Problem 84 - Calculus - AP Edition

Question

Writing The "co" in "cosine" comes from "complementary," since the cosine of an angle is the sine of the complementary angle, and vice versa: $$ \cos x=\sin \left(\frac{\pi}{2}-x\right) \quad \text { and } \quad \sin x=\cos \left(\frac{\pi}{2}-x\right) $$ Suppose that we define a function \(g\) to be a cofunction of a function \(f\) if $$ g(x)=f\left(\frac{\pi}{2}-x\right) \quad \text { for all } x $$ Thus, cosine and sine are cofunctions of each other, as are cotangent and tangent, and also cosecant and secant. If \(g\) is the cofunction of \(f\), state a formula that relates \(g^{\prime}\) and the cofunction of \(f^{\prime}\), Discuss how this relationship is exhibited by the derivatives of the cosine, cotangent, and cosecant functions.

Step-by-Step Solution

Answer

If \( g \) is the cofunction of \( f \), then \( g'(x) = -f'(\frac{\pi}{2} - x) \). This relationship is confirmed by examining the derivatives of trigonometric functions.

Step 1: Establish the Formula for Cofunctions

Given that the function \( g \) is a cofunction of \( f \), we have the relationship \( g(x) = f\left(\frac{\pi}{2} - x\right) \). This relationship implies that taking the derivative of \( g(x) \) with respect to \( x \) will involve the chain rule, specifically due to the composite function \( \frac{\pi}{2} - x \), which has a derivative of \( -1 \).

Step 2: Derive Formula for the Derivative Relationship

When differentiating \( g(x) = f\left(\frac{\pi}{2} - x\right) \), using the chain rule gives us \( g'(x) = -f'\left(\frac{\pi}{2} - x\right) \). Thus, the derivative of the cofunction \( g'(x) \) is related to the derivative of \( f \) evaluated at \( \frac{\pi}{2} - x \), with a negative sign.

Step 3: Relate This to Trigonometric Functions

For trigonometric cofunctions, the derivative of a function and its cofunction should reflect the chain rule result. For instance, the derivative of \( \cos x \) is \( -\sin x \), and the cofunction \( \sin x \) translates to \( \cos\left(\frac{\pi}{2} - x\right) = \cos x \). Hence, the derivative relationship is consistent with \( g'(x) = -f'\left(\frac{\pi}{2} - x\right) \). Repeat a similar check for cotangent and cosecant.

Step 4: Verify with Cotangent and Cosecant

Consider that the derivative of \( \cot x \) is \( -\csc^2 x \), and \( \tan x \) is the cofunction, where \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \). This is consistent with the derived relationship. Similarly, \( \csc x = \sec\left(\frac{\pi}{2} - x\right) \), and upon differentiating \( \csc x \), we retrieve \( -\csc x \cot x \), matching the expected pattern.

Key Concepts

Trigonometric Functions

Trigonometric functions are foundational in understanding angles and their relationships in various geometric contexts. They include the sine, cosine, tangent, cotangent, secant, and cosecant functions. Each of these functions represents a specific ratio derived from a right triangle's angles and side lengths. Trigonometric functions are periodic, which means they repeat values at regular intervals. This periodicity is crucial in both mathematics and physics for modeling waves and oscillations.
A unique aspect of trigonometric functions is their relationship as cofunctions. Cofunctions are pairs where the function of an angle is equal to the complementary function of its complement. For example:

• The sine of an angle equals the cosine of its complement: \( \sin x = \cos\left(\frac{\pi}{2} - x\right) \)
• Similarly, the cosine of an angle equals the sine of its complement: \( \cos x = \sin\left(\frac{\pi}{2} - x\right) \)

This complementary relationship is the root of the prefix "co" in cosine, cotangent, and cosecant. Understanding these relationships helps solve problems involving angle transformations and identities, making trigonometric functions invaluable in various applications.

Derivatives

Derivatives measure how a function changes as its input changes. They're fundamental in calculus and play a vital role in understanding the behavior of functions. When we talk about the derivative of a function, we're interested in the rate of change or slope of the function at each point. For trigonometric functions, calculating derivatives often provides insight into wave behavior and oscillatory systems.
The derivatives of standard trigonometric functions can be summarized as follows:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).
• The derivative of \( \cot x \) is \(-\csc^2 x \).
• The derivative of \( \sec x \) is \( \sec x \tan x \).
• The derivative of \( \csc x \) is \(-\csc x \cot x \).

Understanding these derivatives is crucial, especially in differentiating trigonometric cofunctions. Derivatives help in determining the behavior of trigonometric functions as their input angles shift, providing insights that are essential in fields like physics and engineering.

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of composite functions. It describes how to differentiate a function that is nested inside another function. This is particularly useful when dealing with trigonometric cofunctions, as seen in expressions involving angles like \( \frac{\pi}{2} - x \).
To apply the chain rule, you multiply the derivative of the outer function by the derivative of the inner function. Mathematically, if you have a function \( y = f(g(x)) \), the derivative, \( y' \), is found as: \[ y' = f'(g(x)) \cdot g'(x) \] For trigonometric cofunctions, consider \( g(x) = f\left(\frac{\pi}{2} - x\right) \). Differentiating \( g(x) \) requires the chain rule, giving: \[ g'(x) = -f'\left(\frac{\pi}{2} - x\right) \] The derivative of the inside function \( \frac{\pi}{2} - x \) is \(-1\), which accounts for the negative sign. This helps relate the derivative of trigonometric functions to their cofunctions, maintaining consistency with the identities and relationships seen in trigonometry.