# 1.1: Trigonometry and Right Angles

**At Grade**Created by: CK-12

## Basic Functions

**The Basics of Functions**

Some students’ eyes may glaze over at the equation used in the “gas mileage” situation, especially if it’s been a while since their last algebra class. Watch for students who look confused and walk them through setting up the equation if necessary.

Another way to explain the definition of **domain** is that it is the set of real \begin{align*}x-\end{align*}values you can plug into the function that will produce a real number as the output. You can show that the domain of \begin{align*}y = 3x\end{align*} is the set of all real numbers because any real number \begin{align*}x\end{align*} can be multiplied by \begin{align*}3\end{align*} to get a real number \begin{align*}y\end{align*}; then you can show that the domain of \begin{align*} y =\sqrt{x}\end{align*} is restricted to the nonnegative real numbers by demonstrating that you get a real value for \begin{align*}y\end{align*} when you plug in a positive or zero value for \begin{align*}x\end{align*}, but not when you plug in a negative value.

Similarly, you can explain that the **range** of a function is the set of \begin{align*}y-\end{align*}values that could be the possible outputs of the function given a real number \begin{align*}x\end{align*} as the input.

The “rounding” example provides a good opportunity to remind students that rounding “up,” in the case of a negative number, means rounding *toward* zero rather than *away from* zero; since \begin{align*}-5\end{align*} is *greater than* \begin{align*}-6\end{align*}, \begin{align*}-5.5\end{align*} should be rounded up to \begin{align*}-5\end{align*}.

**Families of Functions**

In the “families of functions” table, you may need to clarify what “these functions have a highest exponent of \begin{align*}2\end{align*}” means—i.e., the highest power that appears anywhere in the equation is 2. (We can also say that the degree of the function is \begin{align*}2\end{align*}, or that the function is of the “second degree.”) The same holds, of course, for “these functions have a highest exponent of 3,” but you also may want to clarify “the ends of the graph have opposite behavior”: it simply means that one end of the graph goes up while the other goes down. You may also want to graph \begin{align*}y = x^3\end{align*} to show why it has no local maximum or minimum.

And finally, for students who may have forgotten how asymptotes work, it’s worth reminding them that the values of the function approach the asymptote but never actually reach it. Specifically, if a function has a horizontal asymptote, it means that y will get closer and closer to that value as \begin{align*}x\end{align*} approaches infinity, but will never quite reach it; if a function has a vertical asymptote, it means that \begin{align*}y\end{align*} will get closer and closer to infinity as \begin{align*}x\end{align*} gets closer to the given value, but the function is undefined when \begin{align*}x\end{align*} is exactly equal to that value.

A useful way to explain direct and inverse variation is that with direct variation, the dependent variable increases when the independent variable increases, while with inverse variation, the dependent variable decreases when the independent variable increases. This makes the contrast between the two types of function clearer.

You can clarify the definition of a periodic function by explaining that all the values of the function repeat themselves every \begin{align*}p\end{align*} units. It may be useful to demonstrate with the “weather” example above: \begin{align*}p\end{align*} in this case is \begin{align*}12\end{align*}, so you can show that \begin{align*}f(14) = f(2), f(15) = f(3)\end{align*}, and so on.

**Points to Consider**

(You may want to go over these as a group each lesson.)

Using a calculator to graph functions is quicker and more accurate than doing it by hand, but it can be hard to see precisely where the important points on the graph are.

## Angles in Triangles

**Similar Triangles**

It may be useful to note that the proportions that show that the ratios of corresponding sides are equal can be derived directly from the proportions that show that the side ratios within the two triangles are equal. (For example, \begin{align*}\frac{\overline{AB}}{\overline{DE}} = \frac{\overline{BC}}{\overline{EF}}\end{align*} can be derived directly from \begin{align*}\frac{\overline{AB}}{\overline{BC}} = \frac{\overline{DE}}{\overline{EF}}\end{align*} .)

