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The sine and cosine functions

Throughout Chapter, we will develop shortcut derivative rules to help us bypass the limit definition and quickly compute f'(x) from a formula for f(x).

Introduction

Throughout Chapter, we will develop shortcut derivative rules to help us bypass the limit definition and quickly compute \(f'(x)\) from a formula for \(f(x)\). In Section, we stated the rule for power functions, \[\begin{aligned}\end{aligned}\], and the rule for exponential functions, \[\begin{aligned}\end{aligned}\]. Later in this section, we will use a graphical argument to conjecture derivative formulas for the sine and cosine functions. In Preview Activity we use a graphical approach to understand why the rule for exponential functions is plausible.

Exploration

The sine and cosine functions

The sine and cosine functions are among the most important functions in all of mathematics. Sometimes called the circular functions due to their definition on the unit circle, these periodic functions play a key role in modeling repeating phenomena such as tidal elevations, the behavior of an oscillating mass attached to a spring, or the location of a point on a bicycle tire. Like polynomial and exponential functions, the sine and cosine functions are considered basic functions, ones that are often used in building more complicated functions. As such, we would like to know formulas for \(\frac{d}{dx} [\sin(x)]\) and \(\frac{d}{dx} [\cos(x)]\), and the next two activities lead us to that end.

The results of the two preceding activities suggest that the sine and cosine functions not only have beautiful connections such as the identities \(\sin^2(x) + \cos^2(x) = 1\) and \(\cos(x - \frac{\pi}{2}) = \sin(x)\), but that they are even further linked through calculus, as the derivative of each involves the other. We have now added the sine and cosine functions to our library of basic functions whose derivatives we know. The constant multiple and sum rules still hold, of course, as well as all of the inherent meaning of the derivative.

The following rules summarize the results of the activities These two rules may be formally proved using the limit definition of the derivative and the expansion identities for \(\sin(x+h)\) and \(\cos(x+h)\). .

For all real numbers \(x\), \[\begin{aligned}\end{aligned}\].

Summary

For an exponential function \(f(x) = a^x\) \((a \gt 1)\), the graph of \(f'(x)\) appears to be a scaled version of the original function. In particular, careful analysis of the graph of\(f(x) = 2^x\), suggests that \(\frac{d}{dx}[2^x] = 2^x \ln(2)\), which is a special case of the rule we stated in Section: \(\frac{d}{dx}[a^x] = a^x \ln(a)\).
By carefully analyzing the graphs of \(y = \sin(x)\) and \(y = \cos(x)\), and by using the limit definition of the derivative at select points, we found that \(\frac{d}{dx} [\sin(x)] = \cos(x)\) and \(\frac{d}{dx} [\cos(x)] = -\sin(x)\).
We note that all previously encountered derivative rules still hold, but now may also be applied to functions involving the sine and cosine. All of the established meaning of the derivative applies to these trigonometric functions as well.

