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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 9 · pp. 347–404

Chapter 5: Trigonometric Functions of Angles

This chapter starts with circles: distances, circle equations, and where lines meet circles. It then measures rotation with angles in degrees and radians, which leads to arclength, sector area and rotational speed. Next it defines sine and cosine from points on a circle, then the other four trigonometric functions and their identities, and finally applies all of this to right triangles.

Distances and the Pythagorean Theorem

In a right triangle, the longest side is the hypotenuse, which is opposite the right angle. The Pythagorean Theorem says the squares of the two legs add up to the square of the hypotenuse:

$a^2+b^2=c^2$

For legs 3 and 4, the hypotenuse is $\sqrt{9+16}=5$ . (p. 347)

This theorem gives a way to find the distance between two points. Plot the points and draw a right triangle with the segment joining them as its hypotenuse. The horizontal leg is the difference of the x-values, and the vertical leg is the difference of the y-values. For $(-3,2)$ and $(2,5)$ , the legs are 5 and 3, so the distance is $\sqrt{25+9}=\sqrt{34}$ . (pp. 347–348)

Doing this in general gives the distance formula:

$\text{dist}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Because the differences are squared, it does not matter which point you call first. (p. 348)

The Equation of a Circle

A circle is the set of all points at the same distance $r$ from a center $(h,k)$ . So for any point $(x,y)$ on it, the distance formula gives $r=\sqrt{(x-h)^2+(y-k)^2}$ . Squaring both sides gives the standard equation:

$(x-h)^2+(y-k)^2=r^2$

(p. 348)

Example. A circle with center $(-3,2)$ and radius 4 has $h=-3$ , $k=2$ , $r=4$ . Subtracting $-3$ means adding 3, so the equation is $(x+3)^2+(y-2)^2=16$ . (p. 349)

If the center is the origin, the equation shortens to $x^2+y^2=r^2$ . A circle of radius 3 at the origin is $x^2+y^2=9$ . Notice that the right side is $r^2$ , not $r$ . Compared with a circle at the origin, the numbers $h$ and $k$ show how far the circle has moved sideways and up or down. Watch the signs: $(x+3)^2$ means $h=-3$ . (p. 349)

A circle fails the vertical line test, because most vertical lines cross it twice. So you cannot write $y$ as a function of $x$ , or $x$ as a function of $y$ . On $x^2+y^2=25$ , the input $x=3$ gives both $y=4$ and $y=-4$ . (p. 348)

Finding Points on a Circle

Since $y$ is not a function of $x$ here, you find points by solving algebraically.

Known x-value (or y-value). Substitute it into the circle equation and solve for the other variable. For $x^2+y^2=25$ with $x=3$ :

$9+y^2=25 \;\Rightarrow\; y^2=16 \;\Rightarrow\; y=\pm4$

The points are $(3,4)$ and $(3,-4)$ . Expect two answers, one above and one below the center's horizontal line. (pp. 349–350)

x-intercepts. Set $y=0$ and solve for $x$ . For $(x-2)^2+(y-4)^2=36$ , we get $(x-2)^2+16=36$ , so $(x-2)^2=20$ and $x=2\pm\sqrt{20}=2\pm2\sqrt5$ . The intercepts are $(2+2\sqrt5,0)$ and $(2-2\sqrt5,0)$ . For y-intercepts, set $x=0$ instead. (p. 350)

Circles in Real-World Problems

For an applied problem, draw a picture and add a coordinate system. Placing the origin at the center of the circle keeps the equation simple.

Example. A pizza store delivers within 3 miles. Main Street runs east–west, 2 miles north of the store. Put the store at the origin. The edge of the delivery zone is $x^2+y^2=9$ , and Main Street is the line $y=2$ . Substituting gives $x^2+4=9$ , so $x=\pm\sqrt5\approx\pm2.236$ . The store reaches 2.236 miles east and 2.236 miles west of the cross street, so it covers about 4.472 miles of Main Street. (pp. 350–351)

Where a Line Meets a Circle

A vertical or horizontal line is easy, as above. For any other non-vertical line, use the idea that works for two functions: make the $y$ -values equal. Replace $y$ in the circle equation with the line's expression, then solve the resulting quadratic for $x$ . Finally, put each $x$ back into the line to find its $y$ .

