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Section 6.5 The Unit Circle (TR5)
Objectives
Use reference angles, signs and definitions to determine exact values of trigonometric functions.
Subsection 6.5.1 Activities
Definition 6.5.2 .
The
unit circle is the circle of radius
\(1\) centered at the origin on the coordinate plane.
Figure 6.5.3.
Activity 6.5.4 .
Let
\(\theta\) be the angle shown below in standard form. Notice that the terminal side intersects with the unit circle. (Note: We will assume a circle drawn in this context is the unit circle unless told otherwise.) We will label that point of intersection as
\((x,y)\text{.}\)
(a)
What is the length of line segment \(r\text{,}\) whose endpoints are the origin and the point \((x,y)\) ?
\(\displaystyle 1\)
\(\displaystyle 2\)
\(\displaystyle 3\)
cannot be determined
(b)
We will now create a right triangle using the previous line segment
\(r\) as the hypotenuse. Draw in a line segment of length
\(x\) and another of length
\(y\) to create such a triangle.
(c)
Using the triangle youβve just created, find \(\cos \theta\text{.}\)
\(\displaystyle \dfrac{x}{y}\)
\(\displaystyle \dfrac{1}{x}\)
\(\displaystyle \dfrac{x}{1}\)
\(\displaystyle \dfrac{1}{y}\)
\(\displaystyle \dfrac{y}{1}\)
(d)
Using that same triangle, find \(\sin \theta\text{.}\)
\(\displaystyle \dfrac{x}{y}\)
\(\displaystyle \dfrac{1}{x}\)
\(\displaystyle \dfrac{x}{1}\)
\(\displaystyle \dfrac{1}{y}\)
\(\displaystyle \dfrac{y}{1}\)
(e)
Solve for \(x\) in one of the equations youβve found above to determine an expression for the \(x\) -value of the point \((x,y)\) .
\(\displaystyle y\cos \theta\)
\(\displaystyle y\sin \theta\)
\(\displaystyle \cos \theta\)
\(\displaystyle \sin \theta\)
\(\displaystyle \dfrac{1}{\cos \theta}\)
\(\displaystyle \dfrac{1}{\sin \theta}\)
(f)
Solve for \(y\) in one of the equations youβve found above to determine an expression for the \(y\) -value of the point \((x,y)\) .
\(\displaystyle y\cos \theta\)
\(\displaystyle y\sin \theta\)
\(\displaystyle \cos \theta\)
\(\displaystyle \sin \theta\)
\(\displaystyle \dfrac{1}{\cos \theta}\)
\(\displaystyle \dfrac{1}{\sin \theta}\)
Activity 6.5.6 .
Consider each angle
\(\theta\) given below. Find the coordinates
\((x,y)\) for the point at which
\(\theta\) intersects the unit circle.
(a)
\(\theta = \dfrac{\pi}{4}\)
\(\displaystyle \left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\)
\(\displaystyle (0,1)\)
\(\displaystyle (1,0)\)
(b)
\(\theta = 30^\circ\)
\(\displaystyle \left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\)
\(\displaystyle (0,1)\)
\(\displaystyle (1,0)\)
(c)
\(\theta = \dfrac{\pi}{3}\)
\(\displaystyle \left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\)
\(\displaystyle (0,1)\)
\(\displaystyle (1,0)\)
(d)
\(\theta = 0^\circ\)
\(\displaystyle \left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\)
\(\displaystyle (0,1)\)
\(\displaystyle (1,0)\)
(e)
\(\theta = \dfrac{\pi}{2}\)
\(\displaystyle \left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)\)
\(\displaystyle \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\)
\(\displaystyle (0,1)\)
\(\displaystyle (1,0)\)
Activity 6.5.8 .
Letβs consider the angle
\(\theta=150^\circ\text{,}\) drawn below with the unit circle.
