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To find the critical numbers of a function, the first step is to determine its first derivative. This process involves differentiating the function with respect to the variable, in this case, \( \theta \). The first derivative gives us the rate at which the function changes and helps identify where the function's slope is zero or undefined. These points are key to locating critical points that indicate potential maxima, minima, or inflection points.

In our given function, \( f(\theta) = 2 \cos \theta + \sin^2 \theta \), calculating the first derivative starts by applying basic rules of differentiation to each term. Remember:

• The derivative of \( \cos \theta \) is \( -\sin \theta \).
• The derivative of \( \sin^2 \theta \) requires use of the chain rule.

By differentiating accordingly, we find: \ f'(\theta) = -2\sin \theta + 2 \sin \theta \cos \theta.\ This results from effectively applying these derivative rules to both terms in the function individually.

The chain rule is a fundamental tool in calculus used to differentiate composite functions. When a function is nested within another, the chain rule efficiently provides the derivative. For instance, if you have a function \( g(h(x)) \), the chain rule states that the derivative is \( g'(h(x)) \times h'(x) \).

In the function we're working with, \( \sin^2 \theta \) is a composite function involving the basic sine function squared. To find its derivative, you apply the chain rule:

• First, recognize \( \sin^2 \theta \) as \( (\sin \theta)^2 \).
• Differentiating \( (\sin \theta)^2 \) gives \( 2\sin \theta \cdot \cos \theta \), utilizing the inner derivative \( \cos \theta \).

This illustrates effectively breaking down the function using the chain rule, simplifying complex differentiation tasks.

Trigonometric functions like sine and cosine are periodic functions that often appear in calculus problems. These functions are crucial in various fields, including physics and engineering, due to their wave-like properties.

For our exercise, it's important to recall:

• The derivative of \( \cos \theta \) is \( -\sin \theta \).
• The derivative of \( \sin \theta \) is \( \cos \theta \).

These basic derivatives are vital while finding critical numbers or solving differential equations involving trigonometric expressions.

When dealing with trigonometric functions, always pay attention to their unique properties, such as the fact that specific values repeat based on the unit circle's symmetry. These properties often simplify solving equations or understanding function behavior.

The zero-product property is an algebraic principle affirming that if the product of two quantities is zero, at least one of the quantities must be zero. Mathematically, if \( a \cdot b = 0 \), then \( a = 0 \) or \( b = 0 \).

This property is pivotal in solving for critical numbers after obtaining the first derivative of a function. For our derivative: \ f'(\theta) = 2 \sin \theta (\cos \theta - 1) \ setting it to zero results in the equation:\[ 2 \sin \theta (\cos \theta - 1) = 0. \]

By applying the zero-product property, break this into simpler equations:

• \( 2 \sin \theta = 0 \) which implies \( \sin \theta = 0 \).
• \( \cos \theta - 1 = 0 \) results in \( \cos \theta = 1 \).

Solving these straightforward conditions provides the specific values or ranges for \( \theta \), leading us directly to the critical points of the function.