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Topos Uranos

Summary

This lesson studies the sine, the cosine and the tangent as functions of a real variable, measured in radians. It determines their domains, their ranges, their zeros, their extreme values and their parity, and proves that 2π2\pi is the smallest positive period of the sine and the cosine, and π\pi that of the tangent. It constructs the graph of the sine by unrolling the unit circle, obtains that of the cosine by a translation and describes the vertical asymptotes of the tangent with sign charts; from there it deduces the graphs of the secant, the cosecant and the cotangent. It proves, by comparing areas, the inequality sin⁡x<x<tan⁡x\sin x < x < \tan x for 0<x<π20 < x < \frac{\pi}{2}, and studies the sinusoids y=asin⁡(b(x−c))+dy = a \sin(b(x - c)) + d, with which periodic phenomena such as the length of the day, the tides or the motion of a Ferris wheel are modeled.

Learning objectives

  1. Define the sine, the cosine and the tangent as functions of a real variable, and prove their ranges, their zeros, their extrema, their parity and their fundamental periods.
  2. Construct the graphs of the six trigonometric functions from the unit circle, describing the vertical asymptotes by means of sign charts.
  3. Prove the inequality sin⁡x<x<tan⁡x\sin x < x < \tan x for 0<x<π20 < x < \frac{\pi}{2} and the bound ∣sin⁡x∣≤∣x∣\abs{\sin x} \leq \abs{x}, and use them to study functions and equations.
  4. Interpret the amplitude, the period, the phase shift and the vertical shift of a sinusoid as transformations of the graph of the sine, and fit sinusoids to periodic phenomena.

The sine and the cosine as functions

In the lesson on trigonometric ratios the sine and the cosine of an arbitrary angle were defined as the coordinates of the point P(θ)P(\theta) at which the terminal side of the angle, in standard position, meets the unit circle. Now, according to the lesson on angles and their measure, every real number is the radian measure of some oriented angle, and two angles have the same terminal side exactly when their measures differ by an integer multiple of 2π2\pi. If this is so, the sine and the cosine can be regarded as functions defined on all of R\mathbb{R}: to each real xx is assigned a coordinate of the point P(x)P(x).

For each real number xx, let P(x)P(x) be the point of the unit circle corresponding to the angle of xx radians in standard position. The sine function assigns to xx the ordinate of P(x)P(x), and the cosine function, its abscissa. Both have as their domain the set of all real numbers.
P(x)=(cos⁡x,sin⁡x),x∈RP(x) = (\cos x, \sin x), \quad x \in \mathbb{R}

The variable is always measured in radians, and not in degrees. The choice is not an indifferent convention: the results of the second section, such as sin⁡x<x\sin x < x for positive xx, compare the sine with the variable itself, and that comparison makes sense only when xx is the length of the arc traversed on the unit circle, which is what a radian measures.

TheoremProperties of the sine and the cosine

The range of the sine and that of the cosine is the interval [−1,1]\icc{-1}{1}. For every real xx we have sin⁡(x+2π)=sin⁡x\sin(x + 2\pi) = \sin x and cos⁡(x+2π)=cos⁡x\cos(x + 2\pi) = \cos x; the sine is an odd function, and the cosine, an even function. The sine vanishes exactly at the integer multiples of π\pi, equals 11 exactly at the numbers π2+2kπ\frac{\pi}{2} + 2k\pi and equals −1-1 exactly at the numbers −π2+2kπ-\frac{\pi}{2} + 2k\pi, with kk an integer. The cosine vanishes exactly at the numbers π2+kπ\frac{\pi}{2} + k\pi, equals 11 exactly at the numbers 2kπ2k\pi and equals −1-1 exactly at the numbers π+2kπ\pi + 2k\pi.

Everything is read on the unit circle, with the Pythagorean identity and the symmetries proved in the lesson on trigonometric ratios.

Proof

  1. sin⁡2x=1−cos⁡2x≤1→−1≤sin⁡x≤1\sin^2 x = \resaltar{1 - \cos^2 x} \leq 1 \rightarrow \dato{c}{\resaltar{-1 \leq \sin x \leq 1}}

    By the Pythagorean identity, the square of the sine is at most 11, and the sine lies between −1-1 and 11; the same holds for the cosine.

  2. Q=(1−c2,c),(1−c2)2+c2=1→ran⁡(sin⁡)=ran⁡(cos⁡)=[−1,1]Q = \left( \sqrt{1 - c^2}, c \right), \quad \resaltar{\left( \sqrt{1 - c^2} \right)^2 + c^2} = \resaltar{1} \rightarrow \operatorname{ran}(\sin) = \operatorname{ran}(\cos) = \icc{-1}{1}

    Conversely, given cc with −1≤c≤1-1 \leq c \leq 1, the following point QQ lies on the unit circle; by the covering property of the traversal axiom, admitted in the lesson on angles and their measure, it is the point P(x)P(x) of some angle xx, and then sin⁡x=c\sin x = c. In the same way, the point (c,1−c2)\left( c, \sqrt{1 - c^2} \right) shows that the cosine takes the value cc.

