Next Generation Advanced Algebra and Functions Study Guide for the ACCUPLACER Test

Trigonometry

Trigonometry explores the relationships between the angles and side lengths of triangles.

A useful introduction to the topic is the unit circle, which relates the degree measurement of a central angle, so the trigonometric ratios and values associated with that central angle measurement. Much of the unit circle grows out of the Pythagorean theorem and its coupling with the trigonometric ratios for the sine, cosine, and tangent functions.

When learning the unit circle, it is highly recommended to develop an understanding of how the relationships emerge, and how the coordinate points arise. Being able to derive the unit circle is far more practical than memorizing each of the angle measurements and their corresponding coordinate values.

Trigonometric Equations

A trigonometric equation is any equation involving a trigonometric function. All of the methods of solving equations from algebra still apply to trigonometric equations; however, because sine, cosine, etc. are functions, there are a couple of tricks to be learned to arrive at a final answer.

Consider this example:

If \(\cos\;\theta = \frac{-4}{5}\) and \(90^\circ < \theta < 180^\circ\), what is \(\sin\;\theta =\)?

The cosine function, which is equal to the ratio of the adjacent side of an angle to the hypotenuse opposite that angle, is negative in the range given. This informs us that the adjacent side is to the left of the origin. A picture will illustrate:

Based on the diagram, \(\sin\;\theta = \frac{opp}{hyp} = \frac{3}{5}\).

Consider the following equation, for which we will solve for \(x\):

\[2\cos\;x - 1 = 0\]

As always, when isolating a variable, it is necessary to undo the operations performed on that variable, in the reverse order in which those operations are performed on the variable. It is straightforward to solve for \(\cos\;x\):

\[\cos\;x = \frac{1}{2}\]

But here we encounter an issue. How do we undo the cosine function? To undo a trig. function, the inverse of that trig. function is used. On your calculator, this operation appears as \(\sin^{-1}\), which is the same as \(\arcsin\), or \(\cos^{-1}\), which is the same as \(\arccos\), etc.

So, to isolate the \(x\):

\[\cos^{-1}\;(\cos\;x) = \cos^{-1}(\frac{1}{2})\]

which gives

\[x = \cos^{-1}(\frac{1}{2}) = 60^\circ\]

The final calculation can be performed with a calculator or by way of the unit circle.

Right Triangle Trigonometry

Right triangles exhibit a special relationship between their angle measurements and side lengths. Note the following diagram:

The side opposite the right angle is always called the hypotenuse. For any specific angle in question, in this case, \(\theta\), the side opposite \(\theta\) is referred to as such, and the side next to the angle is called adjacent. Notice that if \(\theta\) was switched to the other acute angle, the opposite and adjacent sides would also switch.

The side lengths and angle measurements are related through the following definitions:

\[\sin\;\theta = \frac{opposite}{hypotenuse}\]

(where opposite is the side length opposite angle \(\theta\) and hypotenuse is side length opposite the right angle)

\[\cos\;\theta = \frac{adjacent}{hypotenuse}\] \[\tan\;\theta = \frac{opposite}{adjacent}\]

These relationships can be easily remembered using the mnemonic SOHCAHTOA: sin is opposite over hypotenuse, cos is adjacent over hypotenuse, and tan is opposite over adjacent.

Equivalent Trigonometric Functions

Many trigonometric functions can be rewritten in terms of other trigonometric functions. It is worthwhile to practice enough trigonometry problems such that each of these relationships is easily recalled.

By definition:

\[\sin\;\theta = \frac{o}{h}\] \[\cos\;\theta = \frac{a}{h}\] \[\tan\;\theta = \frac{\sin\;\theta}{\cos\;\theta}\] \[\csc\;\theta = \frac{1}{\sin\;\theta}\] \[\sec\;\theta = \frac{1}{\cos\;\theta}\] \[\cot\;\theta = \frac{\cos\;\theta}{\sin\;\theta}\]

Recall from the Pythagorean theorem that \(a^2 + b^2 = c^2\). This theorem can be written in terms of the trig. functions:

\[\cos^2{\theta} + \sin^2{\theta} = 1\] \[\tan^2{\theta} + 1 = \sec^2{\theta}\] \[1 + \cot^2{\theta} = \csc^2{\theta}\]

The cofunction formulas enable one function to be written as another function:

\[\sin({\frac{\pi}{2} - \theta}) = \cos{\theta}\] \[\cos({\frac{\pi}{2} - \theta}) = \sin{\theta}\] \[\tan({\frac{\pi}{2} - \theta}) = \cot{\theta}\] \[\csc({\frac{\pi}{2} - \theta}) = \sec{\theta}\] \[\sec({\frac{\pi}{2} - \theta}) = \csc{\theta}\]

The double angle formulas enable double angles to be rewritten:

\[\sin(2\theta) = 2\sin\theta\cos\theta\] \[\cos(2\theta) = cos^2\theta - sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta\] \[\tan(2\theta) = \frac{2\tan\theta}{1 - tan^2\theta}\]

The equations presented above, as well as the half-angle formulas, the sum and difference formulas, sum to product, and product to sum formulas all enable fluid transitions from one trigonometric expression to another. Doing so often reduces the complexity of work required to solve a particular problem.

Graphing Trigonometric Relationships

It is worthwhile to commit the general forms of each trigonometric function to memory. Familiarization with each of these general forms combined with an understanding of amplitude and transformations enables one to quickly deduce which trig. function is shown on a graph, or which trig. function would best represent the data.

\(\sin x\) graphs as:

\(\cos x\) graphs as:

\(\tan x\) graphs as:

\(\sec x\) graphs as:

\(\csc x\) graphs as:

\(\cot x\) graphs as:

In addition to these general forms, it is useful to develop an idea of the terminology associated with these functions, as well as the graphs produced from various transformations.