Example. Find where $y=4x$ meets $(x-2)^2+y^2=16$ .

Substitute: $(x-2)^2+(4x)^2=16$ .
Expand: $x^2-4x+4+16x^2=16$ .
Rearrange to zero: $17x^2-4x-12=0$ .
Use the quadratic formula: $x=\frac{4\pm\sqrt{16+816}}{34}$ , so $x\approx0.966$ or $x\approx-0.731$ .
Use $y=4x$ : $y\approx3.864$ and $y\approx-2.923$ .

The intersections are about $(0.966,3.864)$ and $(-0.731,-2.923)$ . (pp. 351–352)

Angles and Their Measure

Many circle problems involve turning, so we need a way to measure rotation. An angle is formed by two rays from the same point. It starts at the initial side and ends at the terminal side. Its measure tells how far the rotation went, so it is a rotational measure, not a length. (p. 357)

The most familiar unit is the degree. One full turn around a circle is $360^\circ$ , so one degree is $\frac{1}{360}$ of a turn. A quarter turn is $360^\circ\div4=90^\circ$ , and a half turn is $180^\circ$ . Always write the word "degrees" or the symbol $^\circ$ . (p. 357)

Unless told otherwise, angles are drawn in standard position. This means the angle starts on the positive horizontal axis and turns counterclockwise. To sketch $30^\circ$ , split the first quarter turn ( $90^\circ$ ) into three equal parts and take one part. (p. 357)

Coterminal and Negative Angles

An angle larger than $360^\circ$ has gone around more than once. Two angles are coterminal if their terminal sides end up pointing the same way after their rotations are done. Subtracting or adding a full turn of $360^\circ$ does not change where the terminal side points. So to find a coterminal angle in a given range, add or subtract $360^\circ$ (or $2\pi$ when using radians) until the angle fits. (p. 358)

Example. For $800^\circ$ : $800-360=440$ , and $440-360=80$ . So $80^\circ$ is coterminal with $800^\circ$ . Seeing the angle as $80^\circ$ makes it easy to tell which way it points. (pp. 358–359)

Negative angles work like negative numbers on a number line: they go the opposite way. Positive angles turn counterclockwise, and negative angles turn clockwise from the positive horizontal axis. The angle $-45^\circ$ is half of a quarter turn clockwise. Adding a full turn gives $-45^\circ+360^\circ=315^\circ$ , which has the same terminal side. (p. 359)

Angles to Memorize

Within one rotation, you will constantly meet the multiples of $30^\circ$ , $45^\circ$ , $60^\circ$ and $90^\circ$ . These include $30^\circ, 60^\circ, 90^\circ, 120^\circ, 135^\circ, 150^\circ, 180^\circ$ , and so on up to $360^\circ$ . Learn where each one sits on the circle, because exact trigonometric values later in the chapter depend on them. (p. 359)

Arclength and why it leads to radians

The arclength $s$ is the length of the piece of a circle that an angle cuts off. It is the part of the circumference between the angle's initial and terminal sides (p. 360).

The whole circumference is $C=2\pi r$ . An angle cuts off a fraction of that. A $45^\circ$ angle is $\frac18$ of a full turn, so it cuts off $\frac18$ of the circumference (p. 360).

Circle of radius 2: $C=4\pi$ , so $s=\frac18(4\pi)=\frac{\pi}{2}$ .
Circle of radius 3: $C=6\pi$ , so $s=\frac18(6\pi)=\frac{3\pi}{4}$ .

Now divide each arclength by its radius. Both give $\frac{\pi}{4}$ . For a fixed angle, $\frac{s}{r}$ is the same on every circle. It depends only on the angle, so it can be used to measure the angle (p. 360).

