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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).

Chapter 10 · pp. 405–462

Periodic Functions

This chapter looks closely at functions that repeat, starting with sine and cosine as sinusoidal graphs and then moving to the other trig functions, their inverses, and trig equations. It ends with modeling real situations such as Ferris wheels and pistons. This first part covers periodic functions, the sine and cosine graphs, their symmetries, and how to transform them into models of motion.

Periodic Functions and the Sine and Cosine Graphs

A function is periodic if shifting it horizontally by some fixed amount $P$ gives the same function back: $f(x+P)=f(x)$ for every $x$ . The smallest positive $P$ that works is the period. A Ferris wheel is a good picture. The London Eye makes one full turn every 30 minutes, so a rider's height repeats every 30 minutes and the period is 30 minutes. (p. 405)

A rider's height matches the $y$ -coordinate of a point moving around a circle, and sine gives exactly that coordinate on the unit circle. Cosine gives the $x$ -coordinate. To build the graphs, list the coordinates at the standard angles and plot them against the angle. At $\theta = 0, \pi/2, \pi, 3\pi/2$ , the sine values are $0, 1, 0, -1$ . Sine is positive from $0$ to $\pi$ (quadrants 1 and 2) and negative from $\pi$ to $2\pi$ (quadrants 3 and 4). Cosine can be tracked through the four quadrants in the same way. (pp. 406–407)

Domain and range. You can find a point on the unit circle for any angle, so sine and cosine are defined for all real numbers: the domain is $(-\infty,\infty)$ . A coordinate on the unit circle never goes above 1 or below $-1$ , so the range is $[-1,1]$ . (p. 407)

Period. Coterminal angles land on the same point, so $\sin(\theta+2\pi)=\sin\theta$ and $\cos(\theta+2\pi)=\cos\theta$ . Both functions have period $2\pi$ . Shifting either graph $2\pi$ units left or right gives the identical graph. Graphs with this wave shape are called sinusoidal. (p. 407)

Symmetry and the Link Between Sine and Cosine

The sine graph is symmetric about the origin, as a cubic is, so sine is odd. The cosine graph is symmetric about the $y$ -axis, as a parabola is, so cosine is even. This gives the negative angle identities:

$\sin(-\theta)=-\sin\theta,\qquad \cos(-\theta)=\cos\theta$

For example, $\sin(-\pi/6)=-\tfrac12$ , while $\cos(-\pi/6)=\cos(\pi/6)$ . (p. 408)

The cofunction identity says $\sin\theta=\cos(\pi/2-\theta)$ . Cosine is even, so $\cos(\pi/2-\theta)=\cos(\theta-\pi/2)$ . Therefore

$\sin\theta=\cos\!\left(\theta-\tfrac{\pi}{2}\right),$

which means the sine graph is the cosine graph shifted right by $\pi/2$ . (p. 408)

Common mistake: the $\pi/2$ shift is not the period. It only shows that each graph is a translation of the other. Both functions still have period $2\pi$ . (p. 408)

The negative angle identities also help with simplifying. To simplify $\dfrac{\sin(-\theta)}{\tan(-\theta)}$ :

Rewrite the numerator as $-\sin\theta$ .
Rewrite the denominator as $-\dfrac{\sin\theta}{\cos\theta}$ .
Invert and multiply: $-\sin\theta\cdot\dfrac{\cos\theta}{-\sin\theta}$ .
Cancel to get $\cos\theta$ . (p. 409)

Amplitude and Midline

The amplitude $A$ is the vertical stretch factor. It is how far the graph rises above, or falls below, its center. A point on a circle of radius 3 has height $y=3\sin\theta$ . Its outputs run between $-3$ and $3$ , so the amplitude is 3. The stretch does not change the period. (p. 409)

The midline is the center value the graph oscillates around. It comes from a vertical shift. For $f(\theta)=A\cos\theta+k$ , the midline is the line $y=k$ . Take a circle of radius 3 ft whose center is 4 ft above the ground. The lowest point on it has height $h(\theta)=-3\cos\theta+4$ . This function starts at its minimum, so we use a reflected cosine. The 3 is the amplitude, and the midline is $y=4$ . The height runs from 1 ft to 7 ft. (p. 410)

Writing a Sinusoid from a Graph

Follow these steps:

Midline: average the maximum and minimum.
Amplitude: half the difference between the maximum and minimum.
Period: measure the distance between two matching points, such as two peaks.
Choose the function. Use cosine if the graph starts at a maximum or minimum. Use sine if it starts on the midline.
Check for a reflection. Add a negative sign if a cosine graph starts at its minimum, or if a sine graph decreases from the midline.

