Understanding the Trigonometric Function and Its Range

The question asks for the range of the function \(A = \cos^2 \theta + \sin^4 \theta\) for all possible values of \(\theta\). To find the range of this trigonometric function, we can express it in terms of a single trigonometric ratio, preferably \(\sin^2 \theta\) or \(\cos^2 \theta\).

Rewriting the Function

We know the identity \(\cos^2 \theta + \sin^2 \theta = 1\). From this, we can write \(\cos^2 \theta = 1 - \sin^2 \theta\). Let's substitute this into the expression for A:

\(A = (1 - \sin^2 \theta) + \sin^4 \theta\)

This function now depends only on \(\sin^2 \theta\). Let's simplify this expression.

Introducing a Substitution

To make the analysis easier, let \(x = \sin^2 \theta\). We know that for any real value of \(\theta\), the value of \(\sin \theta\) is between -1 and 1 (inclusive). Therefore, \(\sin^2 \theta\) will be between 0 and 1 (inclusive).

So, the variable \(x\) is in the interval \([0, 1]\), i.e., \(0 \le x \le 1\).

Substituting \(x = \sin^2 \theta\) into the expression for A, we get:

\(A = (1 - x) + x^2\)

Rearranging the terms, we get a quadratic function in terms of \(x\):

\(A = x^2 - x + 1\)

Finding the Range of the Quadratic Function

We need to find the range of the function \(f(x) = x^2 - x + 1\) for \(x \in [0, 1]\). This is a quadratic function in the form \(ax^2 + bx + c\) with \(a=1\), \(b=-1\), and \(c=1\). The parabola opens upwards since \(a > 0\).

The vertex of the parabola occurs at \(x = -\frac{b}{2a}\).

Calculating the x-coordinate of the vertex:

\(x_{\text{vertex}} = -\frac{-1}{2 \times 1} = \frac{1}{2}\)

Since the vertex at \(x = \frac{1}{2}\) lies within the interval \([0, 1]\), the minimum value of the function occurs at the vertex.

Calculating the minimum value of A:

\(A_{\text{min}} = f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1\)

\(A_{\text{min}} = \frac{1}{4} - \frac{1}{2} + 1\)

\(A_{\text{min}} = \frac{1}{4} - \frac{2}{4} + \frac{4}{4}\)

\(A_{\text{min}} = \frac{1 - 2 + 4}{4} = \frac{3}{4}\)

The maximum value of the quadratic function over a closed interval occurs at one of the endpoints of the interval. The interval for \(x\) is \([0, 1]\).

Calculating the value of A at the endpoints:

• At \(x = 0\): \(A = f(0) = 0^2 - 0 + 1 = 1\)
• At \(x = 1\): \(A = f(1) = 1^2 - 1 + 1 = 1 - 1 + 1 = 1\)

Comparing the values at the endpoints and the vertex, the minimum value is \(\frac{3}{4}\) and the maximum value is 1.

Therefore, the range of the function \(A\) for all values of \(\theta\) is \(\frac{3}{4} \le A \le 1\).

Conclusion on the Range of A

The range of the function \(A = \cos^2 \theta + \sin^4 \theta\) is found by analyzing the quadratic \(A = x^2 - x + 1\) where \(x = \sin^2 \theta\) and \(0 \le x \le 1\). The minimum value is \(\frac{3}{4}\) and the maximum value is 1.

The minimum value is \(\frac{3}{4}\) and the maximum value is 1 within the domain \(x \in [0, 1]\).

Thus, the range of A is \(\frac{3}{4} \le A \le 1\).

Revision Table: Key Concepts

Additional Information: Exploring Function Ranges

Finding the range of trigonometric functions often involves using identities to simplify the expression and then analyzing the resulting function, which is frequently a polynomial or another familiar type of function. For functions involving \(\sin^2 \theta\) and \(\cos^2 \theta\), substituting \(x = \sin^2 \theta\) (or \(x = \cos^2 \theta\)) is a common technique. Remember that \(x\) will always be in the interval \([0, 1]\).

Once the function is expressed in terms of \(x\), we analyze the function \(f(x)\) over the interval \([0, 1]\). For quadratic functions, this involves finding the vertex and evaluating the function at the endpoints of the interval. The minimum and maximum values of \(f(x)\) over \([0, 1]\) give the range of the original trigonometric function A.

This method is applicable to many problems involving powers of \(\sin \theta\) and \(\cos \theta\).