  3. sin⁡(x+2π)=sin⁡x,cos⁡(x+2π)=cos⁡x,sin⁡(−x)=−sin⁡x,cos⁡(−x)=cos⁡x\sin(x + 2\pi) = \sin x, \quad \cos(x + 2\pi) = \cos x, \quad \sin(-x) = -\sin x, \quad \cos(-x) = \cos x

    The angles of xx and of x+2πx + 2\pi radians have the same terminal side, so that P(x+2π)=P(x)P(x + 2\pi) = P(x). Moreover, P(−x)P(-x) is the reflection of P(x)P(x) in the horizontal axis, as was proved in the lesson on trigonometric ratios: the sine is odd and the cosine is even.

  4. sin⁡x=0↔x=kπ,cos⁡x=0↔x=π2+kπ,k∈Z\sin x = 0 \leftrightarrow x = k\pi, \quad \cos x = 0 \leftrightarrow x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}

    The sine vanishes when P(x)P(x) lies on the horizontal axis, that is, when it is (1,0)(1, 0) or (−1,0)(-1, 0); these are the points of the angles 00 and π\pi and of their coterminal angles. The cosine vanishes when P(x)P(x) is (0,1)(0, 1) or (0,−1)(0, -1).

  5. sin⁡x=1→cos⁡2x=1−1=0→P(x)=(0,1)\sin x = 1 \rightarrow \cos^2 x = \resaltar{1 - 1} = \resaltar{0} \rightarrow P(x) = (0, 1)

    If sin⁡x=1\sin x = 1, the Pythagorean identity forces the cosine to be zero, and the point P(x)P(x) is (0,1)(0, 1), which corresponds to the angle π2\frac{\pi}{2}; if sin⁡x=−1\sin x = -1, the point is (0,−1)(0, -1), which corresponds to −π2-\frac{\pi}{2}.

  6. sin⁡x=1↔x=π2+2kπ,sin⁡x=−1↔x=−π2+2kπ,cos⁡x=1↔x=2kπ,cos⁡x=−1↔x=π+2kπ\sin x = 1 \leftrightarrow x = \frac{\pi}{2} + 2k\pi, \quad \sin x = -1 \leftrightarrow x = -\frac{\pi}{2} + 2k\pi, \quad \cos x = 1 \leftrightarrow x = 2k\pi, \quad \cos x = -1 \leftrightarrow x = \pi + 2k\pi

    By the same reasoning, the cosine equals 11 when P(x)=(1,0)P(x) = (1, 0) and equals −1-1 when P(x)=(−1,0)P(x) = (-1, 0). In summary, the extreme values of the sine and the cosine are attained at the following numbers, with kk an integer.

The graph of the sine

The graph of y=sin⁡xy = \sin x is obtained by unrolling the unit circle along the horizontal axis: while the point P(x)P(x) travels around the circle, the arc traversed, of length xx, is carried to the axis of abscissas, and above each abscissa the height of the point is raised. The symmetries from the lesson on trigonometric ratios make it possible to construct the whole graph from its first quarter.

ExampleThe graph of the sine as an unrolled circle

The figure unrolls the unit circle along the horizontal axis, with the same scale on both: the height of the point P(x)P(x) is transferred to the abscissa xx, and the complete graph results from repeating and reflecting the first quarter.

Proof

  1. sin⁡0=0,sin⁡π2=1\sin 0 = 0, \quad \sin \frac{\pi}{2} = 1

    As xx goes from 00 to π2\frac{\pi}{2}, the point P(x)P(x) travels through the first quadrant from (1,0)(1, 0) to (0,1)(0, 1). By the half-turn property of the traversal axiom, its abscissa cos⁡x\cos x decreases strictly from 11 to 00; since its height is the nonnegative root 1−cos⁡2x\sqrt{1 - \cos^2 x}, it rises strictly from 00 to 11. In the lesson on the inverse trigonometric functions the same argument is extended to the whole interval [−π2,π2]\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}.

  2. sin⁡π6=12,sin⁡π4=22,sin⁡π3=32\sin \frac{\pi}{6} = \frac{1}{2}, \quad \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}

    The special values computed in the lesson on trigonometric ratios give three intermediate points of the first arch.