For example, a graph with low $-1$ and high $3$ has midline $y=1$ and amplitude 2. If its peaks are at $x=-2$ and $x=6$ , the period is 8, so $B=2\pi/8=\pi/4$ . It is on the midline at $x=0$ and then decreases, so $f(t)=-2\sin\!\left(\tfrac{\pi}{4}t\right)+1$ . (pp. 410, 414)

Sinusoids in Time

Sine and cosine take angles as inputs. To use time as the input, build an angle function $\theta(t)$ from the angular velocity and compose it with the trig function. A point makes 1 revolution ( $2\pi$ radians) every 2 minutes, so it turns $\pi$ radians per minute and $\theta(t)=\pi t$ . If it starts at $(5,0)$ on a circle of radius 5, its $x$ -coordinate is $x(t)=5\cos(\pi t)$ . The composition compresses the graph horizontally and changes the period. (p. 411)

In general, a point that takes $P$ minutes per revolution has angular velocity $2\pi/P$ , so $\theta(t)=\frac{2\pi}{P}t$ . For $f(t)=\sin(Bt)$ or $\cos(Bt)$ :

$B=\frac{2\pi}{P},\qquad P=\frac{2\pi}{B}$

Think of $B$ as the ratio of the normal period $2\pi$ to the new period. For $f(t)=\sin\!\left(\tfrac{\pi}{6}t\right)$ , $B=\pi/6$ and $P=2\pi\div\tfrac{\pi}{6}=12$ . (pp. 411–412)

Combining all the transformations gives the standard form $f(t)=A\sin(Bt)+k$ or $A\cos(Bt)+k$ :

$A$ is the amplitude. A negative $A$ reflects the graph vertically.
$B$ sets the period through $P=2\pi/B$ .
$k$ is the vertical shift and gives the midline $y=k$ .

For $f(t)=3\sin(2t)+1$ , the amplitude is 3, the period is $2\pi/2=\pi$ , and the midline is $y=1$ . (pp. 412–413)

Other Inputs: Distance Rolled

The input does not have to be time. Suppose a wheel of radius 14 in rolls $x$ inches. The distance rolled is an arclength, and $s=r\theta$ gives $\theta=x/14$ . The bottom-most point starts at height 0, so

$h(x)=-14\cos\!\left(\tfrac{x}{14}\right)+14.$

Here $B=\tfrac1{14}$ , so the period is $2\pi\div\tfrac1{14}=28\pi$ inches. This is the wheel's circumference, which makes sense: the pattern repeats after exactly one full turn. (p. 413)

Modeling a Ferris Wheel

To model a Ferris wheel, match each feature of the situation to a part of the formula. For the London Eye, the capsules sit on a circle of diameter 130 m. Riders board from a platform 5 m above the ground.

Amplitude: the radius, $130/2=65$ m.
Midline: the center height, $65+5=70$ m.
Period: 30 minutes, so $B=2\pi/30=\pi/15$ .
Shape: riders start at the lowest point, so use a negative cosine.

The model is $h(t)=-65\cos\!\left(\tfrac{\pi}{15}t\right)+70$ . (pp. 414–415)

Horizontal Shifts of Sine and Cosine

So far, a sinusoid has been stretched, flipped and moved up or down. It can also slide left or right. The general forms are $f(t)=A\sin(B(t-h))+k$ and $f(t)=A\cos(B(t-h))+k$ . Here $h$ is the horizontal shift. The inside of the function must be factored so that $B$ multiplies the whole $(t-h)$ group. Only then can you read $h$ directly (p. 416).

Example. Take $f(t)=3\sin\left(\frac{\pi}{4}t-\frac{\pi}{4}\right)$ . Factor out $\frac{\pi}{4}$ to get $3\sin\left(\frac{\pi}{4}(t-1)\right)$ . Now read off the features:

The amplitude is 3.
The period is $P=\frac{2\pi}{B}=\frac{2\pi}{\pi/4}=8$ .
The graph is shifted right by 1. It starts on the midline and rises, but it starts at $t=1$ instead of $t=0$ . (p. 416)

A word about "phase shift." Some books call $h$ the phase shift. Other books call the unfactored constant inside the function the phase shift. For the example above, that gives 1 under the first meaning and $\frac{\pi}{4}$ under the second. Because the meanings conflict, we will say "horizontal shift" and always mean the $h$ in the factored form (p. 416).