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Let \(q(x) = 2 - (x-1)^2\), and use a graphing utility to plot this quadratic function. From the graph, at what value(s) of \(x\) is \(q'(x) = 0\)? (Remember: \(q'(x) = 0\) wherever the tangent line to \(q(x)\) is horizontal.) For what values of \(x\) is \(q'(x)\) positive? For what values of \(x\) is \(q'(x)\) negative?
Let \(f(x) = \sin(x)\), and use a graphing utility to plot this trigonometric function. State 3 values of \(x\) for which \(f'(x) = 0\). What is one interval of values of \(x\) on which \(f'(x)\) is positive? What is one interval of values of \(x\) on which \(f'(x)\) is negative?
Let \(g(x) = \cos(x)\), and use a graphing utility to plot this trigonometric function. State 3 values of \(x\) for which \(g'(x) = 0\). What is one interval of values of \(x\) on which \(g'(x)\) is positive? What is one interval of values of \(x\) on which \(g'(x)\) is negative?
Suppose that \(V(t) = 24 \cdot 1.07^t + 6 \sin(t)\) represents the value of a person's investment portfolio in thousands of dollars in year \(t\), where \(t = 0\) corresponds to January 1, 2010.
1. At what instantaneous rate is the portfolio's value changing on January 1, 2012? Include units on your answer.
2. Determine the value of \(V''(2)\). What are the units on this quantity and what does it tell you about how the portfolio's value is changing?
3. On the interval \(0 \le t \le 20\), graph the function \(V(t) = 24 \cdot 1.07^t + 6 \sin(t)\) and describe its behavior in the context of the problem. Then, compare the graphs of the functions \(A(t) = 24 \cdot 1.07^t\) and \(V(t) = 24 \cdot 1.07^t + 6 \sin(t)\), as well as the graphs of their derivatives \(A'(t)\) and \(V'(t)\). What is the impact of the term \(6 \sin(t)\) on the behavior of the function \(V(t)\)?
Sýna svarið
1. January 1, 2012 corresponds to the instant \(t = 2\), so we compute \(V'(2)\). First, we observe by the sum and constant multiple rules that \(V'(t) = 24 \cdot 1.07^t \cdot \ln(1.07) + 6 \cos(t)\). Thus, \(V'(2) = 24 \cdot 1.07^2 \cdot \ln(1.07) + 6 \cos(2) \approx -0.63778\). Since \(V\) is measured in thousands of dollars and \(t\) in years, we know that the portfolio's value is decreasing at a rate of \(637.78\) dollars per year on January 1, 2012.
2. Since \(V'(t) = 24 \cdot 1.07^t \cdot \ln(1.07) + 6 \cos(t)\), it follows that \(V''(t) = 24 \cdot 1.07^t \cdot (\ln(1.07))^2 - 6 \sin(t)\), and thus \(V''(2) = 24 \cdot 1.07^2 \cdot \ln(1.07)^2 - 6 \sin(2) \approx -5.33\), with units thousands of dollars per year per year. This means that around the time that \(t = 2\), the portfolio's value is decreasing at while the value function is concave down. More specifically, we know that \(V'\) is decreasing at the moment \(t = 2\) and expect the derivative's value to decrease by about \(5.33\) thousand dollars per year over the course of the next year. In other words, the portfolio's value is not only decreasing, but we expect it to decrease more rapidly as the year goes on.
3. In the figure below, we see plots of both \(V\) and \(A\) along with their corresponding derivatives. We see that the function \(A\) is a purely exponential function that always increases at an increasing rate. By adding the term \(6\sin(t)\) to \(A\) to create the function \(V\), we add volatility to the value of the portfolio: the oscillating nature of the sine function not only creates variance in the stock's value, but also in its rate of change. We also observe that as time progresses and the value of \(24 \cdot (1.07)^t\) increases, the effect of \(6 \sin(t)\) lessens.
Let \(f(x) = 3\cos(x) - 2\sin(x) + 6\).
1. Determine the exact slope of the tangent line to \(y = f(x)\) at the point where \(a = \frac{\pi}{4}\).
2. Determine the tangent line approximation to \(y = f(x)\) at the point where \(a = \pi\).
3. At the point where \(a = \frac{\pi}{2}\), is \(f\) increasing, decreasing, or neither?