(a)
If we reflect this angle across the
\(y\) -axis, we can obtain an angle
\(\alpha\) in the first quadrant. What is the measure of
\(\alpha\text{?}\)
\(\displaystyle 0^\circ\)
\(\displaystyle 30^\circ\)
\(\displaystyle 45^\circ\)
\(\displaystyle 60^\circ\)
\(\displaystyle 75^\circ\)
(b)
We can find the sine and cosines values of our original angle,
\(\theta=150^\circ\text{,}\) by using the angle
\(\alpha=30^\circ\) to help. Find the point
\((x_1,y_1)\text{,}\) where the terminal side of the
\(30^\circ\) angle intersects the unit circle.
\(\displaystyle \left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\)
\(\displaystyle \left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)\)
\(\displaystyle \left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)\)
\(\displaystyle \left(\dfrac{\sqrt{3}}{2},-\dfrac{1}{2}\right)\)
\(\displaystyle \left(\dfrac{-\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)\)
(c)
How does the point youβve just found compare with the point
\((x,y)\text{,}\) where the terminal edge of
\(\theta=150^\circ\) intersects the unit circle?
The \(x\) -values and the \(y\) -values are switched with each other.
The \(x\) -values will be the same, but the \(y\) -values will have opposite signs.
The \(y\) -values will be the same, but the \(x\) -values will have opposite signs.
The \(x\) -values and the \(y\) -values will both have opposite signs.
(d)
What are the cosine and sine values of
\(\theta=150^\circ\text{?}\)
\(\cos 150^\circ = \dfrac{\sqrt{3}}{2}\) and \(\sin 150^\circ = \dfrac{1}{2}\)
\(\cos 150^\circ = \dfrac{1}{2}\) and \(\sin 150^\circ = \dfrac{\sqrt{3}}{2}\)
\(\cos 150^\circ = \dfrac{1}{2}\) and \(\sin 150^\circ = -\dfrac{\sqrt{3}}{2}\)
\(\cos 150^\circ = -\dfrac{\sqrt{3}}{2}\) and \(\sin 150^\circ = \dfrac{1}{2}\)
\(\cos 150^\circ = \dfrac{\sqrt{3}}{2}\) and \(\sin 150^\circ = -\dfrac{1}{2}\)
Definition 6.5.9 .
The
reference angle for a given angle
\(\theta\) is the angle in the first quadrant obtained from reflecting
\(\theta\text{.}\) Equivalently, it is the smallest angle between the terminal side of
\(\theta\) and the
\(x\) -axis.
Activity 6.5.10 .
Letβs consider the angle
\(\theta=\dfrac{4\pi}{3}\text{,}\) drawn below with the unit circle.
(a)
The angle below represents the reference angle for
\(\theta=\dfrac{4\pi}{3}\text{,}\) which is the smallest angle between the terminal side of
\(\theta\) and the
\(x\) -axis. What is the measure of this reference angle?
\(\displaystyle \dfrac{\pi}{2}\)
\(\displaystyle \dfrac{\pi}{3}\)
\(\displaystyle \dfrac{\pi}{4}\)
\(\displaystyle \dfrac{\pi}{5}\)
\(\displaystyle \dfrac{\pi}{6}\)
(b)
We can find the sine and cosines values of our original angle,
\(\theta=\dfrac{4\pi}{3}\text{,}\) by using the reference angle to help. Find the point
\((x_1,y_1)\text{,}\) where the terminal side of the angle
\(\dfrac{\pi}{3}\) intersects the unit circle.
\(\displaystyle \left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\)
\(\displaystyle \left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)\)
\(\displaystyle \left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)\)
\(\displaystyle \left(\dfrac{\sqrt{3}}{2},-\dfrac{1}{2}\right)\)
\(\displaystyle \left(\dfrac{-\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)\)
(c)
How does the point youβve just found compare with the point
\((x,y)\text{,}\) where the terminal edge of
\(\theta=\dfrac{4\pi}{3}\) intersects the unit circle?
The \(x\) -values and the \(y\) -values are switched with each other.
The \(x\) -values will be the same, but the \(y\) -values will have opposite signs.
The \(y\) -values will be the same, but the \(x\) -values will have opposite signs.