One Graph, Many Formulas

A sinusoid can be described by a sine or a cosine, positive or negative, each with its own shift. All of these versions are equivalent and give the same graph (p. 417).

Example. Suppose a graph has a highest value of 1 and a lowest value of $-5$ , and peaks at $x=1$ and $x=7$ .

Midline: halfway between 1 and $-5$ , which is $-2$ .
Amplitude: the distance from the midline to a peak, which is 3.
Period: $7-1=6$ , so $B=\frac{2\pi}{6}=\frac{\pi}{3}$ .

There are several valid ways to finish:

a cosine shifted 1 right: $3\cos\left(\frac{\pi}{3}(x-1)\right)-2$
a negative cosine shifted 2 left: $-3\cos\left(\frac{\pi}{3}(x+2)\right)-2$
a sine shifted $\frac12$ left
a negative sine shifted 2.5 right

Pick the form with the simplest shift. Here the cosine forms use whole numbers, so they are easier to work with (p. 417).

The Tangent Graph

Tangent can be pictured as the slope of a line through the origin that makes a given angle with the positive $x$ -axis. This matches the definition $\tan\theta=\frac{y}{x}=\frac{\sin\theta}{\cos\theta}$ (p. 422).

At angle 0 the line is flat, so the slope is 0.
As the angle grows toward $\frac{\pi}{2}$ , the line gets steeper and the tangent grows without bound.
At $\frac{\pi}{2}$ the line is vertical, so the tangent is undefined.
Just past $\frac{\pi}{2}$ the line has a steep negative slope. The graph jumps from very large positive to very large negative values.

Because $\tan\theta=\frac{\sin\theta}{\cos\theta}$ , tangent is zero where sine is zero: $-\pi, 0, \pi,\dots$ . It is undefined where cosine is zero: $\pm\frac{\pi}{2}, \pm\frac{3\pi}{2},\dots$ . Those points are the vertical asymptotes. Tangent is positive in quadrants 1 and 3 (p. 422).

An angle and the angle half a circle away give the same slope. So the graph repeats every $\pi$ , and one continuous cycle runs from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ (p. 422).

Features of $\tan\theta$ :

Period: $\pi$
Domain: all reals except $\frac{\pi}{2}+k\pi$ , where $k$ is an integer
Range: $(-\infty,\infty)$ (pp. 422–423)

Writing a Tangent Formula from a Graph

Horizontal and vertical stretches are separate for tangent, just as they are for sine and cosine. The period gives the horizontal stretch. A point on the graph gives the vertical stretch (p. 423).

Example. A tangent graph has one full cycle from $-4$ to $4$ , so its period is 8. Compare the normal period $\pi$ with the new period 8. The inside factor is the ratio $\frac{\pi}{8}$ , so the form is $a\tan\left(\frac{\pi}{8}x\right)$ .

Now use the point $(2,2)$ :

$2=a\tan\left(\frac{\pi}{8}\cdot 2\right)=a\tan\left(\frac{\pi}{4}\right)=a\cdot 1$

So $a=2$ and $f(x)=2\tan\left(\frac{\pi}{8}x\right)$ (p. 423).

Secant and Cosecant

Secant is the reciprocal of cosine: $\sec\theta=\frac{1}{\cos\theta}$ . It is undefined where cosine is 0, so it has vertical asymptotes at $\frac{\pi}{2}, \frac{3\pi}{2},\dots$ . Cosine never has an absolute value above 1. So its reciprocal never has an absolute value below 1, and secant never takes values between $-1$ and $1$ . The secant curves open away from the peaks and troughs of the cosine graph (p. 424).

Cosecant is the reciprocal of sine: $\csc\theta=\frac{1}{\sin\theta}$ . It is undefined where sine is 0, at $0,\pi,2\pi,\dots$ . Since $\sin\theta=\cos\left(\theta-\frac{\pi}{2}\right)$ , we get $\csc\theta=\sec\left(\theta-\frac{\pi}{2}\right)$ . So the cosecant graph is a horizontal shift of the secant graph (p. 424).

