[math]
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[/math]
The formulas for the derivative and integral of the functions [math]\sin[/math]
and [math]\cos[/math] follow in a straightforward way from one fundamental limit theorem. It is
[[math]]
\lim_{t \rightarrow 0} \frac{\sin t}{t} = 1.
[[/math]]
Show Proof
It is convenient first to impose the restriction that [math]t \gt 0[/math] and prove that the limit from
the right equals 1; i.e.,
[[math]]
\begin{equation}
\lim_{t \rightarrow 0+} \frac{\sin t}{t} = 1.
\label{eq6.2.1}
\end{equation}
[[/math]]
Since, in proving (1), we are concerned only with small values of
[math]t[/math], we may assume
that
[math]t \lt \frac{\pi}{2}[/math]. Thus we have
[math]0 \lt t \lt \frac{\pi}{2}[/math] and, as a consequence,
[math]\sin t \gt 0[/math] and
[math]\cos t \gt 0[/math]. Let
[math]S[/math] be the region in the plane bounded by the circle
[math]x^2 + y^2 = 1[/math],
the positive
[math]x[/math]-axis, and the line segment which joins the origin to the point
[math](\cos t, \sin t)[/math]; i.e.,
[math]S[/math] is the shaded sector in Figure 6. Since
the area of the circle is
[math]\pi[/math] and the circumference is
[math]2\pi[/math], the area of
[math]S[/math] is equal to
[math]\frac{t}{2\pi} \cdot \pi = \frac{t}{2}[/math]. Next, consider the right triangle
[math]T_{1}[/math] with vertices (0, 0),
[math](\cos t, \sin t)[/math], and
[math](\cos t, 0)[/math]. Since the area of any triangle is one half the base times the
altitude, it follows that
[math]area(T_{1}) = \frac{1}{2} \cos t \sin t[/math]. The line which passes through (0,0)
and
[math](\cos t, \sin t)[/math] has slope
[math]\frac{\sin t}{\cos t}[/math] and equation
[math]y =\frac{\sin t}{\cos t}x[/math].
Setting
[math]x = 1[/math], we see that it passes through the point
[math]\Bigl(1,\frac{\sin t}{\cos t} \Bigr)[/math], as shown in Figure 6. Hence if
[math]T_{2}[/math] is the right triangle with
vertices (0,0),
[math]\Bigl(1, \frac{\sin t}{\cos t} \Bigr)[/math], and (1, 0), then
[[math]]
area(T_{2}) = \frac{1}{2} \cdot 1 \cdot \frac{\sin t}{\cos t} = \frac{1}{2} \frac{\sin t}{\cos t}.
[[/math]]
Since
[math]T_{1}[/math] is a subset of
[math]S[/math] and since
[math]S[/math] is a subset of
[math]T_{2}[/math], it follows by a fundamental property of area [see (1.3), page 171] that
[[math]]
area(T_{1}) \leq area(S) \leq area(T_{2}).
[[/math]]
Hence
[[math]]
\frac{1}{2} \cos t \sin t \leq \frac{t}{2} \leq \frac{1}{2} \frac{\sin t}{\cos t} .
[[/math]]
If we multiply through by
[math]\frac{2}{\sin t}[/math], we get
[[math]]
\cos t \leq \frac{t}{\sin t} \leq \frac{ 1}{\cos t}.
[[/math]]
Taking reciprocals and reversing the direction of the inequalities, we obtain finally
[[math]]
\begin{equation}
\frac{1}{\cos t} \geq \frac{\sin t}{t} \geq \cos t.
\label{eq6.2.2}
\end{equation}
[[/math]]
With these inequalities, the proof of (1) is essentially finished. Since the function
[math]\cos[/math] is continuous, we have
[math]\lim_{t \rightarrow 0+} \cos t = \cos 0 = 1[/math]. Moreover, the limit of a quotient is the quotient of the limits, and so
[math]\lim_{t \rightarrow 0+} \frac{1}{\cos t} = \frac{1}{1} = 1[/math].
Thus
[math]\frac{\sin t}{t}[/math] lies between two quantities both of which approach 1 as
[math]t[/math] approaches zero from the right. It follows that
[[math]]
\lim_{t \rightarrow 0+} \frac{\sin t}{t} = 1.
