Finding the Maximum Value of sin 2x ⋅ cos 2x

The question asks for the maximum value of the expression \(\sin 2x \cdot \cos 2x\). To find this, we can use a trigonometric identity to simplify the expression.

Using the Double Angle Identity for Sine

Recall the double angle identity for sine:

• \(\sin 2\theta = 2 \sin \theta \cos \theta\)

We have the expression \(\sin 2x \cdot \cos 2x\). This looks very similar to the right side of the identity, \(2 \sin \theta \cos \theta\). If we let \(\theta = 2x\), then \(2\theta = 4x\). Substituting this into the identity, we get:

\(\sin(2 \cdot 2x) = 2 \sin 2x \cos 2x\)

Which simplifies to:

\(\sin 4x = 2 \sin 2x \cos 2x\)

Now we can express the original expression \(\sin 2x \cos 2x\) in terms of \(\sin 4x\) by dividing both sides by 2:

\(\sin 2x \cos 2x = \dfrac{1}{2} \sin 4x\)

Determining the Maximum Value

We need to find the maximum value of \(\dfrac{1}{2} \sin 4x\). The maximum value of the standard sine function, \(\sin \theta\), is 1. The range of \(\sin \theta\) is \([-1, 1]\) for any real value of \(\theta\).

In our expression, the argument of the sine function is \(4x\). As \(x\) varies over all real numbers, \(4x\) also varies over all real numbers. Therefore, the value of \(\sin 4x\) can take any value between -1 and 1, inclusive. The maximum value of \(\sin 4x\) is 1.

To find the maximum value of \(\dfrac{1}{2} \sin 4x\), we multiply the maximum value of \(\sin 4x\) by \(\dfrac{1}{2}\).

Maximum value \(= \dfrac{1}{2} \times (\text{Maximum value of } \sin 4x)\)

Maximum value \(= \dfrac{1}{2} \times 1\)

Maximum value \(= \dfrac{1}{2}\)

Evaluating the Options

Let's compare our calculated maximum value with the given options:

• Option 1: \(\dfrac{1}{2}\)
• Option 2: 1
• Option 3: 2
• Option 4: 4

Our calculated maximum value is \(\dfrac{1}{2}\), which matches Option 1.

Conclusion on Maximum Value

By using the double angle identity for sine, we transformed the expression \(\sin 2x \cos 2x\) into \(\dfrac{1}{2} \sin 4x\). Since the maximum value of the sine function is 1, the maximum value of \(\dfrac{1}{2} \sin 4x\) is \(\dfrac{1}{2} \times 1 = \dfrac{1}{2}\).

Revision Table: Key Concepts

Additional Information on Trigonometric Identities and Maximum Values

Trigonometric identities are equations that are true for all values of the variables for which the expressions are defined. They are crucial for simplifying trigonometric expressions and solving trigonometric equations. The double angle identities are a set of identities that express trigonometric functions of \(2\theta\) in terms of functions of \(\theta\).

• \(\sin 2\theta = 2 \sin \theta \cos \theta\)
• \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta\)
• \(\tan 2\theta = \dfrac{2 \tan \theta}{1 - \tan^2 \theta}\)

Understanding the range of trigonometric functions is essential for finding maximum and minimum values of expressions involving them. The range of \(\sin x\) and \(\cos x\) is \([-1, 1]\), meaning their values are always between -1 and 1, inclusive. The range of \(\tan x\) is \((-\infty, \infty)\).

When an expression is in the form \(A \sin(Bx+C) + D\) or \(A \cos(Bx+C) + D\), the maximum value is \(|A| + D\) and the minimum value is \(-|A| + D\). In our case, \(\dfrac{1}{2} \sin 4x\) is in the form \(A \sin(Bx)\) with \(A = \dfrac{1}{2}\), \(B = 4\), and \(D = 0\). The maximum value is \(|\dfrac{1}{2}| + 0 = \dfrac{1}{2}\).