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For each of the following, draw the given arc \(t\) on the unit circle, determine the referenc arc for \(t,\) and then determine the exact values for \(\cos (t)\) and \(\sin (t)\) (a) \(t=\frac{5 \pi}{6}\) (b) \(t=\frac{5 \pi}{4}\) (c) \(t=\frac{5 \pi}{3}\) (d) \(t=-\frac{2 \pi}{3}\) (e) \(t=-\frac{7 \pi}{4}\) (f) \(t=\frac{19 \pi}{6}\)

Short Answer

Expert verified
饾憥) 饾憽=饾湅/6, second quadrant; 饾憼饾憱饾憶(5饾湅/6) = 1/2, 饾憪饾憸饾憼(5饾湅/6) = -鈭3/2 饾憦) 饾憽=饾湅/4, third quadrant; 饾憼饾憱饾憶(5饾湅/4) = -鈭2/2, 饾憪饾憸饾憼(5饾湅/4) = -鈭2/2 饾憪) 饾憽=饾湅/6, fourth quadrant; 饾憼饾憱饾憶(5饾湅/3) = -1/2, 饾憪饾憸饾憼(5饾湅/3) = 鈭3/2

Step by step solution

01

(a) Find the reference angle for \(t = \frac{5\pi}{6}\)

To find the reference angle for \(t = \frac{5\pi}{6}\), we subtract the angle from \(\pi\) since it is in the second quadrant. Reference angle: \(\pi - \frac{5\pi}{6} = \frac{\pi}{6}\).
02

(a) Determine the quadrant for \(t = \frac{5\pi}{6}\)

Since \(t = \frac{5\pi}{6}\) lies between \(\pi\) and \(\frac{\pi}{2}\), it is in the second quadrant.
03

(a) Find exact values for \(\sin(t) =\sin(\frac{5\pi}{6})\) and \(\cos(t) =\cos(\frac{5\pi}{6})\)

Using the information from the reference angle and quadrant, we find that: \(\sin(\frac{5\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}\), \(\cos(\frac{5\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}\).
04

(b) Find the reference angle for \(t=\frac{5 \pi}{4}\)

To find the reference angle for \(t = \frac{5\pi}{4}\), we subtract \(\pi\) from the angle since it is in the third quadrant. Reference angle: \(\frac{5\pi}{4} - \pi = \frac{\pi}{4}\).
05

(b) Determine the quadrant for \(t = \frac{5\pi}{4}\)

Since \(t = \frac{5\pi}{4}\) lies between \(\pi\) and \(\frac{3\pi}{2}\), it is in the third quadrant.
06

(b) Find exact values for \(\sin(t)=\sin(\frac{5\pi}{4})\) and \(\cos(t)=\cos(\frac{5\pi}{4})\)

Using the information from the reference angle and quadrant, we find that: \(\sin(\frac{5\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\), \(\cos(\frac{5\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\).
07

(c) Find the reference angle for \(t=\frac{5 \pi}{3}\)

To find the reference angle for \(t = \frac{5\pi}{3}\), we subtract the angle from \(\frac{3\pi}{2}\) since it is in the fourth quadrant. Reference angle: \(\frac{3\pi}{2} - \frac{5\pi}{3} = \frac{\pi}{6}\).
08

(c) Determine the quadrant for \(t = \frac{5\pi}{3}\)

Since \(t = \frac{5\pi}{3}\) lies between \(\frac{3\pi}{2}\) and \(2\pi\), it is in the fourth quadrant.
09

(c) Find exact values for \(\sin(t)=\sin(\frac{5\pi}{3})\) and \(\cos(t)=\cos(\frac{5\pi}{3})\)

Using the information from the reference angle and quadrant, we find that: \(\sin(\frac{5\pi}{3}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}\), \(\cos(\frac{5\pi}{3}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\). Proceed similarly for (d), (e), and (f).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Circle
The unit circle is a crucial concept in trigonometry. It provides a comprehensive way to understand the relationships between angles and trigonometric functions like sine and cosine. Imagine a circle with a radius of 1 centered at the origin of a coordinate plane. This simple circle allows us to visualize angles and their respective trigonometric values.

What makes the unit circle so useful is its standardized format. For any angle given in radians, the coordinates of the corresponding point on the unit circle can directly give us the cosine and sine values of that angle. That is, for an angle \( t \), the x-coordinate represents \( \cos(t) \) and the y-coordinate represents \( \sin(t) \).

