91Ó°ÊÓ

Trigonometric functions play a central role in calculus by representing periodic phenomena. This problem requires understanding of the trigonometric functions sine and cosine. The function in question here is a product of these two: \( y = \sin x \cos x \). Both \( \sin x \) and \( \cos x \) vary between -1 and 1, creating an oscillating wave pattern.

The sine function, \( \sin x \), starts from 0 at \( x = 0 \) and reaches its maximum of 1 at \( x = \frac{\pi}{2} \). Meanwhile, the cosine function, \( \cos x \), starts from its maximum of 1 at \( x = 0 \) and becomes 0 at \( x = \frac{\pi}{2} \).

Understanding where these functions intersect or relate helps in analyzing their combined behavior in the product formula.

When differentiating a product of two functions, the product rule is essential. The problem involves finding the derivative of \( y = \sin x \cos x \), which is a multiplication of two trigonometric functions.

The product rule states that if you have two functions, say \( u(x) \) and \( v(x) \), the derivative \((uv)'\) is \( u'v + uv' \). Here, \( u(x) = \sin x \) and \( v(x) = \cos x \).

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

Applying the product rule, these derivatives are combined to get \( y' = \cos x \cos x + \sin x (-\sin x) \). This results in \( y' = \cos^2 x - \sin^2 x \).

This expression of the derivative is key in determining where the graph has a horizontal tangent.

Graphical analysis involves looking at the graph of a function to identify certain characteristics, such as turning points, which can indicate where a tangent line is horizontal. In this problem, you use graphical tools to find such points for \( y = \sin x \cos x \).

Identifying horizontal tangents means finding where the slope \( y' \) of the function is 0. The derivative \( y' = \cos^2 x - \sin^2 x \) needs to be set to zero, simplifying to \( \cos^2 x = \sin^2 x \). This implies points where \( \tan x = 1 \).

You are looking for specific \( x \) values within the range \([0, 2\pi]\) where this condition holds. By both analyzing the derivative and visually examining a plotted graph using a graphing utility, one can confirm these locations visually.

Understanding derivatives and applying them is a foundational skill in calculus. Derivatives reveal numerous properties of functions, such as their rates of change and points of horizontal tangency.

In this exercise, the derivative \( y' = \cos^2 x - \sin^2 x \) provides critical information. It's used to find points where the graph \( y = \sin x \cos x \) levels out to a horizontal line. Setting the derivative equal to zero resolves to \( \tan x = 1 \), leading to solutions \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \) within \([0, 2\pi]\).

• These solutions are derived from known values where \( \tan x = 1 \), indicating specific angles where the sine and cosine functions are equal.

The ability to seamlessly transition from this algebraic understanding to confirming the results graphically is a powerful demonstration of derivative applications.