Let `f(theta)= sin theta - cos^(2) theta-1`, where `theta in R` and `m le f(theta) le M` .
Find all values satisfying the equation
`sqrt((1)/(|m|)+sqrt((1)/(|m|)+sqrt((1)/(|m|)+...............oo)))`, is

Options

A. `1/3`
B. `2/3`
C. `3/3`
D. `4/3`

Answer

To solve the problem, we need to analyze the function \( f(\theta) = \sin \theta - \cos^2 \theta - 1 \) and find the values of \( m \) and \( M \) such that \( m \leq f(\theta) \leq M \). Then, we will evaluate the infinite nested radical expression given in the problem.

### Step 1: Rewrite the function \( f(\theta) \)

We start with the function:
\[ f(\theta) = \sin \theta - \cos^2 \theta - 1 \]
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we can rewrite \( f(\theta) \):
\[ f(\theta) = \sin \theta - (1 - \sin^2 \theta) - 1 \]
This simplifies to:
\[ f(\theta) = \sin \theta + \sin^2 \theta - 2 \]

### Step 2: Analyze the function \( f(\theta) \)

Now, we have:
\[ f(\theta) = \sin^2 \theta + \sin \theta - 2 \]
Let \( x = \sin \theta \). The function can be expressed as:
\[ f(x) = x^2 + x - 2 \]
This is a quadratic function in terms of \( x \).

### Step 3: Find the vertex of the quadratic

The vertex of a quadratic \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 1 \):
\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \]
Now we evaluate \( f(x) \) at the vertex:
\[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 2 = \frac{1}{4} - \frac{1}{2} - 2 = \frac{1}{4} - \frac{2}{4} - \frac{8}{4} = -\frac{9}{4} \]

### Step 4: Determine the range of \( f(\theta) \)

The quadratic opens upwards (since \( a > 0 \)), and we need to find the maximum value of \( f(x) \) for \( x \in [-1, 1] \) (the range of \( \sin \theta \)).

Calculating \( f(1) \):
\[ f(1) = 1^2 + 1 - 2 = 0 \]
Calculating \( f(-1) \):
\[ f(-1) = (-1)^2 + (-1) - 2 = 1 - 1 - 2 = -2 \]

Thus, the minimum value \( m = -\frac{9}{4} \) and the maximum value \( M = 0 \).

### Step 5: Solve the infinite nested radical expression

The expression given is:
\[ y = \sqrt{\frac{1}{|m|} + \sqrt{\frac{1}{|m|} + \sqrt{\frac{1}{|m|} + \ldots}}} \]
Since \( m = -\frac{9}{4} \), we have \( |m| = \frac{9}{4} \):
\[ y = \sqrt{\frac{4}{9} + y} \]

### Step 6: Square both sides

Squaring both sides gives:
\[ y^2 = \frac{4}{9} + y \]
Rearranging this leads to:
\[ y^2 - y - \frac{4}{9} = 0 \]

### Step 7: Solve the quadratic equation

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot \left(-\frac{4}{9}\right)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 1 + \frac{16}{9} = \frac{25}{9} \]
Thus:
\[ y = \frac{1 \pm \frac{5}{3}}{2} \]
Calculating the two possible values:
1. \( y = \frac{1 + \frac{5}{3}}{2} = \frac{8}{6} = \frac{4}{3} \)
2. \( y = \frac{1 - \frac{5}{3}}{2} = \frac{-2}{6} = -\frac{1}{3} \)

### Step 8: Final answer

Since \( y \) must be non-negative, we take:
\[ y = \frac{4}{3} \]