Features of secant and cosecant:

Period: $2\pi$ , the same as cosine and sine.
Range of both: $(-\infty,-1]\cup[1,\infty)$ .
Domain of secant: all reals except $\frac{\pi}{2}+k\pi$ .
Domain of cosecant: all reals except $k\pi$ . (p. 425)

Example: domain and range of $f(x)=2\csc\left(\frac{\pi}{2}x\right)+1$ .

Asymptotes. Basic cosecant has asymptotes at $\pi k$ . The inside factor $\frac{\pi}{2}$ compresses these to $\pi k\div\frac{\pi}{2}=2k$ . The domain is $x\neq 2k$ .
Range. Start with $(-\infty,-1]\cup[1,\infty)$ . The vertical stretch by 2 gives $(-\infty,-2]\cup[2,\infty)$ . The shift up 1 gives $(-\infty,-1]\cup[3,\infty)$ . (p. 425)

It helps to sketch $2\sin\left(\frac{\pi}{2}x\right)+1$ lightly. The cosecant curves touch its peaks and troughs.

Cotangent

Cotangent is cosine over sine, so it is undefined where sine is 0.

Period: $\pi$
Domain: all reals except $k\pi$
Range: all real numbers

Its graph is a horizontal flip and shift of the tangent graph (p. 426).

Negative Angle Identities for the Other Functions

Tangent is odd. This follows from the sine and cosine identities:

$\tan(-\theta)=\frac{\sin(-\theta)}{\cos(-\theta)}=\frac{-\sin\theta}{\cos\theta}=-\tan\theta$

Secant inherits the evenness of cosine, and cosecant inherits the oddness of sine. Altogether:

$\tan(-\theta)=-\tan\theta$
$\cot(-\theta)=-\cot\theta$
$\csc(-\theta)=-\csc\theta$
$\sec(-\theta)=\sec\theta$

For instance, if $\tan x=3$ , then $\tan(-x)=-3$ (pp. 426–427).

Proving an Identity

To show $\tan\theta=-\cot\left(\theta+\frac{\pi}{2}\right)$ , rewrite everything in sine and cosine and use cofunction identities (p. 427):

$\tan\theta=\frac{\sin\theta}{\cos\theta}$ .
Cofunction identities turn this into $\frac{\cos\left(\frac{\pi}{2}-\theta\right)}{\sin\left(\frac{\pi}{2}-\theta\right)}$ , which is $\cot\left(\frac{\pi}{2}-\theta\right)$ .
Factor a negative from the inside: $\cot\left(-\left(\theta-\frac{\pi}{2}\right)\right)$ .
Cotangent is odd, so this equals $-\cot\left(\theta-\frac{\pi}{2}\right)$ .
Cotangent has period $\pi$ , so $\cot\left(\theta-\frac{\pi}{2}\right)=\cot\left(\theta+\frac{\pi}{2}\right)$ . This gives the stated identity.

Why inverse trig functions need restricted domains

Sometimes we know a sine, cosine, or tangent value and need the angle that produced it. That calls for an inverse function. An inverse only exists for a one-to-one function, meaning each output comes from exactly one input. Sine, cosine, and tangent repeat their values, so none of them is one-to-one on all real numbers (p. 432).

The fix is to cut each function down to a piece where it does not repeat. Each piece includes the angle 0:

Sine on $[-\pi/2,\pi/2]$
Cosine on $[0,\pi]$
Tangent on $(-\pi/2,\pi/2)$

On these pieces every output value is reached once, so the inverses are well defined (p. 432).

The inverse functions swap inputs and outputs, so the restricted domain of each original function becomes the range of its inverse:

$\sin^{-1}x$ : domain $[-1,1]$ , range $[-\pi/2,\pi/2]$
$\cos^{-1}x$ : domain $[-1,1]$ , range $[0,\pi]$
$\tan^{-1}x$ : domain all real numbers, range $(-\pi/2,\pi/2)$

These are also called arcsine, arccosine, and arctangent. The output is always an angle (pp. 432–433).

Common misconception. Many angles have sine $1/2$ , so it is tempting to say $\sin^{-1}(1/2)$ has many answers. But an inverse is a function, so it gives exactly one output. It picks the angle inside its range. Both $\pi/6$ and $5\pi/6$ have sine $1/2$ , but only $\pi/6$ lies in $[-\pi/2,\pi/2]$ . So $\sin^{-1}(1/2)=\pi/6$ (p. 433).

Evaluating inverse trig functions

Exact values. Ask: which angle in the inverse's range has this trig value?

