What is the ratio of the greatest to the smallest value of 2 – 2 sin x – sin 2x, 0 ≤ x ≤ (π/2)?
-2
Finding the Ratio of Greatest to Smallest Value
The question asks for the ratio of the greatest value to the smallest value of the expression \(2 - 2 \sin x - \sin^2 x\) over the interval \(0 \le x \le \frac{\pi}{2}\).
Let the given expression be \(E = 2 - 2 \sin x - \sin^2 x\).
Analyzing the Trigonometric Expression and Interval
The expression involves the term \(\sin x\). The given interval for \(x\) is \(0 \le x \le \frac{\pi}{2}\).
We need to determine the range of \(\sin x\) for this interval:
- When \(x = 0\), \(\sin x = \sin 0 = 0\).
- When \(x = \frac{\pi}{2}\), \(\sin x = \sin \frac{\pi}{2} = 1\).
For \(x\) in the interval \(0 \le x \le \frac{\pi}{2}\), the value of \(\sin x\) increases monotonically from 0 to 1. Therefore, the range of \(\sin x\) is \([0, 1]\).
Transforming the Expression into a Quadratic Function
Let \(y = \sin x\). Since the range of \(\sin x\) for the given interval of \(x\) is \([0, 1]\), the variable \(y\) is in the interval \([0, 1]\).
Substitute \(y = \sin x\) into the expression \(E\):
\[ E = 2 - 2y - y^2 \]Let's define a function \(f(y) = 2 - 2y - y^2\) for \(y \in [0, 1]\).
This is a quadratic function of \(y\). We can rewrite it as \(f(y) = -y^2 - 2y + 2\). This is a downward-opening parabola because the coefficient of \(y^2\) is negative (which is -1).
Finding the Vertex and Analyzing the Function's Behavior
The vertex of the parabola \(ay^2 + by + c\) is at \(y = -\frac{b}{2a}\).
For \(f(y) = -y^2 - 2y + 2\), \(a = -1\) and \(b = -2\).
The vertex is at \(y = -\frac{-2}{2(-1)} = -\frac{-2}{-2} = -1\).
The interval for \(y\) is \([0, 1]\). The vertex \(y = -1\) lies outside this interval, specifically to the left of it.
Since the parabola opens downwards and the interval \([0, 1]\) is entirely to the right of the vertex \(y = -1\), the function \(f(y)\) will be strictly decreasing over the interval \([0, 1]\).
Determining the Greatest and Smallest Values
Since \(f(y)\) is decreasing on \([0, 1]\), the greatest value will occur at the smallest value of \(y\) in the interval, which is \(y=0\). The smallest value will occur at the largest value of \(y\) in the interval, which is \(y=1\).
- Greatest Value: Evaluate \(f(y)\) at \(y=0\). \[ f(0) = 2 - 2(0) - (0)^2 = 2 - 0 - 0 = 2 \] This corresponds to \(x=0\), where \(\sin x = 0\).
- Smallest Value: Evaluate \(f(y)\) at \(y=1\). \[ f(1) = 2 - 2(1) - (1)^2 = 2 - 2 - 1 = -1 \] This corresponds to \(x=\frac{\pi}{2}\), where \(\sin x = 1\).
The greatest value of the expression is 2, and the smallest value is -1.
Calculating the Ratio
The question asks for the ratio of the greatest value to the smallest value.
Ratio = \(\frac{\text{Greatest Value}}{\text{Smallest Value}} = \frac{2}{-1} = -2\).
| Value of x | Value of sin x (y) | Value of 2 - 2 sin x - sin2 x (f(y)) | Result Type |
|---|---|---|---|
| 0 | 0 | \(2 - 2(0) - (0)^2 = 2\) | Greatest Value |
| \(\pi/2\) | 1 | \(2 - 2(1) - (1)^2 = 2 - 2 - 1 = -1\) | Smallest Value |
The ratio of the greatest value (2) to the smallest value (-1) is \(\frac{2}{-1} = -2\).
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Trigonometric Function Range | The set of all possible output values of a trigonometric function for a given input domain. | Determining the range of \(\sin x\) for \(0 \le x \le \pi/2\). |
| Quadratic Function | A polynomial function of degree 2, like \(f(y) = ay^2 + by + c\). | Transforming the expression into a quadratic in terms of \(y = \sin x\). |
| Vertex of Parabola | The highest or lowest point on the graph of a quadratic function. Its position helps determine maximum/minimum values. | Locating the vertex of \(f(y) = -y^2 - 2y + 2\) to understand function behavior on the interval \([0, 1]\). |
| Monotonic Function | A function that is either entirely non-increasing or entirely non-decreasing over its domain or a specific interval. | Identifying that \(f(y)\) is decreasing on \([0, 1]\) helps find max/min at endpoints. |
Additional Information: Optimizing Functions
Optimizing a function means finding its maximum or minimum value over a specific interval. For continuous functions on a closed interval, the maximum and minimum values must occur either at the endpoints of the interval or at a critical point within the interval (where the derivative is zero or undefined).
In this problem, we transformed the trigonometric expression into a quadratic function \(f(y) = -y^2 - 2y + 2\) over the interval \(y \in [0, 1]\). Since this is a simple quadratic, we could analyze its parabolic shape and vertex location relative to the interval \([0, 1]\).
Alternatively, we could use calculus:
- Find the derivative of \(f(y)\) with respect to \(y\): \(f'(y) = \frac{d}{dy}(-y^2 - 2y + 2) = -2y - 2\).
- Find critical points by setting the derivative to zero: \(-2y - 2 = 0 \implies -2y = 2 \implies y = -1\).
- The critical point \(y = -1\) is outside the interval \([0, 1]\).
- Therefore, the maximum and minimum values must occur at the endpoints of the interval \([0, 1]\).
- Evaluate \(f(y)\) at the endpoints:
- \(f(0) = 2\)
- \(f(1) = -1\)
The greatest value is 2 and the smallest value is -1, consistent with our previous method.
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