The \(x\) -values and the \(y\) -values will both have opposite signs.
(d)
What are the cosine and sine values of
\(\theta=\dfrac{4\pi}{3}\text{?}\)
\(\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}\) and \(\sin \dfrac{4\pi}{3} = \dfrac{1}{2}\)
\(\cos \dfrac{4\pi}{3} = -\dfrac{1}{2}\) and \(\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}\)
\(\cos \dfrac{4\pi}{3} = \dfrac{1}{2}\) and \(\sin \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}\)
\(\cos \dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}\) and \(\sin \dfrac{4\pi}{3} = \dfrac{1}{2}\)
\(\cos \dfrac{4\pi}{3} = \dfrac{\sqrt{3}}{2}\) and \(\sin \dfrac{4\pi}{3} = -\dfrac{1}{2}\)
Activity 6.5.11 .
Find
\(\sin \theta\) and
\(\cos \theta\) for each angle given.
(a)
\(\theta = \dfrac{\pi}{4}\)
Answer .
\(\sin \theta = \dfrac{\sqrt{2}}{2}\) and
\(\cos \theta = \dfrac{\sqrt{2}}{2}\)
(b)
\(\theta = \dfrac{2\pi}{3}\)
Answer .
\(\sin \theta = \dfrac{\sqrt{3}}{2}\) and
\(\cos \theta = -\dfrac{1}{2}\)
(c)
\(\theta = \dfrac{11\pi}{6}\)
Answer .
\(\sin \theta = -\dfrac{1}{2}\) and
\(\cos \theta = \dfrac{\sqrt{3}}{2}\)
(d)
Answer .
\(\sin \theta = \dfrac{\sqrt{2}}{2}\) and
\(\cos \theta = -\dfrac{\sqrt{2}}{2}\)
Activity 6.5.12 .
Find the following for each angle graphed below.
\(\theta\) in radians and degrees
\(\displaystyle \sin \theta\)
\(\displaystyle \cos \theta\)
\(\displaystyle \tan \theta\)
\(\displaystyle \sec \theta\)
\(\displaystyle \csc \theta\)
\(\displaystyle \cot \theta\)
(a)
Answer .
\(\theta=\dfrac{5\pi}{6}\) or \(150^\circ\)
\(\displaystyle \sin \theta = \dfrac{1}{2}\)
\(\displaystyle \cos \theta = -\dfrac{\sqrt{3}}{2}\)
\(\tan \theta = -\dfrac{\sqrt{3}}{3}\) or \(-\dfrac{1}{\sqrt{3}}\)
\(\sec \theta = -\dfrac{2\sqrt{3}}{3}\) or \(-\dfrac{2}{\sqrt{3}}\)
\(\displaystyle \csc \theta = 2\)
\(\displaystyle \cot \theta = -\sqrt{3}\)
(b)
Answer .
\(\theta=\dfrac{5\pi}{4}\) or \(225^\circ\)
\(\displaystyle \sin \theta = -\dfrac{\sqrt{2}}{2}\)
\(\displaystyle \cos \theta = -\dfrac{\sqrt{2}}{2}\)
\(\displaystyle \tan \theta = 1\)
\(\sec \theta = -\sqrt{2}\) or \(-\dfrac{2}{\sqrt{2}}\)
\(\csc \theta = -\sqrt{2}\) or \(-\dfrac{2}{\sqrt{2}}\)
\(\displaystyle \cot \theta = 1\)
Activity 6.5.14 .
A point
\((x,y)\) lies on the unit circle in Quadrant IV. Its
\(x\) -coordinate is
\(\dfrac{3}{4}\text{.}\)
(a)
Draw a sketch of the angle
\(\theta\) whose terminal side intersects the unit circle as described above.
(b)
What sign will the \(y\) -coordinate be?
positive
negative
(c)
Find the exact value of the \(y\) -coordinate.