Radian measure

The radian measure of an angle is the arclength it subtends divided by the radius:

$\theta=\frac{s}{r}$

This is a ratio of two lengths, so it has no units. You do not need to write "radians." If an angle has no degree symbol, assume it is in radians. On the unit circle ( $r=1$ ), the radian measure is just the arclength (p. 361).

Example. One third of a full turn cuts off $\frac13(2\pi r)$ . Dividing by $r$ gives $\theta=\frac{2\pi}{3}$ (p. 361).

On the unit circle, one rotation has length $2\pi$ . One rotation is also $360^\circ$ . This gives the key facts (p. 361):

Full rotation: $360^\circ=2\pi$ radians
Half rotation: $180^\circ=\pi$ radians
Quarter rotation: $90^\circ=\frac{\pi}{2}$ radians

Converting between degrees and radians

Since $180^\circ=\pi$ radians, use these rules (p. 362):

Degrees to radians: multiply by $\frac{\pi}{180}$ .
Radians to degrees: multiply by $\frac{180}{\pi}$ .

Pick the factor that cancels the unit you start with.

$15^\circ\cdot\frac{\pi}{180}=\frac{\pi}{12}$
$\frac{\pi}{6}\cdot\frac{180}{\pi}=30^\circ$

Common angles in radians

The familiar angles are multiples of $\frac{\pi}{6}$ ( $30^\circ$ ), $\frac{\pi}{4}$ ( $45^\circ$ ), $\frac{\pi}{3}$ ( $60^\circ$ ) and $\frac{\pi}{2}$ ( $90^\circ$ ). Some examples (p. 363):

Degrees	Radians
$120^\circ$	$\frac{2\pi}{3}$
$135^\circ$	$\frac{3\pi}{4}$
$150^\circ$	$\frac{5\pi}{6}$
$210^\circ$	$\frac{7\pi}{6}$
$300^\circ$	$\frac{5\pi}{3}$
$330^\circ$	$\frac{11\pi}{6}$

Learn these well, because later work uses them constantly.

Coterminal angles in radians

In degrees, you add or subtract $360^\circ$ . In radians, you add or subtract $2\pi$ until the angle is in the range you want (p. 363).

Example. Find the angle between $0$ and $2\pi$ coterminal with $\frac{19\pi}{4}$ . Write $2\pi=\frac{8\pi}{4}$ .

$\frac{19\pi}{4}-\frac{8\pi}{4}=\frac{11\pi}{4}$ , which is still larger than $2\pi$ .
$\frac{11\pi}{4}-\frac{8\pi}{4}=\frac{3\pi}{4}$ , which is in range.

The answer is $\frac{3\pi}{4}$ (p. 363).

Arclength formula

Rearranging $\theta=\frac{s}{r}$ gives

$s=r\theta$

where $\theta$ is in radians (p. 364).

Example. Mercury is about 36 million miles from the sun. In one Earth day it travels $0.0114$ of a rotation. In radians, that is $2\pi(0.0114)\approx0.0716$ . Then $s=36(0.0716)\approx2.578$ million miles (p. 364).

Common mistake: using $s=r\theta$ with $\theta$ in degrees. The formula comes from the definition of radians, so it only works in radians. For $215^\circ$ on a circle of radius 10, convert first: $\theta=\frac{215\pi}{180}$ . Then $s=10\cdot\frac{215\pi}{180}\approx37.5$ (pp. 364, 368).

Area of a sector

A sector is a slice of a circle, like a slice of pie. An angle $\theta$ in radians takes up the fraction $\frac{\theta}{2\pi}$ of the circle, because $2\pi$ is a full turn. Multiply that fraction by the circle's area (p. 364):

$A=\frac{\theta}{2\pi}\cdot\pi r^2=\frac12r^2\theta$

Example. A sprinkler sprays 20 ft and turns $30^\circ$ . Convert: $30^\circ=\frac{\pi}{6}$ . Then $A=\frac12(20)^2\cdot\frac{\pi}{6}\approx104.72$ ft $^2$ (p. 365).

As with arclength, $\theta$ must be in radians.