If you want to move through example \begin{align*}4\end{align*} a little more quickly, you can point out that the side lengths in the second triangle are simply half those in the first.

Students may ask whether ASA and SAA are also criteria for determining if two triangles are similar, since criteria like those exist for determining if triangles are congruent. Explain that they are, but for a very simple reason: if two of the angles are congruent, then the third angle must also be congruent, and so the ASA and SAA cases simply reduce to the AAA case.

The HL case, on the other hand, is a special example of SSA: two of the sides are proportional, and an angle that is not between them is congruent. Normally, this would not be enough to determine that two triangles are similar, but what helps us here is that we are dealing with right triangles, which means that we can use two side lengths to determine the third. Once we’ve found that the other leg is proportional too, then instead of looking at this as a case of SSA, we can see it as a case of SSS: all three pairs of sides are proportional. (Incidentally, we could also see it as a case of SAS: the two legs are proportional, and the angle between them (the right angle) is congruent.)

**Points to Consider**

The answer to question #1 can be demonstrated in more than one way. First, you can draw two right angles with the included side between them, and show that the other two sides are now parallel, meaning that they can’t ever meet to make a triangle.

Second, you can point out that two right angles add up to \begin{align*}180^\circ\end{align*}, and since that is the sum of all three angles of a triangle, that would mean the third angle would have to measure \begin{align*}0^\circ\end{align*}, which is not possible.

Similar reasoning holds for question #2. Drawing two obtuse angles with the included side between them demonstrates even more clearly that the other two sides could never meet, and adding together two angle measures greater than \begin{align*}90^\circ\end{align*} each would give you a sum greater than \begin{align*}180^\circ\end{align*}, which is impossible even before you consider the measure of the third angle.

Question #3 has a different answer depending on what situation you are considering. If you just look at an angle in isolation, then in a sense it cannot have a measure greater than \begin{align*}180^\circ\end{align*}, because an angle of, say, \begin{align*}200^\circ\end{align*} could just as easily be described as an angle of \begin{align*}160^\circ\end{align*} viewed from the other side. However, when you are measuring angles of rotation, as the next lesson will cover, an angle can measure more than 180º or even more than \begin{align*}360^\circ\end{align*}. Mentioning this might be a good segue to the next lesson.

## Measuring Rotation

**Measuring Angles**

Some students may need to see example \begin{align*}3\end{align*} worked out in more detail. You may need to spell out that the circumference of the smaller wheel is π meters and that of the larger is \begin{align*}2\pi \;\mathrm{meters}\end{align*}; then, rather than simply explaining that the larger wheel rotates once every time the smaller one rotates twice (and therefore rotates twice when the small one rotates four times), you may need to show that the four rotations of the smaller wheel cause its circumference to travel \begin{align*}4\;\mathrm{meters}\end{align*} along the larger wheel, and that this is equal to \begin{align*}\frac{4\pi}{2\pi}\end{align*} or \begin{align*}2\end{align*} rotations of the larger wheel.

**Angles of Rotation in Standard Position**

Students may not have encountered the terms “initial side” and “terminal side” before. Explain, if necessary, that these terms are specific to this particular situation; when we place an angle in standard position, the initial side is just what we call the side we chose to place along the \begin{align*}x-\end{align*}axis, and the terminal side is simply the other side.

**Co-terminal Angles**

In working through the next example, you may want to take a moment to remind students which quadrant is which (quadrants I through IV proceed counterclockwise starting from the upper right). Knowing the quadrants will be important in upcoming lessons.

Another way to generate the angle \begin{align*}-315^\circ\end{align*}, of course, is to subtract \begin{align*}315^\circ\end{align*} from \begin{align*}360^\circ\end{align*} to get \begin{align*}45^\circ\end{align*}. This method can be faster than rotating clockwise, but students should familiarize themselves with both techniques.