  3. sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x

    Between π2\frac{\pi}{2} and π\pi, the symmetry of supplementary angles says that the height at π−x\pi - x is the same as at xx: the first arch is symmetric about the vertical line x=π2x = \frac{\pi}{2}, and comes down to 00 at π\pi.

  4. sin⁡(x+π)=−sin⁡x\sin(x + \pi) = -\sin x

    Between π\pi and 2π2\pi, the point P(x+π)P(x + \pi) is the opposite of P(x)P(x): the height changes sign, and the graph repeats the preceding arch below the axis, translated by π\pi and reflected.

  5. sin⁡(x+2π)=sin⁡x,sin⁡(−x)=−sin⁡x\sin(x + 2\pi) = \sin x, \quad \sin(-x) = -\sin x

    By periodicity, the portion between 00 and 2π2\pi, called a cycle, repeats to the right and to the left; and since the sine is odd, the graph is symmetric about the origin.

The graph of the sine is a wave that oscillates between the lines y=−1y = -1 and y=1y = 1, crosses the horizontal axis at the multiples of π\pi and attains its maxima at the numbers π2+2kπ\frac{\pi}{2} + 2k\pi and its minima at the numbers −π2+2kπ-\frac{\pi}{2} + 2k\pi, in accordance with the theorem above.

The drawing assumes, moreover, two properties that the unrolled circle suggests, but that the tools of this unit cannot prove: that the graph is a curve without breaks, and that each arch lying above the axis always bends downward. Both are defined by means of limits and are proved in the course Calculus I; here they are admitted. No theorem of the unit depends on them: they are used only in some problems, which say so explicitly.

Admitted fact (continuity of the trigonometric functions and of their inverses). The sine and the cosine are continuous on the whole line; the tangent, the secant, the cosecant and the cotangent are continuous on each interval contained in their domain; and the arcsine, the arccosine and the arctangent, which will be defined in the lesson on the inverse trigonometric functions, are continuous on their domains. Sums, products, quotients with nonzero denominator and compositions of continuous functions are continuous. Consequently, the admitted fact of Course I on the intermediate value theorem, stated in the lesson on rational functions, applies, on every closed interval contained in their domain, to functions such as sin⁡x−x10\sin x - \frac{x}{10}, tan⁡x−x\tan x - x or arcsin⁡x+arctan⁡x\arcsin x + \arctan x: if they change sign at the endpoints, they vanish at some interior point.
Admitted fact (concavity of the sine). The sine is strictly concave on each interval [2kπ,(2k+1)π]\icc{2k\pi}{(2k + 1)\pi}, with kk an integer: if u<x<vu < x < v are points of that interval, the point of the graph with abscissa xx lies strictly above the line segment joining the points of the graph with abscissas uu and vv.
u<x<v→sin⁡x>(v−x)sin⁡u+(x−u)sin⁡vv−uu < x < v \rightarrow \sin x > \frac{(v - x) \sin u + (x - u) \sin v}{v - u}

Periodic functions

A function ff is periodic if there is a number p>0p > 0 such that, for every xx in the domain, x+px + p and x−px - p are in the domain and f(x+p)=f(x)f(x + p) = f(x) holds; such a number is called a period of ff. If pp is a period, so are 2p2p, 3p3p and, by induction, all its natural multiples: f(x+2p)=f((x+p)+p)=f(x+p)=f(x)f(x + 2p) = f((x + p) + p) = f(x + p) = f(x). A periodic function therefore has infinitely many periods, and the most interesting one is the smallest of them, when it exists, which is called the fundamental period. Its graph is obtained by repeating over and over the portion corresponding to an interval of length equal to the period.

LemmaThe periods are multiples of the fundamental period

If qq is the fundamental period of a function ff, the periods of ff are exactly the numbers kqkq, with kk natural.

An arbitrary period is divided by qq, as in division with remainder, and it is shown that the remainder cannot be positive. The figure uses a function with fundamental period qq.

Proof

  1. f(x+kq)=f(x),k∈N\dato{m}{f(x + kq) = f(x)}, \quad k \in \mathbb{N}

    Each natural multiple of qq is a period, as has just been seen by induction.

  2. kq≤p<(k+1)q→0≤r=p−kq<q\resaltar{kq \leq p < (k + 1)q} \rightarrow \dato{r}{\resaltar{0 \leq r = p - kq < q}}

    Conversely, let pp be a period; by the minimality of qq, we have p≥qp \geq q. By the Archimedean property of the reals, proved in the Course I lesson on the decimal expansion of rational numbers, and the well-ordering principle, as in the theorem of reduction to the first turn of the lesson on angles and their measure, there is a natural number k≥1k \geq 1 with kq≤p<(k+1)qkq \leq p < (k + 1)q, and the remainder r=p−kqr = p - kq lies between 00 and qq, the latter excluded.

