substitute $ y = \frac{1}{4} $ back into the quadratic function $ g(y) $: $ g(\frac{1}{4}) = -2(\frac{1}{4})^2 + \frac{1}{4} + 1 $ $ g(\frac{1}{4}) = -2(\frac{1}{16}) + \frac{1}{4} + 1 $ $ g(\frac{1}{4}) = -\frac{2}{16} + \frac{1}{4} + 1 $ $ g(\frac{1}{4}) = -\frac{1}{8} + \frac{2}{8} + \frac{8}{8} $ $ g(\frac{1}{4}) = \frac{-1 + 2 + 8}{8} = \frac{9}{8} $ So, the maximum value the function can attain is $ \frac{9}{8} $.

Checking the Boundary Values

It's important to also check the function's values at the endpoints of the interval $ [-1, 1] $ for $ y $ (which correspond to $ \sin(x) = -1 $ and $ \sin(x) = 1 $), as the maximum could potentially occur there if the vertex was outside the interval.

• When $ y = 1 $ (i.e., $ \sin(x) = 1 $): $ g(1) = -2(1)^2 + 1 + 1 = -2 + 1 + 1 = 0 $
• When $ y = -1 $ (i.e., $ \sin(x) = -1 $): $ g(-1) = -2(-1)^2 + (-1) + 1 = -2(1) - 1 + 1 = -2 $

Comparing the values $ \frac{9}{8} $, $ 0 $, and $ -2 $, we confirm that the absolute maximum value of the function $ f(x) $ is indeed $ \frac{9}{8} $.

Interpreting the Result

The question asks for the "maximum values", and the options are given as coordinate pairs. The highest value the function $ \sin(x) + \cos(2x) $ can reach is $ \frac{9}{8} $. The option $ (\frac{9}{8}, \frac{9}{8}) $ represents this maximum value. It's possible the notation implies the maximum value occurs, and the value itself is repeated in the pair.