Chapter 2: Graphing Trigonometric Functions
 2.1.
Radian Measure
 2.2.
Applications of Radian Measure
 2.3.
Circular Functions of Real Numbers
 2.4.
Translating Sine and Cosine Functions
 2.5.
Amplitude, Period and Frequency
 2.6.
General Sinusoidal Graphs
 2.7.
Graphing Tangent, Cotangent, Secant, and Cosecant
Chapter Summary
Chapter Summary
In this chapter we learned about another way to measure degrees, called radians. Radians are based off of the unit circle, which is a circle with a radius of one. Because all circles are similar, it doesn’t matter which one we use to measure radians, the ratios will always reduce to be the same. Therefore, we can now graph the six trigonometric functions on the \begin{align*}xy\end{align*}
Vocabulary
 Amplitude

A dilation on the \begin{align*}y\end{align*}
y− value of a trigonometric function. \begin{align*}A\end{align*}A is multiplied by the function, to make it stretch or shorten.
 Angular Velocity
 The velocity of rotation, measured in radians.
 Arc Length

The length of a portion of the circumference of a circle. The formula is \begin{align*}\theta \cdot 2\pi r\end{align*}
θ⋅2πr , where \begin{align*}\theta\end{align*}θ is the corresponding central angle, in radians.
 Circular Function
 The collective term for a function that can be defined by the unit circle.
 Critical Angle(s)

Any angle that is a multiple of \begin{align*}30^\circ\end{align*}
30∘ or \begin{align*}45^\circ\end{align*}45∘ .
 Dilation
 A transformation that changes the size of an object or function.
 Frequency

The number of times the graph repeats in \begin{align*}2\pi\end{align*}
2π or \begin{align*}\pi\end{align*}π for tangent and cotangent.
 Period
 The distance it takes a graph to complete one phase.
 Phase Shift

The shift, or translation, in the \begin{align*}x\end{align*}
x− direction of a trigonometric function. Also called a horizontal translation.
 Radian

An alternate way to measure degrees, defined by the arc length on a circle that is equal to the radius. \begin{align*}360^\circ = 2\pi\end{align*}
360∘=2π radians.
 Sector

The area of a portion of a circle. The formula is \begin{align*}\theta \cdot \pi r^2\end{align*}
θ⋅πr2 , where \begin{align*}\theta\end{align*}θ is the central angle, measured in radians.
 Transformation
 Any change made to an object or graph. Transformations can either be dilations or translations.
 Translation
 Either a vertical or horizontal movement of an object or function.
 Vertical Shift
 The vertical translation of a function.
Review Questions
 Convert \begin{align*}160^\circ\end{align*}
160∘ to radians.  Convert \begin{align*}\frac{11\pi}{12}\end{align*}
11π12 to degrees.  Find the exact value of \begin{align*}\cos \frac{3\pi}{4}\end{align*}
cos3π4 .  Find all possible answers in radians, between \begin{align*}0 < \theta < 2\pi: \tan \theta =\sqrt{3}\end{align*}
0<θ<2π:tanθ=3√  This is an image of the state flag of Colorado
It turns out that the diameter of the gold circle is \begin{align*}\frac{1}{3}\end{align*}
13 the total height of the flag (the same width as the white stripe) and the outer diameter of the red circle is \begin{align*}\frac{2}{3}\end{align*}23 of the total height of the flag. The angle formed by the missing portion of the red band is \begin{align*}\frac{\pi}{4}\end{align*}π4 radians. In a flag that is 66 inches tall, what is the area of the red portion of the flag to the nearest square inch?  Suppose the radius of the dial of an electric meter on a house is 7 cm.
 How fast is a point on the outside edge of the dial moving if it completes a revolution in 9 seconds?
 Find the angular velocity of a point on the dial.
 In the figure below, there is a quadrilateral formed by four line segments: the two radii of the circle (in pink), the orange segment (marked as “cotangent”), and the green segment (marked as “tangent”). The tangent (green) segment has been constructed as tangent to the circle (forming a 90degree angle with the radius). How do you know that the number of units that is the length of the cotangent segment is equal to \begin{align*}\frac{x}{y}\end{align*}
xy ? You may assume that the radii shown (pink) are 1 unit.  Graph \begin{align*}y=\sin x\end{align*}
y=sinx and \begin{align*}y=\cos x\end{align*}y=cosx on the same set of axes over the interval \begin{align*}[0, 2\pi]\end{align*}[0,2π] . Where do they intersect?
For questions 912, determine the amplitude, period, frequency, vertical shift, and phase shift. Then, graph each function over the interval \begin{align*}[0, 2\pi]\end{align*}

\begin{align*}y=2+4 \sin 5x\end{align*}
y=−2+4sin5x 
\begin{align*}f(x)=\frac{1}{4} \cos \left(\frac{1}{2}(x\frac{\pi}{3}\right))\end{align*}
f(x)=14cos(12(x−π3)) 
\begin{align*}g(x)=4+\tan \left(2(x+\frac{\pi}{2}\right))\end{align*}
g(x)=4+tan(2(x+π2)) 
\begin{align*}h(x)=36 \cos (\pi x)\end{align*}
h(x)=3−6cos(πx)
For questions 13 and 14, find the equation of the graph below. Only sine and cosine functions will have an amplitude other than 1.
Review Answers