$\sin^{-1}(-\sqrt2/2)$ : the angles $5\pi/4$ and $7\pi/4$ have this sine, but both are outside $[-\pi/2,\pi/2]$ . The coterminal negative angle $-\pi/4$ is in range, so the answer is $-\pi/4$ .
$\cos^{-1}(-\sqrt3/2)$ : we need an angle in $[0,\pi]$ with that cosine, which is $5\pi/6$ .
$\tan^{-1}(1)=\pi/4$ .

(pp. 433–434)

Calculators. A calculator answers in whatever mode it is set to. For example, $\sin^{-1}(0.97)\approx 1.3252$ in radian mode and $\approx 75.93^\circ$ in degree mode. If the angle will be used inside another function, or later in calculus, work in radians (pp. 434, 443).

Angles in right triangles. If you know two sides of a right triangle, choose the ratio that uses those two sides and apply the matching inverse. Suppose the hypotenuse is 12 and the side adjacent to $\theta$ is 9. Then $\cos\theta=\frac{9}{12}$ , so

$\theta=\cos^{-1}\left(\tfrac{9}{12}\right)\approx 0.7227\text{ rad}\approx 41.41^\circ.$

(p. 435)

Compositions with inverse trig functions

Inner trig, outer inverse. Evaluate the inside first. For $\sin^{-1}(\cos(13\pi/6))$ , the angle $13\pi/6$ is coterminal with $\pi/6$ , so $\cos(13\pi/6)=\sqrt3/2$ . Then $\sin^{-1}(\sqrt3/2)=\pi/3$ , which is in the range of inverse sine (p. 435).

Inner inverse, outer trig (using an identity). To find $\sin(\cos^{-1}(4/5))$ , let $\theta=\cos^{-1}(4/5)$ . Then $\cos\theta=4/5$ , and we want $\sin\theta$ . The Pythagorean identity gives

$\sin^2\theta=1-\left(\tfrac45\right)^2=\tfrac{9}{25}.$

The sign comes from the range. Inverse cosine returns angles in $[0,\pi]$ , where sine is never negative. So $\sin\theta=3/5$ (p. 436).

Inner inverse, outer trig (using a triangle). Another method is to sketch a right triangle. For $\sin(\tan^{-1}(7/4))$ , let $\tan\theta=7/4$ . Label the opposite side 7 and the adjacent side 4. The hypotenuse is $\sqrt{4^2+7^2}=\sqrt{65}$ . So $\sin\theta=\frac{7}{\sqrt{65}}$ (pp. 436–437).

Algebraic expressions. The same idea works with a variable. To simplify $\cos(\sin^{-1}(x/3))$ , let $\sin\theta=x/3$ . Then

$\cos^2\theta=1-\frac{x^2}{9}=\frac{9-x^2}{9}.$

The angle $\theta$ is in $[-\pi/2,\pi/2]$ , where cosine is nonnegative, so we take the positive root:

$\cos(\sin^{-1}(x/3))=\frac{\sqrt{9-x^2}}{3},\quad -3\le x\le 3.$

(p. 437)

Solving trig equations with special values

Evaluating a function gives one result, but solving a trig equation can give many. Take $\sin t=\tfrac12$ . On the unit circle, the angles with $y$ -value $\tfrac12$ are $t=\pi/6$ and $t=5\pi/6$ . Any coterminal angle has the same sine, so we add full turns of $2\pi$ . The complete solution set is

$t=\frac{\pi}{6}+2\pi k\quad\text{or}\quad t=\frac{5\pi}{6}+2\pi k,\qquad k\text{ an integer}.$

There are infinitely many solutions (p. 440).

Often you are asked only for solutions in one turn, such as $0\le\theta<2\pi$ . The steps are to isolate the trig function and then match its value to the unit circle. Here is an example. A circle of radius 10 meets the line $x=-5\sqrt2$ . A point on the circle has $x=10\cos\theta$ , so we solve $10\cos\theta=-5\sqrt2$ . Isolating gives $\cos\theta=-\frac{\sqrt2}{2}$ . This is a unit circle value, and the matching angles are $\theta=3\pi/4$ and $\theta=5\pi/4$ (p. 441).

Equations with a transformed input

When the angle is something like $\frac{\pi}{12}t$ , replace it with a temporary variable $u$ . Solve for $u$ first, then substitute back and solve for $t$ .

Example: $4\sin\left(\frac{\pi}{12}t\right)+7=9$ .

