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Chapter 8: Problem 3

Determine the sign of the given functions. $$\tan 135^{\circ}, \sec 50^{\circ}$$

Short Answer

Expert verified
\(\tan 135^{\circ}\) is negative, \(\sec 50^{\circ}\) is positive.

Step by step solution

01

Determine the Quadrant for \(\tan 135^{\circ}\)

The angle \(135^{\circ}\) is located in the second quadrant of the unit circle. In the second quadrant, the tangent function is negative because \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), and while sine is positive, cosine is negative, resulting in a negative tangent value.
02

Determine the Sign of \(\sec 50^{\circ}\)

The angle \(50^{\circ}\) is located in the first quadrant. In the first quadrant, all trigonometric functions, including the secant function (which is the reciprocal of cosine), have positive values because both sine and cosine are positive here, making \(\sec 50^{\circ} = \frac{1}{\cos 50^{\circ}}\) positive.

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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 fundamental concept in trigonometry that helps visualize angles and their corresponding trigonometric values on a coordinate plane. It is a circle with a radius of 1 centered at the origin \(0, 0\). Often, angles are measured in degrees or radians starting from the positive x-axis and moving counterclockwise around the circle.

  • First Quadrant: Angles from \(0^{\circ}\) to \(90^{\circ}\), where all trigonometric functions are positive.
  • Second Quadrant: Angles from \(90^{\circ}\) to \(180^{\circ}\), where sine is positive but cosine is negative.
  • Third Quadrant: Angles from \(180^{\circ}\) to \(270^{\circ}\), where both sine and cosine are negative.
  • Fourth Quadrant: Angles from \(270^{\circ}\) to \(360^{\circ}\), where cosine is positive but sine is negative.
By understanding the unit circle and its quadrants, you can easily determine the signs of trigonometric functions based on the angle's location.
tangent function
The tangent function, often represented as \(\tan(\theta)\), is a crucial trigonometric function that represents the ratio of the length of the opposite side to the adjacent side in a right-angled triangle. It can also be expressed in terms of sine and cosine as \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).

In the context of the unit circle:
  • The tangent function is positive in the first and third quadrants where the signs of both sine and cosine are the same.
  • It is negative in the second and fourth quadrants where sine and cosine have opposite signs.
For the specific case of \(\tan 135^{\circ}\), since \(135^{\circ}\) lies in the second quadrant, \(\tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} \) will be negative, because \(\sin 135^{\circ}\) is positive while \(\cos 135^{\circ}\) is negative.
secant function
The secant function, notated as \(\sec(\theta)\), is the reciprocal of the cosine function. It is expressed mathematically as \(\sec(\theta) = \frac{1}{\cos(\theta)}\). This function is instrumental when you need to determine a value that is directly linked to the reciprocal of the adjacent side relative to the hypotenuse in a right triangle.

When using the unit circle:
  • The secant function is positive in angles where the cosine is positive, specifically in the first and fourth quadrants.
  • It becomes negative when cosine itself is negative, which occurs in the second and third quadrants.
For example, at \(50^{\circ}\), which falls in the first quadrant, both \(\cos 50^{\circ}\) and its reciprocal, \(\sec 50^{\circ}\), are positive.

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Most popular questions from this chapter

solve the given problems. The latitude of Miami is \(26^{\circ} \mathrm{N},\) and the latitude of the north end of the Panama Canal is \(9^{\circ} \mathrm{N}\). Both are at a longitude of \(80^{\circ} \mathrm{W}\). What is the distance between Miami and the Canal? Explain how the angle used in the solution is found. The radius of the Earth is 3960 mi.

Determine the quadrant in which the terminal side of \(\theta\) lies, subject to both given conditions. $$\sec \theta>0, \csc \theta<0$$

For the given values, determine the quadrant(s) in which the terminal side of the angle lies. $$\tan \theta=-2.500$$

Determine the function that satisfies the given conditions. $$\text { Find } \cot \theta \text { when } \sec \theta=6.122 \text { and } \sin \theta<0$$

$$\text {Evaluate the given expressions.}$$ The current \(i\) in an alternating-current circuit is given by \(i=i_{m} \sin \theta,\) where \(i_{m}\) is the maximum current in the circuit. Find \(i\) if \(i_{m}=0.0259 \mathrm{A}\) and \(\theta=495.2^{\circ}\)
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