\begin{align*}160^\circ \cdot \frac{\pi}{180^\circ}=\frac{16\pi}{18}=\frac{8\pi}{9}\end{align*}
160∘⋅π180∘=16π18=8π9 
\begin{align*}\frac{11\pi}{12} \cdot \frac{180^\circ}{\pi}=11 \cdot 15^\circ=165^\circ\end{align*}
11π12⋅180∘π=11⋅15∘=165∘ 
\begin{align*}\cos \frac{3\pi}{4}= \cos 135^\circ=\frac{\sqrt{2}}{2}\end{align*}
cos3π4=cos135∘=−2√2  For \begin{align*}\tan \theta = \sqrt{3}\end{align*}
tanθ=3√ , \begin{align*}\theta\end{align*}θ must equal \begin{align*}60^\circ\end{align*}60∘ or \begin{align*}240^\circ\end{align*}240∘ . In radians, \begin{align*}\frac{\pi}{3}\end{align*}π3 or \begin{align*}\frac{4\pi}{3}\end{align*}4π3 .  There are many difference approaches to the problem. Here is one possibility: First, calculate the area of the red ring as if it went completely around the circle: \begin{align*}A &= A_{total}A_{gold}\\
A &= \pi \left(\frac{2}{3} \times 66 \times \frac{1}{2} \right)^2  \pi \left(\frac{1}{3} \times 66 \times \frac{1}{2} \right)^2\\
A &= \pi \times 22^2  \pi \times 11^2\\
A &= 484 \pi  121 \pi = 363 \pi\\
A &\approx 1140.4 \ in^2\end{align*}
AAAAA=Atotal−Agold=π(23×66×12)2−π(13×66×12)2=π×222−π×112=484π−121π=363π≈1140.4 in2 Next, calculate the area of the total sector that would form the opening of the “\begin{align*}c\end{align*}” \begin{align*}A &= \frac{1}{2} r^2 \theta\\ A &= \frac{1}{2} (22)^2 \left( \frac{\pi}{4} \right)\\ A &\approx 190.1 \ in^2\end{align*} Then, calculate the area of the yellow sector and subtract it from the previous answer. \begin{align*}& A=\frac{1}{2}r^2 \theta \rightarrow A=\frac{1}{2}(11)^2 \left( \frac{\pi}{4} \right) \rightarrow A \approx 47.5 \ in^2\\ & 190.1  47.5 = 142.6 \ in^2\end{align*} Finally, subtract this answer from the first area calculated. The area is approximately \begin{align*}998 \ in^2\end{align*} First find the circumfrence: \begin{align*}2\pi \cdot 7 = 14\pi\end{align*}. This will be the distance for the linear velocity. \begin{align*}v=dt=14 \pi \cdot 9=126 \pi \approx 395.84 \ cm/sec\end{align*}
 \begin{align*}\omega=\frac{\theta}{t}=\frac{2\pi}{9} \approx 0.698 \ rad/sec\end{align*}
 Given such a quadrilateral, and given that the two transverse angles are identified as equal (i.e., both are marked as \begin{align*}\theta\end{align*} in the picture), the orange segment must be parallel to the opposite (pink) radius segment, and this quadrilateral would have to be a square. This means that \begin{align*}\theta\end{align*} must be equal to 45 degrees, and both the tangent and cotangent of 45 degrees are equal to 1. Also, since the radii of the circle are equal to 1 unit, each of the sides of the quadrilateral (including the cotangent segment) are equal to 1 unit. Therefore, since cot \begin{align*}\theta\end{align*} = \begin{align*}\frac{x}{y}\end{align*}, the number of units that is the length of the cotangent segment must be equal to \begin{align*}\frac{x}{y}\end{align*}.
The intersections are \begin{align*}\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)\end{align*} and \begin{align*}\left(\frac{5\pi}{4}, \frac{\sqrt{2}}{2}\right)\end{align*}.
 \begin{align*}y=2+4 \sin 5x, A=4, B=5, p=\frac{2\pi}{5}, C=0, D=2\end{align*}
 \begin{align*}f(x)=\frac{1}{4} \cos \left(\frac{1}{2}(x\frac{\pi}{3} \right)), A=\frac{1}{4}, B=\frac{1}{2}, p=4 \pi, C=\frac{\pi}{3}, D=0\end{align*}
 \begin{align*}g(x)=4+ \tan \left(2(x+\frac{\pi}{2}\right)), A=1, B=2, p=\frac{\pi}{2}, C=\frac{\pi}{2}, D=4\end{align*}
 \begin{align*}h(x)=36 \cos (\pi x), A=6, B=\pi, C=0, D=3\end{align*}
 \begin{align*}y=1+\frac{1}{2} \cos 3x\end{align*}
 \begin{align*}y=\tan 6x\end{align*}
Texas Instruments Resources
In the CK12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9700.