Angular and linear velocity

When a point moves around a circle, two kinds of speed describe it (pp. 365–366):

Angular velocity: $\omega=\frac{\theta}{t}$ , the angle turned per unit time. It can be in radians per second, rotations per minute, degrees per hour, and so on.
Linear velocity: $v=\frac{s}{t}$ , the distance travelled along the circle per unit time.

Example. A water wheel turns once every 5 seconds. One rotation is $2\pi$ radians, so $\omega=\frac{2\pi}{5}\approx1.257$ radians per second (p. 366).

To link the two, substitute $s=r\theta$ into $v=\frac{s}{t}$ :

$v=\frac{r\theta}{t}=r\omega$

This works when $\omega$ is in radians per unit time (p. 366).

From rotation rate to linear speed

Use these steps (p. 367):

Convert revolutions to radians, using $1$ revolution $=2\pi$ radians.
Multiply by the radius.
Convert units if needed.

Example. A bicycle wheel has a 28-inch diameter, so $r=14$ inches. It turns at 180 revolutions per minute.

$\omega=180\cdot2\pi=360\pi$ radians per minute.
$v=r\omega=14\cdot360\pi=5040\pi\approx15{,}834$ inches per minute.
Convert inches to miles and minutes to hours: $5040\pi\cdot\frac{1}{12}\cdot\frac{1}{5280}\cdot60\approx14.99\text{ mph}$

The bike travels at about 15 mph. Radians have no units, so they drop out of the unit work (p. 367).

Sine and cosine from a point on a circle

Start with a circle of radius $r$ centered at the origin. Pick the point $(x,y)$ where the terminal side of angle $\theta$ meets the circle. Drop a vertical segment from that point to the x-axis. This makes a right triangle. The $x$ and $y$ values can be negative, depending on the quadrant. If the point lies on an axis, the triangle collapses, and one coordinate is $0$ while the other is $\pm r$ . (p. 372)

A circle with a different radius gives a similar triangle. The sides change size, but the ratios $\frac{y}{r}$ and $\frac{x}{r}$ stay the same for a given angle. So these ratios depend only on $\theta$ , and we name them:

$\sin\theta=\frac{y}{r},\qquad \cos\theta=\frac{x}{r}$

(pp. 372–373)

Example. The point $(3,4)$ is on a circle of radius $5$ . Then $\cos\theta=\frac35$ and $\sin\theta=\frac45$ . (p. 373)

Axis angles and coordinates from an angle

Some values come straight from the picture. At $90^\circ$ the terminal side points straight up, so the point is $(0,r)$ . Then $\cos90^\circ=\frac0r=0$ and $\sin90^\circ=\frac rr=1$ . At $180^\circ$ the point is $(-r,0)$ , so $\cos180^\circ=-1$ and $\sin180^\circ=0$ . (pp. 373–374, 380)

Solving the two definitions for $x$ and $y$ gives the coordinates of the point at angle $\theta$ :

$x=r\cos\theta,\qquad y=r\sin\theta$

On the unit circle ( $r=1$ ) the point is simply $(\cos\theta,\sin\theta)$ . (p. 374)

Example. On a circle of radius $6$ at $\frac{\pi}{4}$ , the point is $\left(6\cdot\frac{\sqrt2}{2},\,6\cdot\frac{\sqrt2}{2}\right)=(3\sqrt2,\,3\sqrt2)$ . (p. 376)

The Pythagorean Identity

Every point on the circle satisfies $x^2+y^2=r^2$ . Replace $x$ with $r\cos\theta$ and $y$ with $r\sin\theta$ . You get $r^2\cos^2\theta+r^2\sin^2\theta=r^2$ . Divide by $r^2$ :

$\cos^2\theta+\sin^2\theta=1$

This is true for every angle. The notation $\cos^2\theta$ means $(\cos\theta)^2$ . Many calculators and programs do not accept that shorthand, so type the parentheses. (p. 374)

Finding one value from the other. Taking a square root gives two possible answers, $\pm$ . The quadrant tells you which sign to keep. Suppose $\sin\theta=\frac37$ and $\theta$ is in quadrant II. Then