**Points to Consider**

Real-life instances of angles of rotation might include a wheel, a swinging door, a doorknob, or a screw.

**Review Questions**

You may need to walk students through problem 7. First they must find the total distance the car’s inner wheel travels, which is a quarter (\begin{align*}90 \;\mathrm{degrees}\text{'}\end{align*} worth) of the circumference of a circle with a \begin{align*}100\mathrm{m}\end{align*} radius. Then they must find the number of rotations the inner wheel makes in traveling that distance, which takes two steps: first find the circumference of the wheel based on the given diameter of .\begin{align*}6\mathrm{m}\end{align*}, and then divide the total distance traveled by the circumference of the wheel to find the number of rotations it makes. Next, they must find the distance the outer wheel travels. Since the wheels are \begin{align*}2\mathrm{m}\end{align*} apart, the outer wheel follows a curve with radius \begin{align*}2\mathrm{m}\end{align*} greater than the curve the inner wheel follows, so the distance it travels is a quarter of the circumference of a circle with a \begin{align*}102\mathrm{m}\end{align*} radius. Then, dividing that distance by the circumference of the wheel (already found) gives the number of rotations the outer wheel makes. Finally, they must subtract to find how many more rotations the outer wheel makes than the inner.

## Defining Trigonometric Functions

**The Sine, Cosine, and Tangent Functions**

Another way to explain the domain and range of the first three trigonometric functions is as follows: The trigonometric functions take angles as their input, and their output consists of particular ratios of side lengths.

The mnemonic SOH CAH TOA (Sin: Opposite/Hypotenuse; Cosine: Adjacent/Hypotenuse; Tangent: Opposite/Adjacent—pronounced roughly “soak a toe-a”) may help students remember the ratios. Another mnemonic is “SCoTt, Oscar Has A Heap Of Apples”—that is, for Sin, Cos, and Tan respectively, the ratios are \begin{align*}\frac{\mathrm{O}}{\mathrm{H}}, \frac{\mathrm{A}}{\mathrm{H}}\end{align*}, and \begin{align*}\frac{\mathrm{O}}{\mathrm{A}}\end{align*}.

It may be worth stressing that \begin{align*}\sin(x), \cos(x)\end{align*}, and \begin{align*}\tan(x)\end{align*} are abbreviations for types of functions, and do not indicate that anything is being multiplied.

**Secant, Cosecant, and Cotangent Functions**

The fact that the secant, cosecant, and cotangent are reciprocals of the cosine, sine, and tangent functions respectively will be made explicit in a later section of the text, but it may be useful to point it out now, as this may make it easier for students to remember those ratios.

**Trigonometric Functions of Angles in Standard Position**

In example 4, you may need to clarify that although the two legs of the triangle in the diagram are labeled \begin{align*}3\end{align*} and \begin{align*}4\end{align*}, the \begin{align*}x-\end{align*}coordinate we are working with is actually \begin{align*}-3\end{align*}, and so when finding the values of the trig functions, we must plug in \begin{align*}-3\end{align*} as the length of that leg. (This is where trig functions of angles of rotation start to differ from trig functions of angles in right triangles.)

**Points to Consider**

The Pythagorean Theorem is useful in trigonometry in at least two ways: it helps us find the third side of a right triangle when we need to, and it helps establish some important trigonometric identities. However, the latter won’t be covered for a couple more lessons, so you may or may not want to even mention it at this point.

Values of trig functions can be negative when we are dealing with angles of rotation instead of angles in right triangles, because we define the functions in a slightly different way to allow us to describe many more cases. Angles in right triangles must be less than \begin{align*}90^\circ\end{align*}, and when we work out the trig functions for those angles, we always get positive numbers because the triangles’ side lengths are always positive. But when we define the trig functions by reference to \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates and the unit circle, we now have a way of finding their values for angles greater than \begin{align*}90^\circ -\end{align*}and it turns out that some of those values are negative, because the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates we use to find them are sometimes negative. Similarly, trig values can be undefined when we try to find them for quadrantal angles, because some of the coordinates of those angles equal zero.

