Problem 84 - Calculus Early Transcendentals

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

The derivative of the cofunction is the negative of the derivative of the original function evaluated at the complementary angle. This is shown by cosine and sine, cotangent and tangent, and cosecant and secant derivatives.

Step 1: Understand the Cofunction Definition

By definition, if \( g(x) = f \left( \frac{\pi}{2} - x \right) \), then \( g \) is the cofunction of \( f \). This can be extended to their derivatives.

Step 2: Differentiate the Cofunction Relationship

Differentiate both sides of the equation \( g(x) = f \left( \frac{\pi}{2} - x \right) \) with respect to \( x \) using the chain rule. This gives:\[g'(x) = -f' \left( \frac{\pi}{2} - x \right)\]The negative sign comes from differentiating \( \frac{\pi}{2} - x \), which yields \(-1\).

Step 3: Establish the Relationship for Derivatives

The derivative of the cofunction \( g \) results in:\[g'(x) = -f' \left( \frac{\pi}{2} - x \right)\]This shows that the derivative of a cofunction is the negative of the derivative of the original function evaluated at the complementary angle.

Step 4: Apply to Cosine and Sine Derivatives

For \( f(x) = \sin x \), we have \( g(x) = \cos x \). Thus, \( g'(x) = -\sin'(\frac{\pi}{2} - x) = -\cos(x) \). This is verified by the known derivative \( \cos' x = -\sin x \).

Step 5: Apply to Cotangent and Tangent Derivatives

For \( f(x) = \tan x \), we have \( g(x) = \cot x \). So, \( g'(x) = -\tan'(\frac{\pi}{2} - x) = -\sec^2(x) \), which agrees with \( \cot' x = -\csc^2(x) \).

Step 6: Apply to Cosecant and Secant Derivatives

For \( f(x) = \sec x \), we have \( g(x) = \csc x \). Thus, \( g'(x) = -\sec'(\frac{\pi}{2} - x) = -\sec(x)\tan(x) \), matching with \( \csc' x = -\csc(x)\cot(x) \).

Key Concepts

Cosine

The cosine function, often abbreviated as "cos," is one of the fundamental trigonometric functions. It is defined in relation to a right-angled triangle as the ratio of the adjacent side to the hypotenuse. Mathematically, this can be expressed as \[\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}.\] Notably, the "co" in "cosine" stands for "complementary" angle, tying it inherently to the sine function. This connection is expressed through the identity:\[\cos(x) = \sin\left(\frac{\pi}{2} - x\right).\]This indicates that the cosine of an angle is equivalent to the sine of its complementary angle (90 degrees or \(\pi/2\) radians minus the original angle). This linkage can be quite helpful in solving various trigonometry problems, revealing symmetrical properties of the unit circle.
Furthermore, in the context of derivatives, the cosine function acts as the cofunction of the sine. The derivative of the cosine function is: \[\frac{d}{dx} \cos x = -\sin x.\]
This negative sine derivative relates directly to how the cosine changes with respect to the angle, diminishing as the cosine value shifts through the cycle of the unit circle.

Sine

The sine function, or "sin," is another cornerstone of trigonometry, defined as the ratio of the opposite side to the hypotenuse in a right-angled triangle:\[\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}.\]The sine function is deeply linked to the cosine as its cofunction. This is beautifully illustrated by the identity:\[\sin(x) = \cos\left(\frac{\pi}{2} - x\right).\]This means the sine of any angle is equal to the cosine of its complementary angle. The symmetry of these functions across the unit circle showcases their periodic nature and interrelationship.
From a calculus perspective, the derivative of the sine function is:\[\frac{d}{dx} \sin x = \cos x.\]
Here, the positive derivative highlights that a small change in angle will give rise to a proportional increase in the sine value, along the applicable stretch of its wave. In the nature of trigonometric derivatives, understanding this relationship aids in transforming and differentiating more complex functions.

Trigonometric Derivatives

When working with trigonometric functions, derivatives are integral in understanding their changes concerning angles. Trigonometric derivatives are derivatives of the fundamental functions like sine, cosine, tangent, and their cofunctions.
Here’s how some of these derivatives play out:

• For the cosine function, \( \cos(x) \), the derivative is \( -\sin(x) \), highlighting that as the cosine value decreases, its slope is negative.
• For the sine function, \( \sin(x) \), the derivative is \( \cos(x) \), showcasing the direct, equivalent nature of their gradient shifts.
• For the tangent function, \( \tan(x) \), which is not directly a cofunction, the derivative is \( \sec^2(x) \).
• The cofunction, cotangent \( \cot(x) \), has a derivative of \( -\csc^2(x) \), mirroring its complementary nature to the tangent function.

In problems involving these derivatives, the chain rule is often employed to differentiate composite functions or cofunctions accurately. Understanding these basics allows for deeper insights into more complex trigonometric identities and their transformations. Each function and its derivative reveals important insights into the slope and rate of change of these periodic functions on the unit circle.