Problem 53
Question
If \(f(x)=\cos x,\) show 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
Verified Answer
The equation is shown to be correct by substituting the cosine addition formula and rearranging terms.
1Step 1: Apply the Difference Quotient
We start by looking at the difference quotient \( \frac{f(x+h) - f(x)}{h} \). For \( f(x) = \cos x \), we have \( f(x+h) = \cos(x+h) \). Substitute these into the quotient to get: \[ \frac{\cos(x+h) - \cos x}{h}. \]
2Step 2: Use the Cosine Addition Formula
Recall the cosine addition formula: \( \cos(x+h) = \cos x \cos h - \sin x \sin h \). Substituting this into the difference quotient gives: \[ \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}. \]
3Step 3: Factor Out \(\cos x\)
Notice that we can factor \(\cos x \) out from the first and last term in the numerator: \[ \cos x (\cos h - 1) - \sin x \sin h. \] Thus, the expression becomes: \[ \frac{\cos x (\cos h - 1) - \sin x \sin h}{h}. \]
4Step 4: Split the Fraction
Split the difference quotient into two separate fractions: \[ \frac{\cos x (\cos h - 1)}{h} - \frac{\sin x \sin h}{h}. \] Each can be simplified independently.
5Step 5: Write the Final Expression
Now, the expression is written as: \[ \cos x \left( \frac{\cos h - 1}{h} \right) - \sin x \left( \frac{\sin h}{h} \right). \] This matches the given form and confirms the validity of the manipulation.
Key Concepts
Cosine Addition FormulaLimit Definition of DerivativePrecalculusTrigonometric Identities
Cosine Addition Formula
The cosine addition formula is a very useful identity in trigonometry. It helps us express the cosine of a sum of two angles in another form. The formula is stated as follows:\[\cos(x + h) = \cos x \cos h - \sin x \sin h.\]Understanding this formula requires recognizing that it breaks down the cosine of a composite angle into components. This is particularly helpful when we deal with functions like cosine in calculus, as it allows for simplification and derivative calculations. In our exercise, using this formula was essential to rewrite \(\cos(x+h)\) in terms of \(\cos x\) and \(\sin x\). This step turned a seemingly complex trigonometric expression into a simpler form that could be further manipulated to solve the problem.
Limit Definition of Derivative
The limit definition of a derivative is a cornerstone in calculus. It provides a way to calculate the derivative of a function at a specific point. This definition is expressed as:\[\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.\]This mathematical expression helps us understand how a function behaves as it approaches a particular point along its curve. For the function \(f(x) = \cos x\), calculating its derivative involves setting up the difference quotient and simplifying it, much like we see in the exercise. By doing this, we find the rate of change of the function, or its slope, at any given point. In our solution, this meant leveraging the properly manipulated expression to evaluate this limit and confirm our findings.
Precalculus
Precalculus lays the groundwork for many calculus concepts. It covers a range of mathematical ideas that allow students to transition towards more complex calculus studies smoothly. One important aspect of precalculus is understanding how to manipulate functions and expressions, like those involving trigonometric functions.
Precalculus often includes studying functions, limits, and different notation. It ensures familiarity with the algebraic and trigonometric tools necessary to tackle calculus problems. In our specific exercise, the foundation laid in precalculus helps to understand the initial setup of the difference quotient and the application of trigonometric identities. Each algebraic manipulation or identity we use is a skill honed during precalculus studies, making it easier to handle tasks in calculus.
Precalculus often includes studying functions, limits, and different notation. It ensures familiarity with the algebraic and trigonometric tools necessary to tackle calculus problems. In our specific exercise, the foundation laid in precalculus helps to understand the initial setup of the difference quotient and the application of trigonometric identities. Each algebraic manipulation or identity we use is a skill honed during precalculus studies, making it easier to handle tasks in calculus.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values involved. They help simplify and solve trigonometric equations. In trigonometry, identities like the Pythagorean identity, angle sum and difference formulas, and double angle formulas, among others, are crucial tools.
In this problem, employing trigonometric identities, such as the cosine addition formula as explained earlier, allowed us to manipulate the difference quotient efficiently. Understanding these identities not only aids in simplifying the math but also supports solving more complex calculus problems. For instance, recognizing that \(\frac{\sin h}{h} \rightarrow 1\) and \(\frac{\cos h - 1}{h} \rightarrow 0\) as \(h\) approaches 0 are results derived from these identity manipulations in a calculus context, emphasizing their versatility and omnipresence in mathematics.
In this problem, employing trigonometric identities, such as the cosine addition formula as explained earlier, allowed us to manipulate the difference quotient efficiently. Understanding these identities not only aids in simplifying the math but also supports solving more complex calculus problems. For instance, recognizing that \(\frac{\sin h}{h} \rightarrow 1\) and \(\frac{\cos h - 1}{h} \rightarrow 0\) as \(h\) approaches 0 are results derived from these identity manipulations in a calculus context, emphasizing their versatility and omnipresence in mathematics.
Other exercises in this chapter
Problem 53
Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places.
View solution Problem 53
Show that the equation is not an Identity. $$\sqrt{\sin ^{2} t}=\sin t$$
View solution Problem 53
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$\cos \alpha+\sin \alpha=1$$
View solution Problem 54
Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places.
View solution