[[/math]]
It is now a simple matter to remove the restriction [math]t \gt 0[/math]. Since [math]\frac{\sin t}{t} = \frac{- \sin t}{-t} = \frac{\sin(-t)}{-t}[/math], we know that
[[math]]
\begin{equation}
\frac{\sin t}{t} = \frac{\sin |t|}{|t|}.
\label{eq6.2.3}
\end{equation}
[[/math]]
As
[math]t[/math] approaches zero, so does
[math]|t|[/math]; and as
[math]|t|[/math] approaches zero, we have just proved that the right side of (3) approaches 1. The left side, therefore, also a pproaches 1, and so the proof is complete.
■
It is interesting to compare actual numerical values of [math]t[/math] and [math]\sin t[/math].
Table 1 illustrates the limit theorem (2.1) quite effectively.
| [math]t[/math] |
[math]\sin t[/math]
|
| 0.50 |
0.4794
|
| 0.40 |
0.3894
|
| 0.30 |
0.2955
|
| 0.20 |
0.1987
|
| 0.10 |
0.0998
|
| 0.08 |
0.0799
|
| 0.06 |
0.0600
|
| 0.04 |
0.0400
|
| 0.02 |
0.0200
|
A useful corollary of (2.1) is
[[math]]
\lim_{t \rightarrow 0} \frac{1 - \cos t}{t} = 0.
[[/math]]
Show Proof
Using trigonometric identities, we write [math]\frac{1 - \cos t}{t}[/math] in such a form that (2.1) is applicable.
[[math]]
\begin{array}{rcll}
1 &=& & \cos^{2} \frac{t}{2} + \sin^{2}\frac{t}{ 2},\\
\cos t &=& \cos (\frac{t}{2} + \frac{t}{2}) =& \cos^{2} \frac{t}{2} - \sin^{2} \frac{t}{2}.
\end{array}
[[/math]]
Hence
[math]1 - \cos t = 2 \sin^{2} \frac{t}{2}[/math], and
[[math]]
\frac{1 - \cos t}{t} = \frac{t}{2} \sin^{2} \frac{t}{2}
= \Bigl(\frac{\sin \frac{t}{2}}{\frac{t}{2}} \Bigr) \sin \frac{t}{2}.
[[/math]]
As
[math]t[/math] approaches zero,
[math]\frac{t}{2}[/math] also approaches zero, so, by (2.1), the quantity
[[math]]
\frac{\sin \frac{t}{2}}{\frac{t}{2}}
[[/math]]
approaches 1. Moreover,
[math]\sin[/math] is a continuous function, and therefore
[math]\sin \frac{t}{2}[/math] approaches
[math]\sin 0 = 0[/math]. The product therefore approaches
[math]1 \cdot 0 = 0[/math], and the proof is complete.
■
In writing values of the functions [math]\sin[/math] and [math]\cos[/math], we have thus far avoided the letter [math]x[/math] and have not written [math]\sin x[/math] and [math]\cos x[/math] simply because the point on the circle [math]x^{2} + y^{2} = 1[/math] whose coordinates define the value of [math]\cos[/math] and [math]\sin[/math] has nothing to do with, and generally does not lie on, the [math]x[/math]-axis. However, when we study [math]\sin[/math] and [math]\cos[/math] as two real-valued functions of a real variable, it is natural to use [math]x[/math] as the independent variable.
We shall not hesitate to do so from now on.
Example
Evaluate the limits
[[math]]
\mbox{(a)}\;\;\; \lim_{x \rightarrow 0} \frac{\sin 3x}{\sin 7x},\;\;\;
\mbox{(b)}\;\;\; \lim_{x \rightarrow 0} \frac{1 - \cos^{2} x}{x}, \;\;\;
\mbox{(c)}\;\;\; \lim_{x \rightarrow 0} \frac{\cos x}{\sin x}.
[[/math]]
We evaluate the first two limits by writing the quotients in such a form that the fundamental trigonometric limit theorem, [math]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/math], is applicable. For (a),
[[math]]
\frac{\sin 3x}{\sin 7x} = \frac{\sin 3x}{3x} \frac{7x}{\sin 7x} \frac{3}{7}.
