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Derivatives of the Sine and Cosine Functions

We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function \(f \left(\right. x \left.\right) ,\)

\[f^{'} \left(\right. x \left.\right) = \underset{h \rightarrow 0}{\text{lim}} \frac{f \left(\right. x + h \left.\right) - f \left(\right. x \left.\right)}{h} .\]

Consequently, for values of \(h\) very close to 0, \(f^{'} \left(\right. x \left.\right) \approx \frac{f \left(\right. x + h \left.\right) - f \left(\right. x \left.\right)}{h} .\) We see that by using \(h = 0.01 ,\)

\[\frac{d}{d x} \left(\right. \text{sin} x \left.\right) \approx \frac{\text{sin} \left(\right. x + 0.01 \left.\right) - \text{sin} x}{0.01}\]

By setting \(D \left(\right. x \left.\right) = \frac{\text{sin} \left(\right. x + 0.01 \left.\right) - \text{sin} x}{0.01}\) and using a graphing utility, we can get a graph of an approximation to the derivative of \(\text{sin} x\) (Figure 3.25).

The function D(x) = (sin(x + 0.01) − sin x)/0.01 is graphed. It looks a lot like a cosine curve.
Figure 3.25 The graph of the function \(D \left(\right. x \left.\right)\) looks a lot like a cosine curve.

Upon inspection, the graph of \(D \left(\right. x \left.\right)\) appears to be very close to the graph of the cosine function. Indeed, we will show that

\[\frac{d}{d x} \left(\right. \text{sin} x \left.\right) = \text{cos} x .\]

If we were to follow the same steps to approximate the derivative of the cosine function, we would find that

\[\frac{d}{d x} \left(\right. \text{cos} x \left.\right) = − \text{sin} x.\]

Theorem 3.8

The Derivatives of sin x and cos x

The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.

\[\frac{d}{d x} \left(\right. \text{sin} x \left.\right) = \text{cos} x\]
\[\frac{d}{d x} \left(\right. \text{cos} x \left.\right) = − \text{sin} x\]

Proof

Because the proofs for \(\frac{d}{d x} \left(\right. \text{sin} x \left.\right) = \text{cos} x\) and \(\frac{d}{d x} \left(\right. \text{cos} x \left.\right) = − \text{sin} x\) use similar techniques, we provide only the proof for \(\frac{d}{d x} \left(\right. \text{sin} x \left.\right) = \text{cos} x .\) Before beginning, recall two important trigonometric limits we learned in Introduction to Limits:

\[\underset{h \rightarrow 0}{\text{lim}} \frac{\text{sin} h}{h} = 1 \text{and} \underset{h \rightarrow 0}{\text{lim}} \frac{\text{cos} h - 1}{h} = 0 .\]

The graphs of \(y = \frac{\left(\right. \text{sin} h \left.\right)}{h}\) and \(y = \frac{\left(\right. \text{cos} h - 1 \left.\right)}{h}\) are shown in Figure 3.26.

The function y = (sin h)/h and y = (cos h – 1)/h are graphed. They both have discontinuities on the y-axis.
Figure 3.26 These graphs show two important limits needed to establish the derivative formulas for the sine and cosine functions.

We also recall the following trigonometric identity for the sine of the sum of two angles:

\[\text{sin} \left(\right. x + h \left.\right) = \text{sin} x \text{cos} h + \text{cos} x \text{sin} h .\]

Now that we have gathered all the necessary equations and identities, we proceed with the proof.

\[\begin{aligned} \frac{d}{d x} \text{sin} x & = \underset{h \rightarrow 0}{\text{lim}} \frac{\text{sin} \left(\right. x + h \left.\right) - \text{sin} x}{h} & & & \text{Apply the definition} \text{of the derivative}. \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{\text{sin} x \text{cos} h + \text{cos} x \text{sin} h - \text{sin} x}{h} & & & \text{Use trig identity for the sine of the sum of two angles}. \\ & = \underset{h \rightarrow 0}{\text{lim}} \left(\right. \frac{\text{sin} x \text{cos} h - \text{sin} x}{h} + \frac{\text{cos} x \text{sin} h}{h} \left.\right) & & & \text{Regroup}. \\ & = \underset{h \rightarrow 0}{\text{lim}} \left(\right. \text{sin} x \left(\right. \frac{\text{cos} h - 1}{h} \left.\right) + \text{cos} x \left(\right. \frac{\text{sin} h}{h} \left.\right) \left.\right) & & & \text{Factor out} \text{sin} x \text{and} \text{cos} x . \\ & = \text{sin} x \cdot 0 + \text{cos} x \cdot 1 & & & \text{Apply trig limit formulas}. \\ & = \text{cos} x & & & \text{Simplify}. \end{aligned}\]

