Problem 72 - Calculus - AP Edition

Question

Use the derivative formula for \(\sin x\) and the identity $$ \cos x=\sin \left(\frac{\pi}{2}-x\right) $$ to obtain the derivative formula for \(\cos x\)

Step-by-Step Solution

Answer

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

Step 1: Recall the derivative of \( \sin x \)

The derivative of \( \sin x \) with respect to \( x \) is \( \frac{d}{dx}[\sin x] = \cos x \). This is the standard derivative formula for the sine function.

Step 2: Express \( \cos x \) with Sine Identity

Using the given identity, express \( \cos x \) as \( \sin\left(\frac{\pi}{2} - x\right) \). This allows us to differentiate \( \cos x \) using the properties of sine.

Step 3: Differentiate \( \cos x = \sin\left(\frac{\pi}{2} - x\right) \)

Apply the chain rule on \( \sin\left(\frac{\pi}{2} - x\right) \):\[\frac{d}{dx}[\cos x] = \frac{d}{dx}[\sin(\frac{\pi}{2} - x)] = \cos(\frac{\pi}{2} - x) \cdot \frac{d}{dx}[\frac{\pi}{2} - x]\]

Step 4: Compute the derivative of the inner function

Calculate \( \frac{d}{dx}[\frac{\pi}{2} - x] = 0 - 1 = -1 \). The derivative of \( \frac{\pi}{2} \) is zero and \( -x \) is \(-1\).

Step 5: Apply Chain Rule Result

Substitute the derivative of the inner function into the expression:\[\frac{d}{dx}[\cos x] = \cos(\frac{\pi}{2} - x) \cdot (-1)\]\(\Rightarrow \frac{d}{dx}[\cos x] = -\cos(\frac{\pi}{2} - x)\)

Step 6: Simplify Using Trigonometric Identity

Recall that \( \sin(x) = \cos(\frac{\pi}{2} - x) \). Therefore, substitute \( \cos(\frac{\pi}{2} - x) \) with \( \sin x \):\[\frac{d}{dx}[\cos x] = -\sin x\]

Step 7: Conclusion: Derivative of \( \cos x \)

Thus, the derivative of \( \cos x \) with respect to \( x \) is \( -\sin x \).

Key Concepts

Derivative of sine function

The derivative of the sine function, which is often one of the first derivatives learned in calculus, is incredibly important. When you differentiate \( \sin x \) with respect to \( x \), the result is \( \cos x \).
This means that at any point on the curve \( \sin x \), the slope or rate of change of the function is equal to the value of \( \cos x \) at that point.
This fundamental connection between sine and cosine is a cornerstone in calculus and aids significantly in understanding more complex trigonometric derivatives. Here is how this relationship works:

• The derivative \( \frac{d}{dx}[\sin x] = \cos x \) shows how the angle's rate of rotation (slope) is linked to its horizontal projection (cosine).
• This formula provides a straightforward way to predict how sine values change as their angles change.

For instance, at \( x = 0 \), \( \sin x \) is 0, but its rate of change \( \cos x \) is 1. As the angle increases, the change continues to be governed by the cosine function.

Chain rule in calculus

In calculus, the chain rule is crucial when dealing with more complex functions that are composed of other functions. It allows us to take derivatives of composite functions.
A classic example is differentiating \( \sin(a - x) \) where \( a \) and \( x \) are variables or constants. You see this in action when deriving a formula for the derivative of the cosine function using \( \cos x = \sin(\frac{\pi}{2} - x) \).
Let's break down how the chain rule works here:

• Inside Function: In \( \sin(\frac{\pi}{2} - x) \), \( \frac{\pi}{2} - x \) is the inner function.
• Outside Function: \( \sin(u) \) is the outer function, where \( u = \frac{\pi}{2} - x \).
• Derivative of the Inside Function: First, differentiate the inner function: \( \frac{d}{dx}[\frac{\pi}{2} - x] = -1 \), because the derivative of \( \frac{\pi}{2} \) is 0 and \( x \) is \(-1\).
• Applying the Chain Rule: Combine the outer and inner derivatives: \( \cos(\frac{\pi}{2} - x) \cdot (-1) \).

This process yields an elegant result for the derivative of \( \cos x \), confirming that it is \(-\sin x\). The chain rule simplifies the inventive use of identities to find derivatives.

Trigonometric identities

Trigonometric identities are equations involving trig functions that are true for all values of the variables. They are the tools that help in simplifying complex calculations.
A good grasp of these identities is paramount in calculus, particularly when finding derivatives like in this exercise.
In this context, using the identity \( \cos x = \sin(\frac{\pi}{2} - x) \) is extremely useful.

• This identity allows us to transform \( \cos x \) expressions into easier-to-handle sine functions.
• Transforming a cosine into a sine function can be quite beneficial due to our simpler understanding of sine's derivative.

The identity helps us manipulate and simplify derivative calculations using the relationship between sine and cosine. Understanding how these identities interlink illustrates the beauty of trigonometry in calculus, where a deeper understanding of one function's derivative can directly simplify another's.