\(\displaystyle \dfrac{7}{16}\)
\(\displaystyle -\dfrac{7}{16}\)
\(\displaystyle \dfrac{\sqrt{7}}{4}\)
\(\displaystyle -\dfrac{\sqrt{7}}{4}\)
Hint .
Use the Pythagorean Theorem to help.
Answer .
(d)
Find
\(\sin \theta\text{,}\) \(\cos \theta\text{,}\) \(\tan \theta\text{,}\) \(\sec \theta\text{,}\) \(\csc \theta\text{,}\) and
\(\cot \theta\text{.}\)
Answer .
\(\displaystyle \sin \theta = -\dfrac{\sqrt{7}}{4}\)
\(\displaystyle \cos \theta = \dfrac{3}{4}\)
\(\displaystyle \tan \theta = -\dfrac{\sqrt{7}}{3}\)
\(\displaystyle \sec \theta = \dfrac{4}{3}\)
\(\csc \theta = -\dfrac{4}{\sqrt{7}}\) or \(-\dfrac{4\sqrt{7}}{7}\)
\(\cot \theta = -\dfrac{3}{\sqrt{7}}\) or \(-\dfrac{3\sqrt{7}}{7}\)
Activity 6.5.15 .
Let
\(\theta\) be the angle whose terminal side intersects the unit circle at the point described in each situation below. Find
\(\sin \theta\text{,}\) \(\cos \theta\text{,}\) \(\tan \theta\text{,}\) \(\sec \theta\text{,}\) \(\csc \theta\text{,}\) and
\(\cot \theta\text{.}\)
(a)
The point
\(\left(\dfrac{4}{5},y\right)\) that lies on the unit circle in Quadrant I.
Answer .
\(\displaystyle \sin \theta = \dfrac{3}{5}\)
\(\displaystyle \cos \theta = \dfrac{4}{5}\)
\(\displaystyle \tan \theta = \dfrac{3}{4}\)
\(\displaystyle \sec \theta = \dfrac{5}{4}\)
\(\displaystyle \csc \theta = \dfrac{5}{3}\)
\(\displaystyle \cot \theta = \dfrac{4}{3}\)
(b)
The point
\(\left(-\dfrac{35}{37},\dfrac{12}{37}\right)\text{.}\)
Answer .
\(\displaystyle \sin \theta = \dfrac{12}{37}\)
\(\displaystyle \cos \theta = -\dfrac{35}{37}\)
\(\displaystyle \tan \theta = -\dfrac{12}{35}\)
\(\displaystyle \sec \theta = -\dfrac{37}{35}\)
\(\displaystyle \csc \theta = \dfrac{37}{12}\)
\(\displaystyle \cot \theta = -\dfrac{35}{12}\)
(c)
The point
\(\left(x,-\dfrac{\sqrt{11}}{6}\right)\) that lies on the unit circle in Quadrant III.
Answer .
\(\displaystyle \sin \theta = -\dfrac{\sqrt{11}}{6}\)
\(\displaystyle \cos \theta = -\dfrac{5}{6}\)
\(\displaystyle \tan \theta = \dfrac{\sqrt{11}}{5}\)
\(\displaystyle \sec \theta = -\dfrac{6}{5}\)
\(\csc \theta = -\dfrac{6}{\sqrt{11}}\) or \(-\dfrac{6\sqrt{11}}{11}\)
\(\cot \theta = \dfrac{5}{\sqrt{11}}\) or \(\dfrac{5\sqrt{11}}{11}\)
(d)
The point
\(\left(\dfrac{12}{13},-\dfrac{5}{13}\right)\text{.}\)
Answer .
\(\displaystyle \sin \theta = -\dfrac{5}{13}\)
\(\displaystyle \cos \theta = \dfrac{12}{13}\)
\(\displaystyle \tan \theta = -\dfrac{5}{12}\)
\(\displaystyle \sec \theta = \dfrac{13}{12}\)
\(\displaystyle \csc \theta = -\dfrac{13}{5}\)
\(\displaystyle \cot \theta = -\dfrac{12}{5}\)
Subsection 6.5.2 Exercises