Finding the Minimum Value of a Trigonometric Expression

The question asks for the minimum value of the expression \(x=\frac{\sin ^2 A+\sin A+1}{\sin A}\) given the domain \(0 < A ≤ \frac{\pi}{2}\).

Simplifying the Expression

First, let's simplify the given expression by dividing each term in the numerator by \(\sin A\):

\[ x = \frac{\sin^2 A}{\sin A} + \frac{\sin A}{\sin A} + \frac{1}{\sin A} \]

\[ x = \sin A + 1 + \frac{1}{\sin A} \]

We can rearrange the terms to group the \(\sin A\) parts:

\[ x = \left(\sin A + \frac{1}{\sin A}\right) + 1 \]

Analyzing the Domain and \(\sin A\)

The given domain for angle \(A\) is \(0 < A ≤ \frac{\pi}{2}\). In this domain, the value of \(\sin A\) is strictly positive. Specifically, for \(A = \frac{\pi}{2}\), \(\sin A = \sin \frac{\pi}{2} = 1\). As \(A\) decreases towards 0 (but stays greater than 0), \(\sin A\) decreases towards 0. Therefore, the range of \(\sin A\) for \(0 < A ≤ \frac{\pi}{2}\) is \((0, 1]\).

Let \(y = \sin A\). So we need to find the minimum value of \(x = y + \frac{1}{y} + 1\) for \(y \in (0, 1]\).

Finding the Minimum Value of \(y + \frac{1}{y}\) for \(y \in (0, 1]\)

Consider the function \(f(y) = y + \frac{1}{y}\) for \(y \in (0, 1]\). We want to find the minimum value of this function in the specified interval. We can analyze the behavior of this function or use calculus.

Using Calculus (Derivative):

Let \(f(y) = y + y^{-1}\). The derivative with respect to \(y\) is:

\[ f'(y) = \frac{d}{dy}(y + y^{-1}) = 1 - y^{-2} = 1 - \frac{1}{y^2} \]

To find critical points, we set \(f'(y) = 0\):

\[ 1 - \frac{1}{y^2} = 0 \implies 1 = \frac{1}{y^2} \implies y^2 = 1 \]

Since \(y = \sin A\) and \(y \in (0, 1]\), the only relevant critical point is \(y = 1\).

Now let's examine the sign of \(f'(y)\) in the interval \((0, 1)\):

For \(0 < y < 1\), we have \(y^2 < 1\). This means \(\frac{1}{y^2} > 1\). Therefore, \(f'(y) = 1 - \frac{1}{y^2} < 0\) for \(y \in (0, 1)\).

Since the derivative is negative on \((0, 1)\), the function \(f(y) = y + \frac{1}{y}\) is decreasing on the interval \((0, 1]\). A decreasing function on a closed interval achieves its minimum value at the right endpoint of the interval.

The interval for \(y\) is \((0, 1]\). The right endpoint is \(y = 1\). The minimum value of \(f(y)\) in this interval occurs at \(y=1\).

Minimum value of \(y + \frac{1}{y}\) for \(y \in (0, 1]\) is \(f(1) = 1 + \frac{1}{1} = 2\).

Using AM-GM Inequality (for context, but careful application needed for interval):

For positive numbers \(y\) and \(\frac{1}{y}\), the AM-GM inequality states \(\frac{y + \frac{1}{y}}{2} \ge \sqrt{y \cdot \frac{1}{y}} = 1\). This gives \(y + \frac{1}{y} \ge 2\). The equality holds when \(y = \frac{1}{y}\), which means \(y=1\) (since \(y > 0\)). This shows the global minimum for \(y > 0\) is 2 at \(y=1\). Since \(y=1\) is included in our interval \((0, 1]\), the minimum value of \(y + \frac{1}{y}\) in this specific interval \((0, 1]\) is indeed 2, occurring at \(y=1\).

Calculating the Minimum Value of x

We found that \(x = \left(\sin A + \frac{1}{\sin A}\right) + 1\). Let \(y = \sin A\). The minimum value of \(y + \frac{1}{y}\) for \(y \in (0, 1]\) is 2, which occurs when \(y = 1\).

Since the minimum value of \(\sin A + \frac{1}{\sin A}\) is 2, the minimum value of \(x\) is:

\[ x_{min} = \left(\sin A + \frac{1}{\sin A}\right)_{min} + 1 = 2 + 1 = 3 \]

The minimum value of \(x\) is 3, and this occurs when \(\sin A = 1\), which corresponds to \(A = \frac{\pi}{2}\) in the given domain \(0 < A ≤ \frac{\pi}{2}\).

Summary of Steps

• Simplify the given expression for \(x\).
• Identify the range of \(\sin A\) based on the given domain of \(A\).
• Analyze the minimum value of the resulting expression, particularly the term \(\sin A + \frac{1}{\sin A}\), within the relevant range of \(\sin A\).
• Calculate the minimum value of \(x\).

The minimum value of \(x\) is 3.

Revision Table: Key Concepts

Additional Information: Function Optimization on Intervals

When finding the minimum or maximum value of a continuous function \(f(y)\) on a closed interval \([a, b]\), we evaluate the function at the critical points within the interval and at the endpoints \(a\) and \(b\). The smallest value among these is the minimum, and the largest is the maximum.

In this problem, the interval for \(y = \sin A\) is \((0, 1]\), which is not a closed interval at the left end. However, we analyzed the derivative and found that the function \(f(y) = y + \frac{1}{y}\) is strictly decreasing on the open interval \((0, 1)\). For a decreasing function on \((0, 1]\), the minimum value occurs at the right endpoint, which is \(y=1\).

If the function had critical points within \((0, 1)\), we would evaluate the function at those points as well. In this case, the only critical point for \(y > 0\) is at \(y=1\), which is the endpoint of our interval.

Understanding the graph of \(y + 1/y\) for \(y > 0\) also helps. The graph is symmetric about the line \(y=1\). For \(0 < y < 1\), the value \(y + 1/y\) is greater than 2 and decreases as \(y\) approaches 1. For \(y > 1\), the value \(y + 1/y\) is also greater than 2 and increases as \(y\) increases. The minimum is indeed at \(y=1\).

The minimum value of the expression \(x\) depends directly on the minimum value of the component \(\sin A + \frac{1}{\sin A}\) within the allowed range of \(\sin A\).