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Figure 3.27 shows the relationship between the graph of \(f \left(\right. x \left.\right) = \text{sin} x\) and its derivative \(f^{'} \left(\right. x \left.\right) = \text{cos} x .\) Notice that at the points where \(f \left(\right. x \left.\right) = \text{sin} x\) has a horizontal tangent, its derivative \(f^{'} \left(\right. x \left.\right) = \text{cos} x\) takes on the value zero. We also see that where \(f \left(\right. x \left.\right) = \text{sin} x\) is increasing, \(f^{'} \left(\right. x \left.\right) = \text{cos} x > 0\) and where \(f \left(\right. x \left.\right) = \text{sin} x\) is decreasing, \(f^{'} \left(\right. x \left.\right) = \text{cos} x < 0 .\)

The functions f(x) = sin x and f’(x) = cos x are graphed. It is apparent that when f(x) has a maximum or a minimum that f’(x) = 0.
Figure 3.27 Where \(f \left(\right. x \left.\right)\) has a maximum or a minimum, \(f ' \left(\right. x \left.\right) = 0\) that is, \(f ' \left(\right. x \left.\right) = 0\) where \(f \left(\right. x \left.\right)\) has a horizontal tangent. These points are noted with dots on the graphs.

Example 3.39

Differentiating a Function Containing sin x

Find the derivative of \(f \left(\right. x \left.\right) = 5 x^{3} \text{sin} x .\)

Solution

Using the product rule, we have

\[\begin{aligned} f ' \left(\right. x \left.\right) & = \frac{d}{d x} \left(\right. 5 x^{3} \left.\right) \cdot \text{sin} x + \frac{d}{d x} \left(\right. \text{sin} x \left.\right) \cdot 5 x^{3} \\ & = 15 x^{2} \cdot \text{sin} x + \text{cos} x \cdot 5 x^{3} . \end{aligned}\]

After simplifying, we obtain

\[f^{'} \left(\right. x \left.\right) = 15 x^{2} \text{sin} x + 5 x^{3} \text{cos} x .\]

Checkpoint 3.25

Find the derivative of \(f \left(\right. x \left.\right) = \text{sin} x \text{cos} x .\)

Example 3.40

Finding the Derivative of a Function Containing cos x

Find the derivative of \(g \left(\right. x \left.\right) = \frac{\text{cos} x}{4 x^{2}} .\)

Solution

By applying the quotient rule, we have

\[g^{'} \left(\right. x \left.\right) = \frac{\left(\right. − \text{sin} x \left.\right) 4 x^{2} - 8 x \left(\right. \text{cos} x \left.\right)}{\left(\right. 4 x^{2} \left.\right)^{2}} .\]

Simplifying, we obtain

\[\begin{aligned} g^{'} \left(\right. x \left.\right) & = \frac{−4 x^{2} \text{sin} x - 8 x \text{cos} x}{16 x^{4}} \\ & = \frac{− x \text{sin} x - 2 \text{cos} x}{4 x^{3}} . \end{aligned}\]

Checkpoint 3.26

Find the derivative of \(f \left(\right. x \left.\right) = \frac{x}{\text{cos} x} .\)

Example 3.41

An Application to Velocity

A particle moves along a coordinate axis in such a way that its position at time \(t\) is given by \(s \left(\right. t \left.\right) = 2 \text{sin} t - t\) for \(0 \leq t \leq 2 \pi .\) At what times is the particle at rest?

Solution

To determine when the particle is at rest, set \(s^{'} \left(\right. t \left.\right) = v \left(\right. t \left.\right) = 0 .\) Begin by finding \(s^{'} \left(\right. t \left.\right) .\) We obtain

\[s^{'} \left(\right. t \left.\right) = 2 \text{cos} t - 1 ,\]

so we must solve

\[2 \text{cos} t - 1 = 0 \text{for} 0 \leq t \leq 2 \pi .\]

The solutions to this equation are \(t = \frac{\pi}{3}\) and \(t = \frac{5 \pi}{3} .\) Thus the particle is at rest at times \(t = \frac{\pi}{3}\) and \(t = \frac{5 \pi}{3} .\)

Checkpoint 3.27

A particle moves along a coordinate axis. Its position at time \(t\) is given by \(s \left(\right. t \left.\right) = \sqrt{3} t + 2 \text{cos} t\) for \(0 \leq t \leq 2 \pi .\) At what times is the particle at rest?

This lesson is part of:

Derivatives

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