Isolate the sine: $\sin\left(\frac{\pi}{12}t\right)=\tfrac12$ .
Let $u=\frac{\pi}{12}t$ , so $\sin u=\tfrac12$ . Then $u=\frac{\pi}{6}+2\pi k$ or $u=\frac{5\pi}{6}+2\pi k$ .
Substitute back: $\frac{\pi}{12}t=\frac{\pi}{6}+2\pi k$ or $\frac{\pi}{12}t=\frac{5\pi}{6}+2\pi k$ .
Multiply both sides by $\frac{12}{\pi}$ : $t=2+24k$ or $t=10+24k$ .

(p. 442)

The $24k$ is not a coincidence. The solutions repeat once per period of the transformed function, and here the period is $\frac{2\pi}{\pi/12}=24$ . Each full cycle of sine contains two solutions. More solutions come from adding multiples of the period (p. 442).

In a tide model of this kind, the result says the depth reaches 9 feet at 2 and 10 hours after midnight, and the pattern repeats every 24 hours (p. 442).

Solving equations that are not special values

Many trig equations, such as $\sin\theta=0.8$ , do not match a unit circle value. For these, use a calculator and an inverse function. Check the mode first. Degree mode gives degrees and radian mode gives radians. For work with composed functions, or any time you are not solving for an angle, use radians. (p. 443)

An inverse trig function gives only one output, but the equation usually has more solutions. To find the rest, remember what each function means on the unit circle. Sine is the $y$ -value, cosine is the $x$ -value, and tangent is the slope. Angles with the same value share the same reference angle. (p. 443)

Example. Solve $\sin\theta=0.8$ . The calculator gives $\sin^{-1}(0.8)\approx0.927$ . A second angle has the same height on the circle, in quadrant II: $\pi-0.927\approx2.214$ . Adding full turns gives all solutions: $\theta\approx0.927+2\pi k$ and $\theta\approx2.214+2\pi k$ , where $k$ is an integer. (pp. 443–444)

Solutions on a given interval

When the problem gives an interval, find one solution with the inverse and use the reference angle to get the other. Then add or subtract full cycles until both lie in the interval. (p. 444)

Example. Solve $\sin x=-8/9$ on $0\le x<360^\circ$ . The calculator gives $-62.734^\circ$ , which is outside the interval. Adding $360^\circ$ gives $297.266^\circ$ . The reference angle is $62.734^\circ$ , so the quadrant III solution is $180^\circ+62.734^\circ=242.734^\circ$ . (p. 444)

Tangent is different. Sine and cosine usually give two solutions per cycle. Tangent takes each value only once per cycle, so the other solutions come from adding the period $\pi$ . For $\tan x=3$ , the inverse gives $x\approx1.249$ . Adding $\pi$ gives $4.391$ . All solutions are $1.249+k\pi$ . (p. 445)

Inside transformations and the full method

When the input is changed, such as $\cos(3t)$ or $\sin(\pi t)$ , replace the inside with a temporary variable $u$ . Solve for $u$ , then undo the substitution. (p. 446)

Compression over several cycles. Solve $\cos(3t)=0.2$ for $0\le t\le 4\pi/3$ . Let $u=3t$ . Then $u$ runs over $[0,4\pi]$ , which is two cycles. On the first cycle, $u=\cos^{-1}(0.2)\approx1.369$ and $u=2\pi-1.369\approx4.914$ . Add $2\pi$ to get the second cycle: $7.653$ and $11.197$ . Divide each by 3: $t\approx0.456,\,1.638,\,2.551,\,3.732$ . (p. 446)

The method:

Isolate the trig function.
Substitute for the inside.
Use the inverse for one solution.
Use symmetry for a second solution, if there is one.
Add full periods as needed.
Undo the substitution. (pp. 446–447)

Example. For $\sin(\pi t)=-2/3$ , let $u=\pi t$ . Then $u=-0.730+2\pi k$ or $u=3.871+2\pi k$ . Dividing by $\pi$ gives $t=-0.232+2k$ or $t=1.232+2k$ . Notice the period of the solutions is 2, which is the period of $\sin(\pi t)$ . (pp. 446–447)