4. At the point where \(a = \frac{3\pi}{2}\), does the tangent line to \(y = f(x)\) lie above the curve, below the curve, or neither? How can you answer this question without even graphing the function or the tangent line?
Sýna svarið
1. The slope of the line tangent to the graph of \(f\) at \(a = \frac{\pi}{4}\) is given by \(f'\left(\frac{\pi}{4}\right)\). To find \(f'\left(\frac{\pi}{4}\right)\) we first find \(f'(x)\), which is \(f'(x) = -3\sin(x) - 2\cos(x)\). Thus, \[\begin{aligned}f'\left(\frac{\pi}{4}\right) &= -3\sin\left(\frac{\pi}{4}\right) - 2\cos\left(\frac{\pi}{4}\right) \\ & -3\left(\frac{\sqrt{2}}{2}\right) - 3\left(\frac{\sqrt{2}}{2}\right) = -5\left(\frac{\sqrt{2}}{2}\right)\end{aligned}\].
2. The linearization \(L(x)\) to \(f\) at \(a = \pi\) is \((x) = f(\pi) + f'(\pi)(x-\pi)\)L. We calculated \(f'(x)\) in part (a) and so \(f'(\pi) = -3\sin(\pi) - 2\cos(\pi) = 2\). We also have \(f(\pi) = 3\cos(\pi) - 2\sin(\pi) + 6 = 3\). Hence \(L(x) = 3+2(x-\pi)\).
3. A differentiable function \(f\) is increasing at \(x=a\) if and only if \(f'(a) \gt 0\) and decreasing if and only if \(f'(a) \lt 0\). Now \(f'\left(\frac{\pi}{2}\right) = -3\sin\left(\frac{\pi}{2}\right) - 2\cos\left(\frac{\pi}{2}\right) = -3\), so \(f\) is decreasing at \(a = \frac{\pi}{2}\).
4. If a curve defined by a function \(f\) is concave up at \(x=a\), then the tangent line to \(f\) at \(x=a\) lies below the curve and if the graph of \(f\) is concave down at \(x=a\), then the tangent line to \(f\) at \(x=a\) lies above the curve. The concavity of a curve is determined by the second derivative and \(f''(x) = -3\cos(x)+2\sin(x)\). Since \[\begin{aligned}\end{aligned}\], the graph of \(f\) is concave down at \(a = \frac{3\pi}{2}\) and the tangent line to \(f\) lies above the curve at this point.
In this exercise, we explore how the limit definition of the derivative more formally shows that \(\frac{d}{dx}[\sin(x)] = \cos(x)\). Letting \(f(x) = \sin(x)\), note that the limit definition of the derivative tells us that \[\begin{aligned}\end{aligned}\].
1. Recall the trigonometric identity for the sine of a sum of angles \(\alpha\) and \(\beta\): \(\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\). Use this identity and some algebra to show that \[\begin{aligned}\end{aligned}\].
2. Next, note that as \(h\) changes, \(x\) remains constant. Explain why it therefore makes sense to say that \[\begin{aligned}\end{aligned}\].
3. Finally, use small values of \(h\) to estimate the values of the two limits in (c): \[\begin{aligned}\end{aligned}\].
4. What do your results in (b) and (c) thus tell you about \(f'(x)\)?
5. By emulating the steps taken above, use the limit definition of the derivative to argue convincingly that \(\frac{d}{dx}[\cos(x)] = -\sin(x)\).
Sýna svarið
1. By definition, we know that \[\begin{aligned}\end{aligned}\]. By the sum of two angles identity for the sine function, \(\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)\), and thus \[\begin{aligned}\end{aligned}\].. Combining the first and last terms and removing a factor of \(\sin(x)\), \[\begin{aligned}\end{aligned}\]..
2. Continuing our work in (a), we divide each part of the numerator by \(h\) and see it follows \[\begin{aligned}\end{aligned}\]. Next, we note that \(\sin(x)\) does not depend on \(h\), and likewise \(\cos(x)\) is independent of \(h\). It thus makes sense to view those constants as being outside the limit being taken, and also that the limit of a sum is the sum of the limits. Hence we write \[\begin{aligned}\end{aligned}\].
3. Using small positive and negative values of \(h\) to estimate the desired limits, we find that it appears \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\] It suffices in each case to consider \(h\)-values of \(h = \pm 0.01, \pm 0.001, \pm 0.0001\).
4. Combining our results in (b) and (c), we have found that \[\begin{aligned}\end{aligned}\].