  3. f(x+r)=f(x+r+kq)=f(x+p)=f(x)f(x + r) = f(x + \resaltar{r + kq}) = f(x + \resaltar{p}) = f(x)

    If rr were positive, it would be a period: advancing by rr and then by kqkq amounts to advancing by pp, and both kqkq and pp are periods.

  4. r=0→p=kqr = 0 \rightarrow p = kq

    But a positive period smaller than qq contradicts the minimality of qq. Hence the remainder is zero, and pp is a natural multiple of qq.

TheoremFundamental period of the sine and the cosine

The number 2π2\pi is the fundamental period of the sine and the cosine.

By the properties theorem, 2π2\pi is a period of both functions. It remains to see that no number between 00 and 2π2\pi is one, and for that it suffices to look at where the maximum is attained.

Proof

  1. sin⁡(π2+p)=sin⁡π2=1\dato{m}{\sin \left( \frac{\pi}{2} + p \right) = \sin \frac{\pi}{2} = 1}

    Suppose that pp, with 0<p<2π0 < p < 2\pi, is a period of the sine. Evaluating the equality sin⁡(x+p)=sin⁡x\sin(x + p) = \sin x at x=π2x = \frac{\pi}{2}, the number π2+p\frac{\pi}{2} + p would also be a maximum point.

  2. π2+p=π2+2kπ→p=2kπ\frac{\pi}{2} + p = \frac{\pi}{2} + 2k\pi \rightarrow p = 2k\pi

    By the properties theorem, the sine equals 11 only at the numbers π2+2kπ\frac{\pi}{2} + 2k\pi: then pp would be an integer multiple of 2π2\pi, which is impossible between 00 and 2π2\pi. In other words, two consecutive maxima of the sine are exactly 2π2\pi apart.

  3. cos⁡p=cos⁡0=1→p=2kπ\cos p = \cos 0 = 1 \rightarrow p = 2k\pi

    For the cosine one argues in the same way at x=0x = 0: a period pp would give cos⁡p=cos⁡0=1\cos p = \cos 0 = 1, and the cosine equals 11 only at the integer multiples of 2π2\pi.

By the lemma above, the periods of the sine and the cosine are exactly the numbers 2kπ2k\pi with kk natural.

The cosine as a shifted sine

The graph of the cosine does not require a new construction, because it is that of the sine shifted.

TheoremThe cosine is a shifted sine

For every real xx we have cos⁡x=sin⁡(x+π2)\cos x = \sin \left( x + \frac{\pi}{2} \right). Consequently, the graph of y=cos⁡xy = \cos x is that of the sine translated π2\frac{\pi}{2} units to the left.

The symmetries of complementary and supplementary angles are combined, and then the transformations theorem of the Course I lesson on function operations.

Proof

  1. cos⁡x=sin⁡(π2−x)\cos x = \dato{c}{\sin \left( \frac{\pi}{2} - x \right)}

    By the symmetry of complementary angles, the cosine of xx is the sine of π2−x\frac{\pi}{2} - x.

  2. cos⁡x=sin⁡(π−(π2−x))=sin⁡(x+π2)\cos x = \sin \left( \resaltar{\pi - \left( \frac{\pi}{2} - x \right)} \right) = \dato{t}{\sin \left( \resaltar{x + \frac{\pi}{2}} \right)}

    By the symmetry of supplementary angles, sin⁡(π−u)=sin⁡u\sin(\pi - u) = \sin u; with u=π2−xu = \frac{\pi}{2} - x, the angle π−u\pi - u is π2+x\frac{\pi}{2} + x.

  3. y=sin⁡(x−(−π2))y = \sin \left( x - \left( -\frac{\pi}{2} \right) \right)

    By the transformations theorem, the graph of y=f(x−h)y = f(x - h) is that of ff translated hh units horizontally. Here ff is the sine and h=−π2h = -\frac{\pi}{2}: each point of the graph of the sine moves π2\frac{\pi}{2} to the left. The maximum point (π2,1)\left( \frac{\pi}{2}, 1 \right) goes to (0,1)(0, 1).

  4. cos⁡(−x)=cos⁡x,cos⁡0=1,cos⁡π=−1\cos(-x) = \cos x, \quad \cos 0 = 1, \quad \cos \pi = -1

    On the translated graph the properties of the cosine can be read: it is symmetric about the vertical axis, because the cosine is even, it crosses the horizontal axis at the numbers π2+kπ\frac{\pi}{2} + k\pi, and it attains the maximum 11 at the even multiples of π\pi and the minimum −1-1 at the odd ones.