Using equations in real situations

Time above a height. A Ferris wheel rider has height $h(t)=-65\cos(\frac{\pi}{15}t)+70$ . To find the time above 100 m, solve $h(t)=100$ . This gives $\cos u=-30/65$ , where $u=\frac{\pi}{15}t$ . The inverse gives $u=2.051$ , and the symmetric solution is $2\pi-2.051=4.230$ . Then $t=9.793$ and $t=20.197$ minutes. The graph shows the rider is above 100 m between these times, so the time is $20.197-9.793\approx10.404$ minutes. (pp. 447–448)

Angles in right triangles. If you know two sides, pick the ratio that uses them and apply the inverse. An airfield 300 miles east and 200 miles north gives $\tan\alpha=200/300$ , so $\alpha\approx33.7^\circ$ north of east. A ladder whose base is 1 unit from the wall for every 4 units of length gives $\cos\theta=1/4$ , so $\theta\approx75.52^\circ$ . (pp. 451–452)

Rotating a point. Treat the point as lying on a circle about the origin. For $(-3,5)$ , $r=\sqrt{34}$ . From $x=r\cos\theta$ , the angle is $\cos^{-1}(-3/\sqrt{34})\approx120.964^\circ$ . This is the quadrant II angle, which is correct because $y$ is positive. Turning 20° clockwise gives $100.964^\circ$ . Then $x=\sqrt{34}\cos(100.964^\circ)\approx-1.109$ and $y=\sqrt{34}\sin(100.964^\circ)\approx5.725$ . (p. 452)

Modeling with sinusoids

To build a model from a high and a low:

Midline = average of the high and low.
Amplitude = half their difference.
Period = length of one cycle, which gives $B=2\pi/P$ .
Use cosine if the start is a maximum, flipped cosine if it is a minimum, and so on. (p. 453)

Seattle daylight goes from 8.5 to 16 hours over 12 months. The midline is $12.25$ , the amplitude is $3.75$ , and $B=\pi/6$ . January is the low, so $h(t)=-3.75\cos(\frac{\pi}{6}t)+12.25$ . (p. 453)

To answer a question, set the model equal to the target and solve. For 14 hours, $\cos(\frac{\pi}{6}t)=-1.75/3.75$ . The first solution is $t\approx3.927$ months, near the end of April. Choose the solution that fits the context. A garden planted in spring needs only the first one. (pp. 453–454)

Frequency, springs, and pistons

Frequency is cycles per unit time, and it is the reciprocal of the period. An object that oscillates 3 times per second has period $1/3$ s, so $B=2\pi/(1/3)=6\pi$ . (p. 454)

A spring has natural length 20 cm and is pulled 8 cm past it. It starts at a maximum, so $x(t)=8\cos(6\pi t)+20$ . Setting $x(t)=27$ gives $\cos(6\pi t)=7/8$ , so $t\approx0.505/(6\pi)\approx0.0268$ s. By symmetry the object is beyond 27 cm for $2(0.0268)=0.0536$ s each cycle. (pp. 454–455)

Piston. A wheel of radius 4 cm turns 5 times per second, which is $10\pi$ rad/s. Point A is $(4\cos(10\pi t),4\sin(10\pi t))$ . A 10 cm rod joins A to B on the horizontal axis. The right triangle has height $4\sin(10\pi t)$ and hypotenuse 10. Its base is $\sqrt{100-16\sin^2(10\pi t)}$ . So B is at $x(t)=4\cos(10\pi t)+\sqrt{100-16\sin^2(10\pi t)}$ . (pp. 455–456)

To find when $x(t)=12$ , isolate the root: $12-4\cos(10\pi t)=\sqrt{100-16\sin^2(10\pi t)}$ . Square both sides and simplify. The $\cos^2$ and $\sin^2$ terms combine to 1 by the Pythagorean identity, leaving $\cos(10\pi t)=60/96$ . Then $u\approx0.896$ or $2\pi-0.896=5.388$ , so $t\approx0.0285$ s and $0.1715$ s. Both repeat every $1/5$ s. (pp. 456–457)

Check yourself

Solve $\cos\theta=-0.3$ on $0\le\theta<2\pi$ and explain how you found the second angle.
Why does $\tan x=2$ have one solution per cycle, while $\sin x=0.5$ has two?
State the domain and range of $\sin^{-1}x$ and explain why $\sin^{-1}(1/2)$ has just one answer.
A wheel's height is $A\cos(Bt)+k$ . Explain how amplitude, period, and midline determine $A$ , $B$ , and $k$ .
Use odd and even properties to simplify $\sin(-\theta)/\tan(-\theta)$ .