Finding the Least Value of a Trigonometric Expression

The problem asks for the least value of the function \(m(\theta) = \cot^2\theta + n^2\tan^2\theta + 2n\), where \(n\) is a fixed positive real number.

The expression contains trigonometric terms \(\cot^2\theta\) and \(\tan^2\theta\), and a constant term \(2n\).

Let's analyze the components of the expression:

• The term \(2n\) is a constant because \(n\) is a fixed positive real number. Its value does not change with \(\theta\).
• The terms \(\cot^2\theta\) and \(\tan^2\theta\) vary with \(\theta\). We know the relationship between them is \(\cot\theta = \frac{1}{\tan\theta}\), so \(\cot^2\theta = \frac{1}{\tan^2\theta}\).

Let's make a substitution to simplify the varying part of the expression. Let \(x = \tan^2\theta\).

Since \(\tan\theta\) can take any real value except for specific points where it is undefined, \(\tan^2\theta\) can take any non-negative value. However, if \(\tan^2\theta = 0\), then \(\cot^2\theta\) would be undefined. Therefore, for the expression \(m(\theta)\) to be defined, we must have \(\tan\theta \ne 0\), which means \(\tan^2\theta > 0\). So, \(x > 0\).

The expression for \(m(\theta)\) can be rewritten in terms of \(x\):

\(m(\theta) = \frac{1}{x} + n^2 x + 2n\)

We need to find the minimum value of this expression for \(x > 0\).

Consider the part that varies with \(x\), which is \(\frac{1}{x} + n^2 x\). Both \(\frac{1}{x}\) and \(n^2 x\) are positive since \(x > 0\) and \(n > 0\).

We can use the AM-GM (Arithmetic Mean - Geometric Mean) inequality to find the minimum value of the sum of two positive terms. The AM-GM inequality states that for any non-negative real numbers \(a\) and \(b\), the arithmetic mean is greater than or equal to the geometric mean: \(\frac{a+b}{2} \ge \sqrt{ab}\). This can be rearranged as \(a+b \ge 2\sqrt{ab}\).

Applying the AM-GM inequality to the terms \(\frac{1}{x}\) and \(n^2 x\):

\(\frac{1}{x} + n^2 x \ge 2\sqrt{\left(\frac{1}{x}\right)(n^2 x)}\)

\(\frac{1}{x} + n^2 x \ge 2\sqrt{\frac{n^2 x}{x}}\)

\(\frac{1}{x} + n^2 x \ge 2\sqrt{n^2}\)

Since \(n\) is a positive real number, \(\sqrt{n^2} = n\).

\(\frac{1}{x} + n^2 x \ge 2n\)

The equality (and thus the minimum value) occurs when the two terms are equal:

\(\frac{1}{x} = n^2 x\)

\(1 = n^2 x^2\)

\(x^2 = \frac{1}{n^2}\)

Since \(x = \tan^2\theta\), \(x\) must be positive. Taking the positive square root:

\(x = \sqrt{\frac{1}{n^2}} = \frac{1}{n}\) (since \(n > 0\))

So, the minimum value of \(\frac{1}{x} + n^2 x\) is \(2n\), and this minimum occurs when \(x = \frac{1}{n}\).

Now, substitute this minimum value back into the expression for \(m(\theta)\):

\(m(\theta) = \left(\frac{1}{x} + n^2 x\right) + 2n\)

The minimum value of \(m(\theta)\) is the minimum value of \(\left(\frac{1}{x} + n^2 x\right)\) plus the constant term \(2n\).

Least value of \(m(\theta) = (\text{minimum of } \frac{1}{x} + n^2 x) + 2n\)

Least value of \(m(\theta) = 2n + 2n\)

Least value of \(m(\theta) = 4n\)

This minimum occurs when \(\tan^2\theta = \frac{1}{n}\). Since \(n\) is positive, \(\frac{1}{n}\) is also positive, and there exist real values of \(\theta\) for which \(\tan^2\theta = \frac{1}{n}\).

Therefore, the least value of \(m(\theta)\) is \(4n\).

Revision Table: Key Concepts for Finding Least Values

Additional Information: AM-GM Inequality in Optimization

The Arithmetic Mean - Geometric Mean (AM-GM) inequality is a fundamental tool in mathematical inequalities and optimization problems, especially when dealing with sums and products of positive numbers. It provides a lower bound for the sum of positive numbers in terms of their geometric mean.

For two positive numbers \(a\) and \(b\), the inequality \(a+b \ge 2\sqrt{ab}\) is particularly useful. The key insight is that the minimum value of the sum \(a+b\) is achieved when \(a=b\).

In this problem, we had the sum \(\frac{1}{x} + n^2 x\). By setting \(a = \frac{1}{x}\) and \(b = n^2 x\), their product \(ab = \left(\frac{1}{x}\right)(n^2 x) = n^2\), which is a constant. This makes the AM-GM inequality very effective because the right side \(2\sqrt{ab} = 2\sqrt{n^2} = 2n\) is also a constant. Thus, the minimum value of the sum is directly given by \(2n\).

The condition for equality, \(a=b\), tells us the value of \(x\) at which this minimum occurs: \(\frac{1}{x} = n^2 x\), leading to \(x = \frac{1}{n}\). This confirms that the minimum value is attainable for a valid value of \(x = \tan^2\theta\).

AM-GM is often applied when an expression involves a term and its reciprocal, or terms whose product is constant, like \(y + \frac{k}{y}\).