(All of this will be covered in more detail in the next lesson.)

The unit circle is useful because it gives us an easy way to calculate the trig functions for any given angle; then, because of similar triangles, we know that those values will be the same when we see that same angle in any triangle, even if the hypotenuse of the triangle is not \begin{align*}1\end{align*}. For example, the unit circle tells us that the cosine of \begin{align*}50^\circ\end{align*} is about \begin{align*}0.6428\end{align*}, so whenever we see a right triangle with a \begin{align*}50^\circ\end{align*} angle in it, we know (because the triangles are similar) that the ratio of the adjacent leg to the hypotenuse will always be \begin{align*}0.6428\end{align*}, without having to measure the sides.

\begin{align*}\frac{\text{adjacent}}{\text{hypotenuse}} & = 0.6428 \\ \text{adjacent}& = \text{hypotenuse} * 0.6428 \\ 15.2 & = \text{hypotenuse} * 0.6428 \\ \text{hypotenuse} & = \frac{15.2}{0.6428} \approx 23.6465\end{align*}

## Trigonometric Functions of Any Angle

**Reference Angles and Angles in the Unit Circle**

After example 1, you may need to show more explicitly how we know the ordered pair for 150° based on the ordered pair for \begin{align*}30^\circ\end{align*}. Remind students that a \begin{align*}150^\circ\end{align*} angle is the reflection across the \begin{align*}y-\end{align*}axis of a \begin{align*}30^\circ\end{align*} angle (refer to the earlier diagram), and remind them (and demonstrate visually) that when we reflect a point across the \begin{align*}y-\end{align*}axis, its \begin{align*}y-\end{align*}coordinate stays the same and its \begin{align*}x-\end{align*}coordinate changes sign.

Example 2 provides another opportunity to make this clear. Each angle that has \begin{align*}60^\circ\end{align*} as its reference angle is simply the angle we get if we reflect a \begin{align*}60^\circ\end{align*} angle across the \begin{align*}x-\end{align*}axis (putting it in the fourth quadrant), the \begin{align*}y-\end{align*}axis (putting it in the second quadrant), or both (putting it in the third quadrant). If we reflect it across the \begin{align*}x-\end{align*}axis, its \begin{align*}x-\end{align*} coordinate stays the same and its \begin{align*}y-\end{align*}coordinate changes sign; if we reflect it across the \begin{align*}y-\end{align*}axis, its \begin{align*}y-\end{align*}coordinate stays the same and its \begin{align*}x-\end{align*}coordinate changes sign; and if we do both, both coordinates change sign. So we can easily find the coordinates for any angle once we know its reference angle and which quadrant it is in.

**Trigonometric Function Values in Tables**

When you arrive at the table of trig function values, you may want to encourage students to compare the values of the trig functions for pairs of supplementary angles (like \begin{align*}85^\circ\end{align*} and \begin{align*}95^\circ\end{align*}, or \begin{align*}125^\circ\end{align*} and \begin{align*}55^\circ\end{align*}). The table makes it clear that cosines of supplementary angles are equal, and sines and tangents of supplementary angles are opposites. Thinking in terms of reference angles will make it clearer why this happens: an angle between \begin{align*}90^\circ\end{align*} and \begin{align*}180^\circ\end{align*} has a reference angle that is equal to its supplement, so the values of the trig functions for that angle are closely related to the values for its supplementary angle.

Because of this fact, there is another way to find the answer to example 6a; challenge students to figure out what it is. (Hint: what is the reference angle of \begin{align*}130^\circ\end{align*}?)

You may also want to encourage students to compare the sine and cosine values for pairs of complementary angles, like \begin{align*}35^\circ\end{align*} and \begin{align*}55^\circ\end{align*}; the table shows that the sine of an angle is equal to the cosine of its complement. This fact will be useful later, and the reason for it will be clearer when we study the unit circle in more detail.