$\cos^2\theta=1-\frac{9}{49}=\frac{40}{49},\qquad \cos\theta=\pm\frac{2\sqrt{10}}{7}.$

In quadrant II, $x$ is negative, so $\cos\theta=-\frac{2\sqrt{10}}{7}$ . (p. 375)

Exact values at the special angles

$45^\circ$ ( $\frac{\pi}{4}$ ). This angle cuts the first quadrant in half, so $x=y$ . That means $\cos\theta=\sin\theta$ . The identity becomes $2\cos^2\theta=1$ , so $\cos\theta=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}$ after rationalizing. The unit circle point is $\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)$ . (pp. 375–376)

$30^\circ$ ( $\frac{\pi}{6}$ ). Draw the triangles for $30^\circ$ and $-30^\circ$ together. They form one triangle with all angles $60^\circ$ , so it is equilateral. Its vertical side is $2y$ and it equals $r$ , so $y=\frac r2$ and $\sin30^\circ=\frac12$ . The identity then gives $\cos^2 30^\circ=1-\frac14=\frac34$ , so $\cos30^\circ=\frac{\sqrt3}{2}$ . (pp. 376–377)

$60^\circ$ ( $\frac{\pi}{3}$ ). Reflect the $30^\circ$ triangle over the line $y=x$ . This swaps the coordinates, so $\cos60^\circ=\frac12$ and $\sin60^\circ=\frac{\sqrt3}{2}$ . (p. 377)

The first-quadrant table, written as $(\cos\theta,\sin\theta)$ :

$0$ : $(1,0)$
$\frac{\pi}{6}$ : $\left(\frac{\sqrt3}{2},\frac12\right)$
$\frac{\pi}{4}$ : $\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)$
$\frac{\pi}{3}$ : $\left(\frac12,\frac{\sqrt3}{2}\right)$
$\frac{\pi}{2}$ : $(0,1)$

Across the quadrant, cosine falls from $1$ to $0$ and sine rises from $0$ to $1$ . Memorize this table. (p. 378)

Reference angles and the other quadrants

For an angle in another quadrant, there is a first-quadrant angle with the same $x$ or $y$ size. Mirror images of the point across the axes keep the same $|x|$ and $|y|$ . They change only the signs. (p. 378)

The reference angle is the smallest angle between the terminal side and the horizontal axis. It is always between $0$ and $\frac{\pi}{2}$ ( $90^\circ$ ). An angle has the same sine and cosine as its reference angle, except possibly for sign. (pp. 378–379)

Method for a special angle:

Find the reference angle.
Write the first-quadrant sine and cosine values for it.
Use the quadrant to set the signs. $x$ gives the sign of cosine and $y$ gives the sign of sine.

Example. $150^\circ$ is $30^\circ$ short of $180^\circ$ , so its reference angle is $30^\circ$ . It is in quadrant II, where $x<0$ and $y>0$ . So $\cos150^\circ=-\frac{\sqrt3}{2}$ and $\sin150^\circ=\frac12$ . (p. 379)

Example. $\frac{7\pi}{6}$ is in quadrant III, and its reference angle is $\frac{\pi}{6}$ . Both coordinates are negative. On a circle of radius $12$ :

$x=12\left(-\frac{\sqrt3}{2}\right)=-6\sqrt3,\qquad y=12\left(-\frac12\right)=-6$

The point is $(-6\sqrt3,-6)$ . Checking that the signs match the quadrant is a good habit. (p. 380)

Locating a point from a distance and direction

When a problem gives a distance and a direction, treat the distance as $r$ and the direction as $\theta$ . Then use $x=r\cos\theta$ and $y=r\sin\theta$ . (p. 372)

Example. A boat is $20$ miles from a marina at $225^\circ$ . The reference angle is $45^\circ$ and the angle is in quadrant III, so

$x=20\cos225^\circ=-10\sqrt2\approx-14.14,\qquad y=20\sin225^\circ\approx-14.14.$

The boat is about $14.14$ miles west and $14.14$ miles south of the marina. (p. 381)