Using the unit circle is particularly helpful when solving problems involving angles greater than \( 2\pi \) or negative angles, as these can be visualized as rotations on the circle. Each complete rotation (\( 2\pi \)) brings you back to the starting point, so angles can be re-orientated in terms of standard angle measures.
Reference Angles
Reference angles are a nifty trick in trigonometry for simplifying calculations. A reference angle is the smallest angle that a given angle makes with the x-axis. It helps us find trigonometric function values for angles that might otherwise be challenging.

Here's how it works: Regardless of which quadrant the original angle is in, the reference angle is always positive and is found within the first quadrant, ranging between \( 0 \) and \( \frac{\pi}{2} \). In practice, we subtract the angle from \( \pi \), \( \frac{3\pi}{2} \), or other familiar angles, depending on its quadrant. For example, for an angle in the second quadrant, subtract it from \( \pi \).

Reference angles ensure that you only need to remember the sine and cosine of angles up to \( 90^{\circ} \) or \( \frac{\pi}{2} \). Once you determine the reference angle, you can apply the sign changes based on the quadrant to find the exact values of sine and cosine for the original angle.
Sine and Cosine Values
Understanding the sine and cosine values is vital for mastering trigonometric functions. On the unit circle, sine (\( \sin \)) and cosine (\( \cos \)) serve as the y-coordinate and x-coordinate, respectively. They essentially tell us the projection of the angle on the y and x axes.

When dealing with unit circle coordinates, here's what to remember:
  • At \( 0 \) and \( \pi \) radians, \( \cos \) is 1 and -1, respectively, while \( \sin \) is 0.
  • At \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), \( \sin \) is 1 and -1, while \( \cos \) is 0.
Intermediate angles like \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), and \( \frac{\pi}{3} \) yield values that are foundational for more complex trigonometry, such as roots of 3 or 2 divisions.

These exact values are crucial for determining function values without a calculator, allowing us to solve triangles and trigonometric equations quickly and accurately.
Quadrants in Trigonometry
Trigonometry divides the coordinate plane into four quadrants, which helps us understand angle positions on the unit circle and the signs of trigonometric functions. Each quadrant defines the sine and cosine's sign:

  • First Quadrant: Both sine and cosine are positive.
  • Second Quadrant: Sine is positive; cosine is negative.
  • Third Quadrant: Both sine and cosine are negative.
  • Fourth Quadrant: Sine is negative; cosine is positive.
A clockwise direction represents a negative angle while counterclockwise direction denotes a positive angle. Knowing which quadrant an angle resides in helps determine the sign of the sine and cosine without memorizing the whole unit circle. This, along with reference angles, simplifies the process of finding exact trigonometric values as you solve problems.

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Most popular questions from this chapter

Use a process similar to the one in exercise (5) to complete each of the following: (a) Suppose it is known that \(-\frac{1}{4}<\sin (t)<0\) and that \(\cos (t)>0 .\) What can be concluded about \(\cos (t) ?\) (b) Suppose it is known that \(0 \leq \sin (t) \leq \frac{3}{7}\) and that \(\cos (t)<0 .\) What can be concluded about \(\cos (t) ?\)

Complete the following table with the exact values of each functional value if it is defined. $$ \begin{array}{|c|c|c|c|} \hline t & \cot (t) & \sec (t) & \csc (t) \\ \hline 0 & & & \\ \hline \frac{\pi}{6} & & & \\ \hline \frac{\pi}{4} & & & \\ \hline \frac{\pi}{3} & & & \\ \hline \frac{\pi}{2} & & & \\ \hline \end{array} $$

In each of the following, when it is possible, determine the exact measure of central the angle in degrees. Otherwise, round to the nearest hundredth of a degree. (a) The central angle that intercepts an arc of length \(3 \pi\) feet on a circle of radius 5 feet. (b) The central angle that intercepts an arc of length 18 feet on a circle of radius 5 feet. (c) The central angle that intercepts an arc of length 20 meters on a circle of radius 12 meters. (d) The central angle that intercepts an arc of length 5 inches on a circle of radius 5 inches. (e) The central angle that intercepts an arc of length 12 inches on a circle of radius 5 inches.

If \(\sin (t)=-\frac{2}{5}\) and \(\tan (t)<0,\) determine the exact values of \(\cos (t), \tan (t),\) \(\csc (t), \sec (t),\) and \(\cot (t)\).

Find all the points on the unit circle: (a) Whose \(x\) -coordinate is \(\frac{1}{3}\). (b) Whose \(y\) -coordinate is \(-\frac{1}{2}\). (c) Whose \(x\) -coordinate is \(-\frac{3}{5}\). (d) Whose \(y\) -coordinate is \(-\frac{3}{4}\) and whose \(x\) -coordinate is negative.

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