**Points to Consider**

Here’s one way to explain the difference between the measure of an angle and its reference angle: when you start at the positive \begin{align*}x-\end{align*}axis and rotate counterclockwise to get to the terminal side of the angle, the distance you’ve traveled is the angle measure. When you start at the terminal side of the angle and travel by the quickest route to the closest portion of the \begin{align*}x-\end{align*}axis, the distance you’ve traveled is the reference angle. (Demonstrate this visually with at least one angle that is not in the first quadrant. For example, with a \begin{align*}240^\circ\end{align*} angle, you can show that \begin{align*}240^\circ\end{align*} is the clockwise distance from the positive \begin{align*}x-\end{align*}axis, but \begin{align*}60^\circ\end{align*} is the shortest distance to the closest part of the \begin{align*}x-\end{align*}axis.)

An angle is the same as its reference angle only when it is between \begin{align*}0^\circ\end{align*} and \begin{align*}90^\circ\end{align*}.

The simplest way to answer question #2 is by considering how we find values of trig functions on the unit circle. The values of sine and cosine there are simply equal to the \begin{align*}y-\end{align*}coordinate and \begin{align*}x-\end{align*}coordinate, respectively, of the ordered pair that defines the angle, so the angles that have the same (or opposite) sine (or cosine) value will simply be the ones with the same (or opposite) \begin{align*}y-\end{align*}coordinate (or \begin{align*}x-\end{align*}coordinate).

**Review Questions**

The function in problem 12 is fairly complex (although it can be simplified) and it shouldn’t be immediately obvious what the graph will look like. Students should be able to simplify the expression under the square root sign based on their conjecture from the previous problem; after that, the best they can do is figure out what the function’s values will be for a few key angles (\begin{align*}30^\circ, 45^\circ, 60^\circ\end{align*}, and so on), plot the points, and sketch a graph based on those points, and then graph the function on a calculator to compare it with their sketch.

## Relating Trigonometric Functions

**Reciprocal Identities**

After explaining that we obtain the reciprocal of a fraction by flipping the fraction, you might need to clarify how to obtain the reciprocal of something that is not a fraction. For example, the reciprocal of \begin{align*}x\end{align*} is \begin{align*}\frac{1}{x}\end{align*}, and that’s because \begin{align*}x\end{align*} is equivalent to the fraction \begin{align*}\frac{x}{1}\end{align*}, which we can flip to find the reciprocal. Similarly, a trig value like cos \begin{align*}\theta\end{align*} is equivalent to \begin{align*}\frac{\cos\ \theta}{1}\end{align*} , so if \begin{align*}\sec\ \theta\end{align*} is the reciprocal of \begin{align*}\cos\ \theta\end{align*}, that means it is equivalent to \begin{align*}\frac{1}{\cos\ \theta}\end{align*} .

The identity \begin{align*}1 = \sin^2\ x + \cos^2\ x\end{align*} may not be immediately obvious; it hasn’t previously been explicitly mentioned, although students were encouraged to discover it for themselves in Review Question 11 of the previous lesson. If you wish, you can demonstrate how to derive it from the definitions of sine and cosine and the Pythagorean Theorem: In right triangle ABC where c is the hypotenuse, by definition \begin{align*}\sin\ A = \frac{a}{c}\end{align*} and \begin{align*}\cos\ A = \frac{b}{c}\end{align*} ; therefore \begin{align*}\sin^2\ A + \cos^2\ A = \frac{a^2}{c^2} + \frac{b^2}{c^2}\end{align*} . Then if \begin{align*}\frac{a^2}{c^2} + \frac{b^2}{c^2} =1\end{align*}, multiplying through by \begin{align*}c^2\end{align*} yields \begin{align*}a^2 + b^2 = c^2\end{align*}, which we know is true because it is simply the Pythagorean Theorem.