Using a calculator: degrees or radians

A calculator reads the number you type in whichever mode it is set to. Typing $\cos(30)$ in degree mode gives the cosine of $30^\circ$ . In radian mode it gives the cosine of $30$ radians, which is a very different angle. Most computer software works only in radians. (pp. 381–382)

If you have no degree mode, convert first by multiplying by $\frac{\pi}{180}$ . For example, $\cos20^\circ=\cos\left(20\cdot\frac{\pi}{180}\right)\approx0.9397$ . (p. 382)

The six trigonometric functions

A right triangle drawn inside a circle has three sides, so there are six possible side ratios. Sine and cosine are two of them. The other four are built from the same point $(x,y)$ on a circle of radius $r$ at angle $\theta$ :

$\tan\theta=\frac{y}{x}$
$\sec\theta=\frac{r}{x}$
$\csc\theta=\frac{r}{y}$
$\cot\theta=\frac{x}{y}$

Secant, cosecant and cotangent are the reciprocals of cosine, sine and tangent. Any of these functions is undefined where its denominator is $0$ . (p. 385)

Tangent is a slope. Slope is rise over run, which is $\frac{y}{x}$ for the segment from the origin to the point. So $\tan\theta$ is the slope of that line. A $45^\circ$ line has slope $1$ , and indeed $\tan 45^\circ=1$ . (pp. 385–386)

Reciprocal and quotient identities

Dividing the side ratios gives a link back to sine and cosine. For example, $\frac{y/r}{x/r}=\frac{y}{x}$ . The results are:

$\tan\theta=\frac{\sin\theta}{\cos\theta},\quad \sec\theta=\frac{1}{\cos\theta},\quad \csc\theta=\frac{1}{\sin\theta},\quad \cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}$

These let you evaluate any of the six functions if you know sine and cosine. Example: $\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2}$ , so $\sec\frac{5\pi}{6}=\frac{1}{-\sqrt3/2}=-\frac{2\sqrt3}{3}$ . Another: $\sin\frac{7\pi}{6}=-\frac12$ , so $\csc\frac{7\pi}{6}=-2$ . (pp. 385–386)

An identity is an equation that holds for every input where both sides are defined. Identities come from definitions and earlier facts, such as the Pythagorean Identity. (p. 386)

Simplifying and proving identities

Simplifying. Rewrite every function using sine and cosine, combine the fractions, and cancel common factors. Take $\frac{\sec\theta}{\tan\theta}$ . It becomes $\frac{1/\cos\theta}{\sin\theta/\cos\theta}$ . Invert and multiply, the cosines cancel, and you get $\frac{1}{\sin\theta}=\csc\theta$ . This simplification has also shown a new identity. (p. 387)

Proving. A "prove the identity" problem is a simplification where you already know the answer. Start with one side, usually the more complicated one, and use algebra and known identities until it matches the other side. The other side is a useful guide. If it has no fractions, try to remove them. If it uses only sine and cosine, convert to those. (pp. 387–388)

Example: show $\frac{1+\cot\theta}{\csc\theta}=\sin\theta+\cos\theta$ . Rewrite the left side as $\frac{1+\cos\theta/\sin\theta}{1/\sin\theta}$ . Invert and multiply to get $\left(1+\frac{\cos\theta}{\sin\theta}\right)\sin\theta$ . Distribute to get $\sin\theta+\cos\theta$ . (pp. 387–388)

Common mistake: do not treat an identity like an equation to solve by doing the same thing to both sides. That assumes the two sides are equal, which is what you must show. It is like proving $(x-1)^2=x^2-2x+1$ by expanding only the left side. (p. 387)

Pythagorean identities

Move a term in $\cos^2\theta+\sin^2\theta=1$ and you get $\sin^2\theta=1-\cos^2\theta$ and $\cos^2\theta=1-\sin^2\theta$ . These swap sine and cosine inside an expression. (pp. 388–389)

Example: show $\frac{\cos^2\theta}{1+\sin\theta}=1-\sin\theta$ . Replace $\cos^2\theta$ with $1-\sin^2\theta=(1-\sin\theta)(1+\sin\theta)$ . Cancel the factor $1+\sin\theta$ . (pp. 388–389)

Two more versions come from dividing the Pythagorean Identity:

By $\cos^2\theta$ : $1+\tan^2\theta=\sec^2\theta$ .
By $\sin^2\theta$ : $1+\cot^2\theta=\csc^2\theta$ .