[[/math]]
As [math]x[/math] approaches zero, so does [math]3x[/math] and so does [math]7x[/math]. Hence [math]\frac{\sin 3x}{3x}[/math] approaches 1, and [math]\frac{7x}{\sin 7x} = \Bigl(\frac{\sin 7x}{7x} \Bigr)^{-1}[/math] approaches [math]1^{-1} = 1[/math].
We conclude that
[[math]]
\lim_{x \rightarrow 0} \frac{\sin 3x}{\sin 7x} = 1 \cdot 1 \cdot \frac{3}{7} = \frac{3}{7}.
[[/math]]
To do (b), we use the identity [math]\cos^{2} x + \sin^{2} x = 1[/math]. Thus
[[math]]
\frac{1 - \cos^{2}x}{x} = \frac{\sin^{2} x}{x} = \sin x \frac{\sin x}{x}.
[[/math]]
As [math]x[/math] approaches zero, [math]\sin x[/math] approaches [math]\sin 0 = 0[/math], and [math]\frac{\sin x}{x}[/math] approaches 1. Hence
[[math]]
\lim_{x \rightarrow 0} \frac{1 - \cos^{2}x}{x} = 0 \cdot 1 = 0.
[[/math]]
For (c), no limit exists. The numerator approaches 1, and the denominator approaches zero. Note that we cannot even write the limit as [math]+\infty[/math] or [math]-\infty[/math] because [math]\sin x[/math] may be either positive or negative. As a result, [math]\frac{\cos x}{\sin x}[/math] takes on both arbitrarily large positive values and arbitrarily large negative values as [math]x[/math] approaches zero.
We are now ready to find [math]\frac{d}{dx} \sin x[/math]. The value of the derivative at an arbitrary number [math]a[/math] is by definition
[[math]]
\Bigl(\frac{d}{dx} \sin x \Bigr) (a) = \lim_{t \rightarrow 0} \frac{\sin (a + t) - \sin a}{t}.
[[/math]]
As always, the game is to manipulate the quotient into a form in which we can see what the limit is. Since [math]\sin(a + t) = \sin a \cos t + \cos a \sin t[/math], we have
[[math]]
\begin{eqnarray*}
\frac{\sin(a + t) - \sin a}{t} &=& \frac{\sin a \cos t + \cos a \sin t - \sin a}{t}\\
&=& \cos a \frac{\sin t}{t} - \sin a \frac{1 - \cos t}{t}.
\end{eqnarray*}
[[/math]]
As [math]t[/math] approaches 0, the quantities [math]\cos a[/math] and [math]\sin a[/math] stay fixed. Moreover, [math]\frac{\sin t}{t}[/math] approaches 1, and [math]\frac{1 - \cos t}{t}[/math] approaches 0. Hence, the right side of the above equation approaches
[math](\cos a) \cdot 1 - (\sin a) \cdot 0 = \cos a[/math]. We conclude that
[[math]]
\Bigl (\frac{d}{dx} \sin x \Bigr) (a) = \cos a, \;\;\;\mbox{for every real number}\; a.
[[/math]]
Writing this result as an equality between functions, we get the simpler form
[[math]]
\frac{d}{dx} \sin x= \cos x.
[[/math]]
The derivative of the cosine may be found from the derivative of the sine using the Chain Rule
and the twin identities [math]\cos x = \sin \Bigl(\frac{\pi}{2} - x \Bigr)[/math] and [math]\sin x = \cos \Bigl(\frac{\pi}{2} - x \Bigr)[/math] [see (1 6), page 286].
[[math]]
\begin{eqnarray*}
\frac{d}{dx} \cos x = \frac{d}{dx} \sin \Bigl(\frac{\pi}{2} - x \Bigr)
&=& \cos \Bigl(\frac{\pi}{2} - x \Bigr) \frac{d}{dx} \Bigl(\frac{\pi}{2} - x \Bigr) \\
&=& \cos \Bigl(\frac{\pi}{2} - x \Bigr) (-1) = - \sin x.
\end{eqnarray*}
[[/math]]
Writing this result in a single equation, we have
[[math]]
\frac{d}{dx} \cos x = - \sin x.
[[/math]]
Example
Find the following derivatives.
