Problem 35 - Contemporary Precalculus

Question

If \(f(x)=\cos x\) and \(h\) is a fixed nonzero number, prove that: \(\frac{f(x+h)-f(x)}{h}=\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)\).

Step-by-Step Solution

Answer

#Answer# Using the sum-to-product formula and simplifying the expression, we have shown that for \(f(x) = \cos x\) and a fixed nonzero number \(h\), the following equation holds: $$\frac{f(x+h)-f(x)}{h} = \cos x \left(\frac{\cos h - 1}{h}\right) - \sin x \left(\frac{\sin h}{h}\right)$$

Step 1: Write down the given expressions

Let \(f(x) = \cos x\) and let \(h\) be a fixed nonzero number. Our goal is to prove that: $$\frac{f(x+h)-f(x)}{h}=\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)$$ Notice that we may consider the given equation as the definition of the average rate of change of \(f(x)\) over an interval of length \(h\).

Step 2: Apply the sum-to-product formula to \(f(x+h)\)

To do this, we need to find an expression for \(f(x + h)\). Since \(f(x) = \cos x\), this means: $$f(x + h) = \cos(x + h)$$ Now, we will apply the sum-to-product formula for cosine: $$\cos(x + h) = \cos x \cdot \cos h - \sin x \cdot \sin h$$

Step 3: Substitute \(f(x+h)\) and \(f(x)\) into the given expression

Now that we have an expression for \(f(x + h)\), we can substitute it and \(f(x)\) into the given expression: $$\frac{f(x+h)-f(x)}{h} = \frac{\cos x \cdot \cos h - \sin x \cdot \sin h - \cos x}{h}$$

Step 4: Simplify the expression

Now let's simplify the expression further by factoring out \(\cos x\) from the first two terms: $$\frac{f(x+h)-f(x)}{h} = \frac{\cos x (\cos h - 1) - \sin x \cdot \sin h}{h}$$ Now, we will divide both terms in the numerator by \(h\): $$\frac{f(x+h)-f(x)}{h} = \cos x \left(\frac{\cos h - 1}{h}\right) - \sin x \left(\frac{\sin h}{h}\right)$$ This is the expression we wanted to prove.

Key Concepts

Average Rate of Change

The average rate of change of a function is a fundamental concept in calculus. It measures how much a function's output changes per unit change in the input over a specified interval. Think of it as a way to gauge the 'speed' of change.

For any function \(f(x)\), the average rate of change from \(x\) to \(x + h\) is given by the formula:

• \( \frac{f(x+h) - f(x)}{h} \)

This expression essentially calculates the slope of the secant line connecting two points on the graph of the function.

In the exercise provided, \(f(x) = \cos x\) and \(h\) is a small nonzero interval. The goal is to express how the cosine function changes over this interval. This helps in understanding the behavior of the function over small distances, which is a stepping stone to the derivative concept.

Trigonometric Functions

Trigonometric functions like \(\cos(x)\) and \(\sin(x)\) are essential in mathematics, particularly calculus. These functions relate the angles of a triangle to its side lengths.

Here are a few facts about these functions:

• Cosine (\(\cos\)): It represents the adjacent side over the hypotenuse in a right triangle. It is periodic, repeating every \(2\pi\) radians.
• Sine (\(\sin\)): This function signifies the ratio of the opposite side to the hypotenuse.

Trigonometric functions are used extensively in calculus to model oscillatory behaviors, such as sound waves and light patterns. These concepts are crucial for solving various problems, including proving identities and evaluating limits, as seen in the exercise.

Sum-to-Product Formula

The sum-to-product formulas are incredibly useful trig identities that convert sums into products. This is particularly helpful when simplifying expressions in calculus.

For instance, the sum-to-product formula for cosine, which states:

• \( \cos(A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B \)

helps rearrange complex expressions into more manageable forms. In the provided exercise, we utilize this formula to express \(\cos(x + h)\) in terms of \(\cos x\) and \(\sin x\).

Understanding and applying these formulas can simplify solutions and enable progress in more complex calculus concepts like integration and differentiation.

Difference Quotient

The difference quotient is pivotal in calculus, serving as the foundation for defining derivatives. It measures the change in a function's value relative to the change in its variable.

The difference quotient is written as:

• \( \frac{f(x+h) - f(x)}{h} \)

and gets closer to the derivative as \(h\) approaches zero.

In the exercise, we're essentially calculating the difference quotient for the function \(f(x) = \cos x\). By simplifying the expression \(\cos x \left(\frac{\cos h - 1}{h}\right) - \sin x \left(\frac{\sin h}{h}\right)\), we demonstrate the initial steps that lead to deriving \(f'(x)\), the derivative of \(f(x)\). This understanding aids in mastering how functions behave over infinitesimally small intervals.