First, terminology:

Given the graphs presented above, it should be no surprise that trigonometric functions repeat, or cycle; they are periodic. With these periods comes a set of related terms:

The amplitude is the magnitude of the difference between the highest or lowest point and the center line passing through the function. If the highest point is thought of as a peak, and the lowest point is thought of as a trough, the amplitude is the magnitude of the difference between the peak and the trough divided by \(2\).

A period is the amount of time or length required for the function to repeat itself.

Frequency is inversely related to period.

A phase shift is a horizontal translation from the usual graph of the function.

Likewise, a vertical shift is a vertical translation from the usual graph of the function.

Each of these terms shows up trigonometrically in the algebraic representation of a trigonometric function. Generally:

\(y = A\sin {Bx + C} + D\), where A is the amplitude, \(\frac{2\pi}{B}\) is the period, \(-\frac{C}{B}\) is the phase shift, and D is the vertical shift.

For example:

\(y = 1.5\sin{8x + 3} - 5\), has an amplitude of \(1.5\), a period of \(\frac{2\pi}{8} = \frac{\pi}{4}\), a phase shift of \(-\frac{3}{8}\), and a vertical shift of \(-5\). The function graphs as follows:

Arc Length

Consider this circle, with a central angle measured \(\theta\), and arc length \(s\):

Using the definition of the circumference, \(C = \pi \cdot d = 2 \cdot \pi \cdot r\), a relationship between arc length, \(s\), and \(\angle{\theta}\) can be derived.

Given that \(C = 2\pi r\), \(2\pi = \frac{C}{r}\).

We also know that \(2\pi\) is equal to \(360^\circ\). From this relationship, we can produce a generalized relationship between the central angle, the radius, and the arc length of a circle:

\(\theta = \frac{s}{r}\), where \(\theta\) is the central angle, \(s\) is the arc length, and \(r\) is the radius.

Radian Measures

An angle measurement can be made in degrees or in radians. It is useful to be able to convert between the two. We will use a unit circle, a circle with a radius of one, to clarify the ideas of degrees and radians.

As you know, a \(90^\circ\) angle is the result of two perpendicular axes. Multiplying this value by four generates the total angle measurement of a circle:

\(4 \cdot 90^\circ = 360^\circ\).

And, from the definition of the circumference of a circle, with a radius of \(1\), the distance around the circle can be computed as:

\(C = 2 \cdot \pi \cdot r\), or \(C = 2 \cdot \pi\)

An equivalence can be drawn between the total angle measurement and total distance around a circle as follows:

\[360^\circ = 2 \cdot \pi\]

A radian can be thought of as the ratio of the total degree measurement of a circle to the circumference of that circle. Using the equivalence described above, we can define a radian as:

\(1\;rad = \frac{360^\circ}{2 \cdot \pi}\), which reduces to:

\[1\;rad = \frac{180^\circ}{\pi}\]

This equivalence can be used as a conversion factor when converting degrees to radians, or radians to degrees. For example, \(30^\circ\) can be converted to radians as follows:

\[30^\circ \cdot \frac{\pi}{180^\circ} = \frac{\pi}{6}\;rad\]

In a similar manner, a radian measurement, like \(\frac{\pi}{3}\) can be converted to degrees as follows:

\[\frac{\pi}{3} \cdot \frac{180^\circ}{\pi} = 60^\circ\]

Law of Sines

The law of sines describes a proportionality among the ratios of a triangle’s side length to the sine of the side’s corresponding angle; it is applicable to every triangle. Stated algebraically:

\(\frac{a}{sin{\;A}} = \frac{b}{sin{\;B}} = \frac{c}{sin{\;C}}\), where \(a\), \(b\), and \(c\) are the sides of the triangle, and \(A\), \(B\), and \(C\) are the corresponding angles of the triangle.

Consider this triangle:

(picture not to scale)

Using the law of sines, \(B\) and \(C\) can be found:

\(\frac{9.05}{\sin{\;32.2}}=\frac{9.08}{\sin{\;B}}\), and solving for \(\sin{\;B}\),

\(\sin{\;B} = 9.08(\frac{\sin{\;32.2}}{9.05})\), and evaluating the inverse sine of both sides to isolate \(B\):

\(B = \arcsin{(\;9.08(\frac{\sin{\;32.2}}{9.05}}))\), using a calculator,

\[B \approx 32.32^\circ\]

\(\frac{9.05}{\sin{\;32.2}}=\frac{7.42}{\sin{\;C}}\), solving for \(\sin{\;C}\):

\(\sin{\;C} = 7.42(\frac{\sin{\;32.2}}{9.05})\), and evaluating arcsine of both sides:

\(C = \arcsin{(\;7.42(\frac{\;sin{\;32.2}}{9.05})}\), using a calculator,

\[C \approx 25.9^\circ\]

Law of Cosines

Similar to the law of sines, the law of cosines can be used to solve a triangle when two sides and the angle between them is known, or when three sides of a triangle are known.

In the case of a triangle like this:

the law of cosines is stated in the following equivalent forms:

\[a^2 = b^2 + c^2 - 2bc \cos\;A\] \[b^2 = a^2 + c^2 - 2ac \cos\;B\]

\(c^2 = a^2 + b^2 - 2ab \cos\;C\), where \(a\), \(b\), and \(c\) are side lengths, and \(A\), \(B\), and \(C\) are their corresponding angles.

We will use the following triangle to solve for the side length of \(a\):

\(a^2 = 9^2 + 13^2 - 2(9)(13) \cos\;32^\circ\), which simplifies to

\(a^2 = 81 + 169 - 234(0.848)\), and \(a \approx \sqrt{51.56} = 7.18\)