They are handy when you know a tangent, secant, cotangent or cosecant value. (p. 389)

Finding other values from one value and the quadrant

Suppose $\tan\theta=\frac27$ and $\theta$ is in quadrant III. There are two methods. (pp. 389–390)

Triangle. In quadrant III both $x$ and $y$ are negative, so use the point $(-7,-2)$ . Then $r=\sqrt{49+4}=\sqrt{53}$ and $\cos\theta=-\frac{7}{\sqrt{53}}=-\frac{7\sqrt{53}}{53}$ .
Identity. $\sec^2\theta=1+\left(\frac27\right)^2=\frac{53}{49}$ , so $\sec\theta=\pm\frac{\sqrt{53}}{7}$ . Cosine is negative in quadrant III, so secant is too: $\sec\theta=-\frac{\sqrt{53}}{7}$ . This gives the same cosine.

Always choose the sign of the square root from the quadrant.

Right triangle trigonometry

For an acute angle $\theta$ in a right triangle, label the sides relative to $\theta$ :

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

The memory aid is SohCahToa. These match the circle definitions, with $x$ as adjacent, $y$ as opposite and $r$ as hypotenuse. If the adjacent side is $15$ and the hypotenuse is $17$ , then $\cos\theta=\frac{15}{17}$ . (pp. 395–396)

You can use either acute angle, but "opposite" and "adjacent" swap when you change angles. In a 3-4-5 triangle, one angle has $\sin=\frac45$ and $\cos=\frac35$ . The other angle has $\sin=\frac35$ and $\cos=\frac45$ . (pp. 396–397)

Cofunction identities. The two acute angles add to $\frac{\pi}{2}$ ( $90^\circ$ ), and the side opposite one is adjacent to the other. So

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

For example, $\cos30^\circ=\sin60^\circ$ . (p. 397)

Solving right triangles and applications

Solving a triangle. Knowing one acute angle and one side is enough. Choose the ratio that connects the known side to the unknown one. With a $30^\circ$ angle and hypotenuse $7$ , call the opposite side $b$ . Then $\sin30^\circ=\frac{b}{7}$ , so $b=3.5$ . Find the last side with $a^2+b^2=7^2$ . The other acute angle is $180^\circ-90^\circ-30^\circ=60^\circ$ . (pp. 397–398)

Height from an angle of elevation. Stand $30$ ft from a tree and see its top at $57^\circ$ . Then $\tan57^\circ=\frac{h}{30}$ , so $h=30\tan57^\circ\approx46$ ft. Sketch and label the triangle first. (p. 398)

Two triangles. Solve the triangle that has enough information first. From a $100$ ft roof, the angle of depression to the base of a skyscraper is $20^\circ$ . In that lower triangle, $\tan20^\circ=\frac{100}{x}$ , so $x=\frac{100}{\tan20^\circ}\approx274.7$ ft. The upper triangle has a $42^\circ$ angle and the same horizontal distance $x$ . So $a=x\tan42^\circ\approx247.4$ ft. Add the roof height: $247.4+100\approx347$ ft. (pp. 398–399)

Check yourself

Write the equation of a circle with center $(3,-2)$ and radius $5$ . Then find the points on it where $x=3$ .
Convert $150^\circ$ to radians. Then find the arclength and sector area for that angle on a circle of radius $4$ .
Find the reference angle of $\frac{5\pi}{4}$ and give $\sin\frac{5\pi}{4}$ and $\cos\frac{5\pi}{4}$ .
If $\sin\theta=-\frac35$ and $\theta$ is in quadrant IV, find $\cos\theta$ , $\tan\theta$ and $\csc\theta$ .
Prove that $\tan\theta\cos\theta=\sin\theta$ , and explain why you should not add the same quantity to both sides while proving it.