[[math]]
\begin{array}{ll}
\mbox{(a)}\;\;\; \frac{d}{dx} \sin(x^{2} + 1), &\;\;\; \mbox{(c)}\;\;\; \frac{d}{dt} \sin e^{t}, \\
\mbox{(b)}\;\;\; \frac{d}{dx} \cos 7x, &\;\;\; \mbox{(d)}\;\;\; \frac{d}{dx} \ln (\cos x)^2.
\end{array}
[[/math]]
These are routine exercises which combine the basic derivatives with the Chain Rule.
For (a) we have
[[math]]
\frac{d}{dt} \sin(x^2 + 1 ) = \cos(x^2 + 1 ) \frac{d}{dx} (x^{2} + 1 ) = 2x \cos(x^{2} + 1 ).
[[/math]]
The solution to (b) is
[[math]]
\frac{d}{dx} \cos 7x = - \sin 7x \frac{d}{dx} 7x = - 7 \sin 7x.
[[/math]]
For (c),
[[math]]
\frac{d}{dt} \sin e^{t} = \cos e^{t} \frac{d}{dt} e^{t} = e^{t} \cos e^{t},
[[/math]]
and for (d),
[[math]]
\begin{eqnarray*}
\frac{d}{dx} \ln (\cos x)^2 &=& \frac{1}{(\cos x)^{2}} \frac{d}{dx} (\cos x)^2 \\
&=& \frac{1}{(\cos x)^{2}} 2 \cos x \frac{d}{dx} \cos x \\
&=& \frac{-2\cos x \sin x}{(\cos x)^{2}} = -\frac{2 \sin x}{\cos x}.
\end{eqnarray*}
[[/math]]
Every derivative formula has its corresponding integral formula. For the trigonometric
functions [math]\sin[/math] and [math]\cos[/math], they are
[[math]]
\begin{eqnarray*}
\int \sin x dx &=& -\cos x + c, \\
\int \cos x dx &=& \sin x + c.
\end{eqnarray*}
[[/math]]
The proofs consist of simply verifying that the derivative of the proposed integral is the integrand. For example,
[[math]]
\frac{d}{dx} (-\cos x + c) = - \frac{d}{dx} \cos x = \sin x.
[[/math]]
Example
Find the following integrals.
[[math]]
(a)\; \int \sin 8x dx, \;\;\;(b)\; \int x \cos(x^2) dx, \;\;\;(c)\; \int \cos^{5}x \sin x dx.
[[/math]]
The solutions use only the basic integral formulas and the fact that if [math]F' = f[/math], then [math]\int f(u) \frac{du}{dx} = F(u) + c[/math]. Integral (a) is simple enough to write down at a glance:
[[math]]
\int \sin 8x dx = - \frac{1}{8} \cos 8x + c.
[[/math]]
To do (b), let [math]u = x^2[/math]. Then [math]\frac{du}{dx} = 2x[/math], and
[[math]]
\begin{eqnarray*}
\int x \cos(x^2) dx &=& \frac{1}{2}(\cos(x^2))2x dx \\
&=& \frac{1}{2} \int (\cos u) \frac{du}{dx}dx \\
&=& \frac{1}{2} \sin u + c \\
&=& 2 \sin (x^2) + c.
\end{eqnarray*}
[[/math]]
For (c), we let [math]u = \cos x[/math]. Then [math]\frac{du}{dx} = -\sin x[/math]. Hence
[[math]]
\begin{eqnarray*}
\int \cos^{5} x \sin x dx &=& - \int \cos^{5} x (- \sin x) dx \\
&=& - \int u^{5} \frac{du}{dx} dx \\
&=& - \frac{1}{6} u^{6} + c \\
&=& - \frac{1}{6} \cos^{6} x + c.
\end{eqnarray*}
[[/math]]
The graphs of the functions [math]\sin[/math] and [math]\cos[/math] are extremely interesting and important curves.