5. The argument for showing \(\frac{d}{dx}[\cos(x)] = -\sin(x)\) is almost identical. Do note that the sum of two angles identity for \(\cos(\alpha + \beta)\) is \(\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)\).
Suppose that \(V(t) = 24 \cdot 1.07^t + 6 \sin(t)\) represents the value of a person's investment portfolio in thousands of dollars in year \(t\), where \(t = 0\) corresponds to January 1, 2010.
1. At what instantaneous rate is the portfolio's value changing on January 1, 2012? Include units on your answer.
2. Determine the value of \(V''(2)\). What are the units on this quantity and what does it tell you about how the portfolio's value is changing?
3. On the interval \(0 \le t \le 20\), graph the function \(V(t) = 24 \cdot 1.07^t + 6 \sin(t)\) and describe its behavior in the context of the problem. Then, compare the graphs of the functions \(A(t) = 24 \cdot 1.07^t\) and \(V(t) = 24 \cdot 1.07^t + 6 \sin(t)\), as well as the graphs of their derivatives \(A'(t)\) and \(V'(t)\). What is the impact of the term \(6 \sin(t)\) on the behavior of the function \(V(t)\)?
Sýna svarið
1. January 1, 2012 corresponds to the instant \(t = 2\), so we compute \(V'(2)\). First, we observe by the sum and constant multiple rules that \(V'(t) = 24 \cdot 1.07^t \cdot \ln(1.07) + 6 \cos(t)\). Thus, \(V'(2) = 24 \cdot 1.07^2 \cdot \ln(1.07) + 6 \cos(2) \approx -0.63778\). Since \(V\) is measured in thousands of dollars and \(t\) in years, we know that the portfolio's value is decreasing at a rate of \(637.78\) dollars per year on January 1, 2012.
2. Since \(V'(t) = 24 \cdot 1.07^t \cdot \ln(1.07) + 6 \cos(t)\), it follows that \(V''(t) = 24 \cdot 1.07^t \cdot (\ln(1.07))^2 - 6 \sin(t)\), and thus \(V''(2) = 24 \cdot 1.07^2 \cdot \ln(1.07)^2 - 6 \sin(2) \approx -5.33\), with units thousands of dollars per year per year. This means that around the time that \(t = 2\), the portfolio's value is decreasing at while the value function is concave down. More specifically, we know that \(V'\) is decreasing at the moment \(t = 2\) and expect the derivative's value to decrease by about \(5.33\) thousand dollars per year over the course of the next year. In other words, the portfolio's value is not only decreasing, but we expect it to decrease more rapidly as the year goes on.
3. In the figure below, we see plots of both \(V\) and \(A\) along with their corresponding derivatives. We see that the function \(A\) is a purely exponential function that always increases at an increasing rate. By adding the term \(6\sin(t)\) to \(A\) to create the function \(V\), we add volatility to the value of the portfolio: the oscillating nature of the sine function not only creates variance in the stock's value, but also in its rate of change. We also observe that as time progresses and the value of \(24 \cdot (1.07)^t\) increases, the effect of \(6 \sin(t)\) lessens.
Let \(f(x) = 3\cos(x) - 2\sin(x) + 6\).
1. Determine the exact slope of the tangent line to \(y = f(x)\) at the point where \(a = \frac{\pi}{4}\).
2. Determine the tangent line approximation to \(y = f(x)\) at the point where \(a = \pi\).
3. At the point where \(a = \frac{\pi}{2}\), is \(f\) increasing, decreasing, or neither?
4. At the point where \(a = \frac{3\pi}{2}\), does the tangent line to \(y = f(x)\) lie above the curve, below the curve, or neither? How can you answer this question without even graphing the function or the tangent line?
Sýna svarið
1. The slope of the line tangent to the graph of \(f\) at \(a = \frac{\pi}{4}\) is given by \(f'\left(\frac{\pi}{4}\right)\). To find \(f'\left(\frac{\pi}{4}\right)\) we first find \(f'(x)\), which is \(f'(x) = -3\sin(x) - 2\cos(x)\). Thus, \[\begin{aligned}f'\left(\frac{\pi}{4}\right) &= -3\sin\left(\frac{\pi}{4}\right) - 2\cos\left(\frac{\pi}{4}\right) \\ & -3\left(\frac{\sqrt{2}}{2}\right) - 3\left(\frac{\sqrt{2}}{2}\right) = -5\left(\frac{\sqrt{2}}{2}\right)\end{aligned}\].