**Domain, Range, and Signs of Functions**

It’s much easier to remember which trig functions are positive and negative in which quadrants if we simply note that in the first quadrant they are all positive; in the second quadrant only sin and its reciprocal, csc, are positive; in the third quadrant only tan and its reciprocal are positive; and in the fourth quadrant only cos and its reciprocal are positive. This can be summarized by the mnemonic “All Students Take Calculus”: All positive, Sin positive, Tan positive, Cos positive.

(The second-best way to figure out which functions are positive where is to think about which functions depend on the \begin{align*}x-\end{align*}coordinate, which ones depend on the \begin{align*}y-\end{align*}coordinate, and which ones depend on both, and then figure out which coordinates are negative depending on which quadrant we are considering. For example, in the third quadrant, where both coordinates are negative, sin and cos will be negative (and therefore so will sec and csc), but tan (and therefore cot also) will be positive because \begin{align*}\frac{x}{y}\end{align*} is positive when both x and y are negative. But this method takes longer, and is more prone to error, than simply using the mnemonic above.)

**Points to Consider**

It can often be easier to tell if an equation is *not* an identity: simply plug in a couple of values for the variables in the equation and see if the equation holds true. If it doesn’t, the equation is not an identity; if it does, the equation might or might not be.

Similarly, you can verify the domain and range of a function by plugging in numbers that are inside or outside the domain or range and seeing what happens. If a number is in the domain of the function, it should yield a sensible result when you substitute it for \begin{align*}x\end{align*}; if it’s not in the domain, it should not yield any real-number result. If a number is in the range of the function, you should be able to get it as a result when you plug in some value of \begin{align*}x\end{align*}; if it’s not in the range, you shouldn’t be able to for any value of \begin{align*}x\end{align*}.

## Applications of Right Triangle Trigonometry

**Solving Right Triangles**

If you are going through example 2 as a class, you may want to have students stop and think about how they might solve the triangle before walking them through the solution presented in the text. First, they should assess how much information they already have available—in this case, one side and two angles. (They may only notice the one angle measure that is written out in numerals—remind them that they also know the measure of the right angle.) Then they should think about what strategies they know for finding the other sides and angles (Pythagorean theorem, trig ratios), and which numbers they would need to plug in for each of those strategies. Finally, considering what numbers they actually have available to plug in should give them some idea of which strategies they could effectively use.

To answer the question posed in the solution: Using the tangent to find the third side is better than using the sine because using the tangent allows us to plug in the side we were given at the beginning, whose length we know precisely, rather than the side we just found, whose length we only know approximately. It’s always best to base our calculations on the most precise information we have available, so that rounding errors don’t accumulate.

**Angles of Elevation and Depression**

Clinometers are tools for measuring angles of elevation and depression; theodolites can measure horizontal angles as well. More information about both of these can be found on the Internet or in an encyclopedia.

For extra precision when measuring angles of elevation and depression, you should of course subtract several inches from your total height to estimate the distance from your eyes to the ground.

**Other Applications of Right Triangles**

An explanation of how we know the information given in example 7: We know the distance between the moon and the earth based on calculations that will be explained later in the book. We know the angle between the moon and the sun at a given time because we can measure it directly from our vantage point on the earth. And finally, we know that the moon makes a right angle with the earth and sun at the first quarter (when the moon is halfway full) because exactly half of the portion of the moon that we can see is lit up by the sun, meaning that the sun must be shining exactly “sideways” on the moon.

**Points to Consider**

In addition to the situations described in this lesson, we also might use right triangles to determine the shortest distance between two points on a grid (like a grid of city blocks), or to determine how long a ladder we need to reach a certain height on a building.

Any right triangle can be solved if we have enough information; at minimum, we need to know the length of at least two sides, or one side and one angle besides the right angle.

Trigonometry can solve problems at any scale because the trig ratios are the same for any size triangle as long as the angles are the same.

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