To begin with, let us consider the graph of [math]\sin x[/math] only for [math]0 \leq x \leq \frac{\pi}{2}[/math]. A few isolated points can be plotted immediately (see Table 2).
| [math]x[/math] |
[math]y = \sin x[/math]
|
| 0 |
0
|
| [math]\frac{\pi}{6}[/math] |
[math]\frac{1}{2}[/math]
|
| [math]\frac{\pi}{4}[/math] |
[math]\frac{1}{2} \sqrt 2[/math] = 0.71 (approximately)
|
| [math]\frac{\pi}{3}[/math] |
[math]\frac{1}{2} \sqrt 3[/math] = 0.87 (approximately)
|
| [math]\frac{\pi}{2}[/math] |
1
|
The slope of the graph is given by the derivative, [math]\frac{d}{dx} \sin x = \cos x[/math].
At the origin it is [math]\cos 0 = 1[/math], and, where [math]x = \frac{\pi}{2}[/math] the slope is [math]\cos \frac{\pi}{2} = 0[/math].
Since
[[math]]
\frac{d}{dx} \sin x = \cos x \gt 0 \;\;\;\mbox{if}\; 0 \lt x \lt \frac{\pi}{2},
[[/math]]
we know that [math]\sin x[/math] is a strictly increasing function on the open interval [math]\Bigl(0, \frac{\pi}{2} \Bigr)[/math]. In addition, there are no points of inflection on the open interval and the curve is concave downward there because
[[math]]
\frac{d^2}{dx^2} \sin x = \frac{d}{dx} \cos x = -\sin x \lt 0 \;\;\; \mbox{if}\; 0 \lt x \lt \frac{\pi}{2}.
[[/math]]
On the other hand, the second derivative changes sign at [math]x = 0[/math], and so there is a point of inflection at the origin. With all these facts we can draw quite an accurate graph. It is shown in Figure 7.
It is now a simple matter to fill in as much of the rest of the graph of [math]\sin x[/math] as we like. For every
real number [math]x[/math], the points [math]x[/math] and [math]\pi - x[/math] on the real number line are symmetrically located about the point [math]\frac{\pi}{2}[/math]. The midpoint between [math]x[/math] and [math]\pi - x[/math] is given by [math]\frac{x + (\pi - x)}{2} = \frac{\pi}{2}[/math]. As [math]x[/math] increases from 0 to [math]\frac{\pi}{2}[/math] the number [math]\pi - x[/math] decreases from [math]\pi[/math] to [math]\frac{\pi}{2}[/math]. Moreover,
[[math]]
\begin{eqnarray*}
\sin(\pi - x) &=& \sin \pi \cos x - \cos \pi \sin x \\
&=& 0 \cdot \cos x - (-1) \cdot \sin x \\
&=& \sin x.
\end{eqnarray*}
[[/math]]
It follows that the graph of [math]\sin x[/math] on the interval [math]\Bigl[\frac{\pi}{2}, \pi \Bigr][/math] is the mirror image of the graph on [math]\Bigl[0, \frac{\pi}{2} \Bigr][/math] reflected across the line [math]x = \frac{\pi}{2}[/math] . This is the dashed curve in Figure 7. Now, because [math]\sin x[/math] is an odd function, its graph for [math]x \leq 0[/math] is obtained by reflecting the graph for [math]x \geq 0[/math] about the origin (i.e., reflecting first about one coordinate axis and then the other). This gives us the graph for [math]-\pi \leq x \leq \pi[/math]. Finally, since [math]\sin x[/math] is a periodic function with period [math]2\pi[/math], its values repeat after intervals of length [math]2\pi[/math]. It follows that the entire graph of [math]\sin x[/math] is the infinite wave, part of which is shown in Figure 8.
The graph of [math]\cos x[/math] is obtained by translating (sliding) the graph of [math]\sin x[/math] to the left a distance [math]\frac{\pi}{2}[/math]. This geometric assertion is equivalent to the algebraic equation [math]\cos x = \sin \Bigl(x + \frac{\pi}{2} \Bigr).[/math] But this follows from the trigonometric identity
[[math]]
\begin{eqnarray*}
\sin \Bigl(x + \frac{\pi}{2} \Bigr) &=& \sin x \cos \frac{\pi}{2} + \cos x \sin \frac{\pi}{2}\\
&=& (\sin x) \cdot 0 + (\cos x) \cdot 1\\
&=& \cos x.
\end{eqnarray*}
[[/math]]
The graphs of [math]\cos x[/math] and [math]\sin x[/math] are shown together in Figure 9.
General references
Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.