2. The linearization \(L(x)\) to \(f\) at \(a = \pi\) is \((x) = f(\pi) + f'(\pi)(x-\pi)\)L. We calculated \(f'(x)\) in part (a) and so \(f'(\pi) = -3\sin(\pi) - 2\cos(\pi) = 2\). We also have \(f(\pi) = 3\cos(\pi) - 2\sin(\pi) + 6 = 3\). Hence \(L(x) = 3+2(x-\pi)\).
3. A differentiable function \(f\) is increasing at \(x=a\) if and only if \(f'(a) \gt 0\) and decreasing if and only if \(f'(a) \lt 0\). Now \(f'\left(\frac{\pi}{2}\right) = -3\sin\left(\frac{\pi}{2}\right) - 2\cos\left(\frac{\pi}{2}\right) = -3\), so \(f\) is decreasing at \(a = \frac{\pi}{2}\).
4. If a curve defined by a function \(f\) is concave up at \(x=a\), then the tangent line to \(f\) at \(x=a\) lies below the curve and if the graph of \(f\) is concave down at \(x=a\), then the tangent line to \(f\) at \(x=a\) lies above the curve. The concavity of a curve is determined by the second derivative and \(f''(x) = -3\cos(x)+2\sin(x)\). Since \[\begin{aligned}\end{aligned}\], the graph of \(f\) is concave down at \(a = \frac{3\pi}{2}\) and the tangent line to \(f\) lies above the curve at this point.
In this exercise, we explore how the limit definition of the derivative more formally shows that \(\frac{d}{dx}[\sin(x)] = \cos(x)\). Letting \(f(x) = \sin(x)\), note that the limit definition of the derivative tells us that \[\begin{aligned}\end{aligned}\].
1. Recall the trigonometric identity for the sine of a sum of angles \(\alpha\) and \(\beta\): \(\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\). Use this identity and some algebra to show that \[\begin{aligned}\end{aligned}\].
2. Next, note that as \(h\) changes, \(x\) remains constant. Explain why it therefore makes sense to say that \[\begin{aligned}\end{aligned}\].
3. Finally, use small values of \(h\) to estimate the values of the two limits in (c): \[\begin{aligned}\end{aligned}\].
4. What do your results in (b) and (c) thus tell you about \(f'(x)\)?
5. By emulating the steps taken above, use the limit definition of the derivative to argue convincingly that \(\frac{d}{dx}[\cos(x)] = -\sin(x)\).
Sýna svarið
1. By definition, we know that \[\begin{aligned}\end{aligned}\]. By the sum of two angles identity for the sine function, \(\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)\), and thus \[\begin{aligned}\end{aligned}\].. Combining the first and last terms and removing a factor of \(\sin(x)\), \[\begin{aligned}\end{aligned}\]..
2. Continuing our work in (a), we divide each part of the numerator by \(h\) and see it follows \[\begin{aligned}\end{aligned}\]. Next, we note that \(\sin(x)\) does not depend on \(h\), and likewise \(\cos(x)\) is independent of \(h\). It thus makes sense to view those constants as being outside the limit being taken, and also that the limit of a sum is the sum of the limits. Hence we write \[\begin{aligned}\end{aligned}\].
3. Using small positive and negative values of \(h\) to estimate the desired limits, we find that it appears \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\] It suffices in each case to consider \(h\)-values of \(h = \pm 0.01, \pm 0.001, \pm 0.0001\).
4. Combining our results in (b) and (c), we have found that \[\begin{aligned}\end{aligned}\].
5. The argument for showing \(\frac{d}{dx}[\cos(x)] = -\sin(x)\) is almost identical. Do note that the sum of two angles identity for \(\cos(\alpha + \beta)\) is \(\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)\).

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The sine and cosine functions

What is a graphical justification for why \frac{d}{dx}[a^x] = a^x \ln(a)?
What do the graphs of y = \sin(x) and y = \cos(x) suggest as formulas for their respective derivatives?
Once we know the derivatives of \sin(x) and \cos(x), how do previous derivative rules work when these functions are involved?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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